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ATOMIC·STRUCTURE

2.1 INTRODUCTION through gases at low pressure in a discharge tube. [A common

discharge tube consists of a hard glass cylindrical tube (about
The word atom is a Greek word meaning indivisible, i. e. , 'an 50 cm long) with two metal electrodes sealed on both the ends. It
ultimate particle which cannot be further subdivided. The idea is connected to a side tube through which it can be evacuated to
that all matter ultimately consists of extremely small particles any desired pressure with the help of a vacuum pump.] Air was
was conceived by ancient Indian and Greek philosophers. The almost completely removed from the discharge tube (pressure
old concept was put on firm footing by Jou Dalton in the form about 10--4 atmosphere). When a high voltage of the order of
of atomic theory which he developed in the years 1803-1808. 10,000 volts or more was impressed across the electrodes, some
This theory was a landmark in the history of chemistry. sort of invisible rays moved from the negative electrode to the
According to this theory, atom is the smallest indivisible part of positive electrode (Fig. 2.1). Since, the negative electrode is
matter which takes part in chemical reactions. Atom is neither referred to as cathode, these rays were called cathode rays.
created nor destroyed. Atoms of the same element are similar in
size, mass and characteristics; however, atoms of different Gas at low pressure
elements have different size, mass and characteristics .
. In 1833, Michael Faraday showed that there is a relationship
between matter· and electricity. This was the frrst major
breakthrough to suggest that atom was not a simple indivisible
particle of all matter but was made up of smaller particles. Vacuum pump Cathode rays
Discovery of electrons, protons and neutrons discarded the
indivisible nature of the atom proposed by John Dalton. Fig. 2.1 Production of cathode raya
The complexity of the atom was further revealed when the
following discoveries were made in subsequent years: Further investigations were made by W. Crookes, 1. Perrin, 1.1.
(i) Discovery of cathode rays. Thomson and others. Cathode rays' possess the following
(ii) Discovery of positive rays. properties:
(iii) Discovery of X-rays. ( . I They travel in straight lines a~ray from the cathode with
(iv) qiscovery of radioactivity. very high velocities ranging from 1{)9 lOll cm per second. A
I Discovery of isotopes and isobars. shadow of m~tal1ic object placed in the path is cast on the wall
(vi) Discovery of quarks and the new atomic model. opposite to the cathode.
During the past 100 years, scientists have made contributions . (L) They produce a green glow when strike the. glass wall
which helped in the development of modem theory of atomic beyond the anode. Light is emitted when they strike the zinc
. structure. The works of I"'. 1'IIonnoII and Ental Iludaerfonl sulphide screen.
actually laid the foundation of the modern picture of the atom. It I They produce heat energy when they collide with the
is now believed that the atom consists of several particles called matter. It shows that cathode rays possess kinetic energy which is
subatomic particles like electron, proton, neutron, positron, converted into heat energy when stopped by matter.
neutrino, meson, etc. Out of these particles, the electron, the (iv) They are deflected by the electric and magnetic fields.
proton and the neutron are called fundamental particles and are When· the rays are passed between two electrically charged
the building blocks of the atoms. plates; these are deflected towards the positively charged plate.
They discharge a positively charged gola leaf electroscope. It
2.2· CATHODE RAYS-DISCOVERY OF· shows that c1\thode rays carry negative charge.
ELECTRON (v) They possess kinetic energy. It is. shown by the
experiment that when a small p,in wheel is placed in their path,
The nature and existence of electron was established by
. the blades of the wheel are set in motion. Thus, the cathode rays
experiments on conduction of electricity through gases. In 1859,
consist of material particles which have mass and velocity•
...... I'IIIck« started the stUdy of conduction .of electricity

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68 G,FtB, PHY$ICAL CHEMISTR ",' COMPETITIONS

These particles carrying negative charge were called negatrQns -,J.6622x 10-19 coulomb. Since,
by Thomson. ch~}~~: known, it was, thus, de . rge.
Thename 'negatron' was changed to 'electron' by Stoney. ~Illl of the electron: Th be
(vi) Cathode rays produce X-rays. When these rays fallon a calculated from the value of eI m
material having high atomic mass, new type of penetrating rays
of very small wavelength are emitted which, are calle4 X-rays. e
m -
(vii) These rays affect the photographlc plate. ' elm
(viii) These rays can penetrate through thin foils of solid
= 9.1096 x 10-
28 gor9.1096x
materials and cause ionisation in gases through which they pass.
(ix) The nature of the cathode rays is independent of: This is termed as the rest mass of the electron, i. e. , mass of the
(a) the nature of the cathode and electron when moving with low speed. The mass of a moving
'.. 'c:''J£ (brtheg~s·lh th~Odlscharge tube. electron may be calculated by applying the following formula:,
In 1897, J. J. Thomson determined the elm value
. I' rest mass of electron
M f
(charge/mass) of the electron by studying the deflections of "". 0 movmg e eclron = - ~1- (~)'
cathode rays in electric and magnetic fields. The value of e / m has
been found to be -1.7588 x 108 coulomb/g.
[The path of an electron in an electric field is parabolic, given
as: where v is the velocity of the electron and c is the velocity of light.
eE 2 When v becomes equal to c, mass of the moving electron becomes
y=--x infinity and when the 'velocity of the electron becomes greater
2
2mv than c, mass of the electron becomes imaginary.
where, y = deflection in the path of electron in y-direction
Mass of the el~trgn relative to that of hydrogen atom~
e = charge on electron
E intensity of applied electric field Mass of hydrogen atom = 1.008 amu
m = mass of electron
v velocity of electron
=1.008 x 1.66 x 10-24 g (since 1 amu =1.66 x 10-24 g)
x =distance between two parallel electric plates = 1.673 X 10-24 g
within which electron is moving. 24
The path of an electron in a magnetic field is circular with Mass of hydrogen atom 1.673 x 10-
radius 'r given as: Mass of the electron 9.1 096 x 10-28
mv
r= 1837
eB
, 1
where, m= mass of electron Thus, Mass of an electron =-- x mass of hydrogen atom
v = velocity of electron 1837
e = charge on electron 1.008
B.= intensity qf applied magnetic field 1837
The radius of the path is prQPortional to momentum.] = 0.000549 amu
By performing a series of experiments, Thomson proved that An electron can, thus, be defined as a subatomic particle
whatever be the gas taken in the discharge tube and whatever which carries charge -1.60 x 10-19 coulomb, i.e., one unit
be the material of the electrodes, the value of e / m,ts always
negative charge and has mass 9.1 x 10-28 g, i. e. , _l_th mass of
the same. Electrons are thus common universal constituents of , 1837
all atoms. the hydrogen atom (0.000549 amu).
J.J. Thomson gave following relation to calculate [Millikan's oil drop method is used to determine the charge on
c~arge/mass ratio: an electron by measuring the' terminal velocity of a charged
e E spherical oil drop which is made stationary between two
m= rB2 electrodes on which a very high potential is applied.
where the terms have USual significance given before 61c 11r
. = 1.7588x lOll Ckg- 1
Charge on an electron' q' = E
(VI + v2 )

Electrons are also produced by the action of ultraviolet light or , where, 11 = cQefficient of viscosity of the gas medium
X-rays on a metal and i)'0111 heated filaments. J)-particles emitted 'VI' v 2 termlnal velocities
by radioactive materials are also electrons. E = field strength
The first precise measurement of the charge on an electron V
was made by Robert ,A. Millikan in 1909 by oil drop r =radius of the oil drop = 11 911 J
V2(/ cr)g
experiment. The charge on, the electron was found to be
(f density of oil; cr = density of gas; g gravitational force)]

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A roM1CSTRUC!LfFl~ 69

2~3 POSITIVE RAYS-DISCOVERY OF present on the electron. Hence, for a smaller value of e / m, it is
PROTON definite that positive particles possess high mass.
Ox) e / m value is dependent on the nature of the gas taken in
With the discovery of electrons, scientists started looking for
the discharge tube, i. e. ,positive particles are different in different
positively charged particles .which were naturally expected gases.
because matter is electrically neutral under ordinary conditions.
Accurate. measurements of the charge and the mass of the
The first experiment that led to the discovery of the positive
particles obtained in the discharge tube containing hydrogen, the
particle was conducted by Goldstein in 1886. He used a
lightest of all gases, were made by lJ. Thomson in 1906. These
perforated cathode in the modified cathode ray tube (Fig. 2.2). It 4
particles were found to have the e / m value as + 9.579 x 10
was observed that when a high potential difference was applied
between the t':lectrodes, not only cathode rays were produced but coulomb/g. This was the maximum value of elm observed for
also a new type of rays were produced simultaneously from any positive particle. It· was thus assumed that the positive
anode moving towards cathode and passed through the holes or particle given by hydrogen represents a fundamental particle of
canals of the cathode. These rays were termed canal rays since positive charge. This particle was named proton by Rutherford
these passed through the canals of the cathode.. These were also in 1911. Its charge was found to be equal in magnitude but
named anode rays as these originated from anode. When the opposite in sign to that of electron.
properties of these rays were studied by Thomson, he observed Thus, proton carries a charge + 1.602 x 10-19 coulomb,
that these rays consisted of positively charged particles and e.,
i. one unit positive charge.
named them as positive rays. The mass of the proton, thus, can be calculated.
19
e 1.602x 10-
Mass of the proton = =- ---- 4
elm 9.579x 10
Positive
rays = 1.672 X 10-24 g
+
or = 1.672 x 10-:27 kg

-Cathode . 1.672 X 10- 24

Mass 0 f th e proton In amu = = 1.0072 amu
1.66 x 10- 24
Fig. 2.2
A proton is defined as a subatomic particle which has a mass
The following characteristics of the positive rays were
recognised: nearly 1 amu and a charge of +1 unit (+ 1.602 x 10 -19 coulomb).
(i) The rays travel in straight lines and cast a shadow of the Protons are produced in a number of nuclear reactions. On the
object placed in their path. basis of such reactions, proton has been recognised as a
(it) Like cathode rays, these rays also rotate the wheel placed fundamental building unit of the atom.
in their path and also have heating effect. Thus, the rays possess
kinetic energy, i. e., mass particles are present. tll;l RUTHERFORD EXPERIMENT--
(iii) The rays produce flashes oflight on zinc sulphide screen . DISCOVERY OF NUCLEUS
. (iv) The rays are deflected by electric and magnetic fields in a After the discovery of electron and proton, the question arose
direction opposite to that of cathode rays. These rays are attracted how these charged particles are distributed in an atom. The
towards the negatively charged plate showing thereby that these answer was given by lJ. Thomson in the form Of first model of
rays carry positive charge. the atom.
(v) These rays can pass through thin metal foils. He proposed that the positive charge is spread over a sphere in
(vi) These rays can produce ionisation in gases. which the electrons are embedded to make the atom as a whole
(vii) These rays are capable of producing physical and neutr~1. This model was much like raisins in a pudding and is also
chemical changes. known as Thomson s plum pudding model. This model was
(viii) Positive particles in these rays have elm values much discarded as it was not consistent with the results of further
smaller than that of electron. This means either m is high or the investigations such as.scattering ofa-particles by thin metal foils.
value of charge i~ small in comparison to electron. Since, positive In 1911, Ernest Rutherford and his co-workers carried out a
particle is fOrnle~ by the loss of electron or electrons, the charge series of experiments using a-particles* (Fig. 2.3 and 2.4). A
on the positive particle must be an integral multiple of the charge beam of a-particles was directed against a thin foil of about
0.0004 cm thickness of gold, platinum, silver or copper
*The radiationS emitted by radioactive substances consist ofa-particles. These particles are positively charged. These particles are actually helium.
atoms from which electrons have been removed. Each a-particle consists of a mass equal to about 4 times that of hydrogen atom and carries a positive
charge of two units~ It is represented by the symbol a or ~He. .
He ~ He2 + + 2e
Helium atom a-particle Electron
a-particles are usually obtained from a natural isotope of polonium-214.

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70 I G.R. B. , PHYSI9Al CHEMISTRY FOR COMPETITIONS

respectively. The foil was surrounded by a circular fluorescent

.: Electrostatic PE
zinc sulphide screen. Whenever an a-partiCle struck the screen, it 41tEo r
produced a flash of light. ' . .
1 4Ze 2
. The following observations were made: ro = - - - -
2
(i) Most of the a-particles (nearly 99%) went straight without 41tEo mv
2
suffering any deflection, 1 . . 2Ze
(ii) A few of them got deflected thi-oughsmall angles.
(iii) A very few (about one in 20,000) did not pass through the where, E K is the original kinetic energy of the a-particles.
foil ataB but suffered large deflections (more than 90°) or even
· came back in more or less the direction from which they have Here, .1 9 X 10 9 Nm2 C- 2 (MKS)
•
41tEo
· come, i. e. ,'a deflection of 1800 •
= 1(CGS)
Movable
screen The distance of closest approach is of the order oflO- 14 m. So,
the radius of the nucleus should be less than 10- 14 m.
Following conclusions were drawn from the above observations:

~r~L~~~\ ';:~;~E:
(i) Since, most of the a:"particles went straight through the
metal foil undeflected, it means that there must be very large
empty space within the atom or the atom is extraordinarily
hollow.

/l\j/":~~:
(ii) A few of the a-particles were deflected from their original
paths through moderate angles; it was concluded that whole of
RadioaCtive"
substance;'
(Polonium)
~</
I_._~
.
'\ " the positive charge is concentrated and the space occupied by this
positive charge is very small in the atom. When a-particles come
closer to this point, they suffer a force of repulsion and deviate
Lead plate from their paths.
Deflected The positively charged heavy mass which occupies only a
a-particles small volume in an atom is called nucleus. It is supposed to be
Fig. 2.3 present at the centre of the atom.
(iii) A very few of the
Largely deflected Slightly deflected
a7particles a-particles a-particles suffered strong ~
deflections or even returned on ~ t
their path 'indicating that the 9- "0
nucleus is rigid an.d a-particles /') I!?
recoil due to direct collision with 'O:Ill
the heavy positivelycbarged mass. .z ~
oal
The graph between angle of
scattering and the number of - Scattering angle
a-particles scattering in the
Fig. 2.4 (b)
corresponding direction is as
shown in Fig. 2.4 (b).
Information of Rutherford's scattering equation can be
memorised by the following relations:
Slightly deflected (a) Kinetic energy of a-particles:
a-particles N=KI/[(1I2)mv2]2
Fig. 2.4(8) (b) Scattering angle '8':
Consider an a-particle of mass' m' moving directly towards a N =K 2 /[sin 4 (8/2)]
nucleus with velocity 'v' at any given time. As this a-particle (c) Nuclear charge:
approaches the nucleus, its velocity and hence kinetic energy
N =K 3 (Ze)2
continues to decrease. At a certain distance ro from the nucleus,
the a-particle will stop and then start retracing its depicted path. Here, N = Number of a-particles striking the screen and
This distance is called the distance of closest approach. At this K 1 , K 2 and K 3 are the constants.
distance, the kinetic energy of the a-particle is transformed into
electrostatic potential energy. If Z be the .atomic number of the
;,~~fit MOSELEY EXPERIMENT-ATOMIC
· nucleus, then
./'-
1 2 1 x2e NUMBER
'-mv
2 Roentgen, in 1895, discovered that when high energy electrons in
a discharge tube collide with the anode, penetrating radiations are
produced which he named X-rays (Fig. 2.5).

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ATOMIC. STRUCTURE .1 71

Cathode rays van den Broek (1913) pointed outthat the atomic number of
an element is equal to the to~l positive charge contained in the
nucleus of its atom. Rutherford was also having the same opinion
that the atomic number of an element repli.esents the number of
+ protons in the nucleus of its atom. Thus, .
Cathode· Anode Atomic number of the element
= Serial number of the element in periodic table
Detracted
X-rays ~,"--~"" = Charge on the nucleus of the atom of the element
, , '
= Number of protons present in the nucleus of the
'!I." ~\ ",
",,~\ / / L X-rays beam atom of the element
"" - = NU)11ber of extranuclear electrons present in tha
,
" ' atom of the element
2::6
Defracted unit
2.6 DISCOVERY OF NEUTRON
Fig. 2.5 The discovery of neutron was actually made about 20 years after
X-rays are electromagnetic radiations of very small the structure of atom was elucidated. by Rutherford. Atomic
wavelengths (0.1-20 A). X-rays are diffracted by diffraction masses of different atoms could not be explained if it was
gratings like ordinary light rays and X-ray spectra are, thus, accepted that atoms consisted only of protons and electrons .
. produced. Each such spectrum is a characteristic property of the Thus, Rutherford (1920) suggested that in an atom, there must be
element used as anode. present at least a third type of fundamental particles which should
Moseley (1912-1913), investigated the X-ray spectril of 38 be electrically neutral and possess mass nearly equal to that of
different elements, starting from aluminium and ending in proton. He proposed the name for such fundamental particle as
gold. He measured the frequency of principal lines of a particular neutron. In 1932, Cbadwlek bombarded beryllium with a
series (thea-lines in the Kseries) of the spe.ctra. It was observed stream Of a-particles. He observed that penetrating radiations
that the frequency of a particular spectral line gradually increased were produced which were not affected by electric and magnetic
with the increase of atomic mass of the element. But, it was soon fields. These radiations consisted of neutral particles, which were
realised that the frequency of the particular spectral line was called neutrons. The nuclear reaction can be shown as:
more precisely related with the serial number of the element in
the periodic table which he termed as atomic number (Z). He ~Be + iHe ~ I~C + ~n
presented the following relationship: B~ryllium a-particle .Carbon Neutron
~.

.JV = a(Z - b) The mass of the neutron was determined. It was

where, v = frequency of X-rays, Z = atomic number, a and bare 1.675 x 10-24 g, i. e:, nearly equal to the mass of proton.
constants. When the values of square root of the frequency were Thus, a neutron is a subatolllic particle which has a mass
plotted against atomic numbers of the elements producing 1.675 x 10-24 g, approximately 1 amu, or nearly equal to the
X-rays, a straight line was obtained (Fig. 2.6).
mass of proton or hydrogen atom and carrying no electrical
charge. The e / m value of a neutron is thus zero.

21;'~r RUTHERFORD MODEL

20
.On the basis of scattering experiments, Rutherford proposed a
model of the atom which is known as nuclear .atomic model.
According to this model:
. (0 An atom consists of a heavy positively charged nucleus
w here all the protons and neutrons are. present. Protons and
neutrons. are collectively referred to as nucleons. Almost whole
of the mass of the atom is contributed by these nucleons. The
magnitude of the positive charge on the nucleus is different for
different atoms.
(ii) The volume of the nucleus is very small and is only a
minute fraction of the total volume of the atom. Nucleus has a
20 30 40 diameter of the order of 10- 12 to 10- 13 cm and the atom has a
Atomic number (Z) diameter of the order of 10-8 cm.
8
Diameter of the atom = 10- = 105
Fig. 2.6 . D"iameter of thenl,lcleus 10- 13

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72 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Thus, diameter (size) of an atom is 100,000 times the diameter (iii) In the solar syst~!!l, there is only one planet which
of the nucleus.* . revolves in any particular orbit, but in the nuclear atomic model
The radius of a nucleus is proportional to the cube root of the more than one electron may rotate in any particular orbit.
number of nucleons within it.
R RoAI13 em Drawbacks of Rutherford Model
(i) According to classical electromagnetic theory, when a
where, Ro = 1.33 X 10- cm; A = mass number; R ::::: Radius of
13
charged particle moves under the influence of attractive force, it
the nucleus. ' .
loses energy continuously in the form of electromagnetic
Rutherford and Marsden calculated the density of the
radiations. Thus, when the electron (a charged particle) moves in
hydrogen nucleus containing only one proton. an attractive field (created by protons present in the nucleus), it
d = Mass [A x 1.66 X 10-24 g] must emit radiations. As a result of this, the electron should lose
. VoJume energy at every tum and move closer and closer to the nucleus
.( following a spiral path (Fig. 2.7). The ultimate result will be that
it will fall into the nucleus, thereby making the atom unstable.
A x 1.66 X 10-24 Bohr made calculations and pointed out that an atom would
=--~~~------------~
4 .' 8
collapse in 10- seconds. Since, the Nucleus
x 3.14 x (1.33 x 10- 13 )3 X A
3 . atom is quite stable, it means the
=1.685 X 1014 gf cm 3 electrons do not fall into the nucleus,
thereby this model does not explain
(iii) There is an empty space around the nucleus called
the stability ofthe atom.
extranuclear part. In this part, electrons are present. The number
(ii) If the electrons lose energy
of electrons in an atom is always equal to number of protons
continuously, the observed spectrum
present in the nucleus. As the nucleus part of the atom is
should be continuous but the actual
responsible for the mass of the atom, ~he extranuclear part is
observed spectrum consists of well
responsible for its volume. The volume of an atom is about 1015
defined lines of definite frequencies.
times the volume of the nucleus. Electron
8 24 Hence, the loss of energy by the
Volume of the atom = (10- )3 = 10- = 1015 electrons is not continuous in an atom. Fig. 2.7
Volume of the nucleus (l0-13)3 10-39
(iv) Electrons revolve round the nucleus in closed orbits with 2.8 ELECTROMAGNETIC RADIATIONS
high speed. The centrifugal force acting on the revolving These are energy radiations which do not need any medium for
electrons is being counterbalanced by the force of attraction propagation, e.g., visible, ultraviolet, infrared, X-rays, y-rays,
between the electrons and the nucleus. etc. An electromagnetic radiation is generated by osci1lations of a
This model was similar to the solar system, the nucleus repre- charged body in a magnetic field or a magnet in an electrical
sentingthe sun and revolving electrons as planets. The electrons field. The frequency of a. wave is the frequency of oscillation of
are,.therefore, generally referred to as planetary electrons. the oscillating charged particle. These radiations or waves have
electrical and magnetic ·fields associated with them and travel at
Dissimilarities between Nuclear Atomic Model and
right angle to these fields. The following are thus the important
Solar System
characteristics of electromagnetic radiations:
(i) The sun and the planets are very big bodies and uncharged • All electromagnetic radiations travel with the velocity of
while the nucleus and electrons are very sinall objects and charged. light.
(ii) The revolution of the planets in the solar system is • These consist of electric and magnetic fields that oscillate
governed by gravitational forces, while the revolution of in directions perpendicular to each other and perpendicular
electrons around the nucleus is governed by electrostatic forces. to the direction in which the wave is trave1ling.
[
S.No. Name Wavelength (A) . Frequency (Hz) Source
14 1 5 9
l. Radio wave 3xlO -3x10 I x 10 .,.;Ix 10 Alternating current of high frequency
2. Micro wave ,3x101 -6xld' 1 x 109 - 5 X 1011 Klystron tube
3. Infrared (IR) 6x 10 6 -7600 5 x 10" - 3.95 X 1016 Incandescent objects
16 14
4. , Visible 7600-3800 3.95 x 10 -7.9 X 10 Electric bulbs, sun rays
5. I
Ultraviolet (UV) 3800-150 7.9 x 10 14
2 X 10 16
Sun rays, arc lamps with mercuryvapours
6. X-Rays 150-0.1 2x lOll> - 3 X 1019 Cathode rays striking metal plate
7. !y-RaYS O.H).OI 1
19
3 x 10 -3 X 10 20
Secondary effect of radioactive. decay
8. ~osmic rays· ! O.ot-Zero ~ ~ 102°-Infinity .-.
Outer space

"'The diameter of various atoms lies in the range of 0.74 to 4.70 A(1 A :; 10-8 cm),

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ATOMIC STRUCTURE 73
"
A wave is always characterized by the following six (vi) Time period: Time taken by the wave for one complete
characteristics: cycle or vibration is called time period. It is denoted by T.
(i) Wavelength: The distance between two nearest crests or I
T==-
nearest troughs is called the wavelength. It is denoted by A
Unit: Second per cycle.
v
(lambda) and is measured in terms of centimetre (em), angstrom
(A), micrometre ().un) or nanometre (nm).
·2~9 EMISSION SPECTRA-HYDROGEN
SPECTRUM
Spectrum is the impression produced on a screen when radiations
of particular wavelengths are analysed through a prism or
• Energy
diffraction grating. It is broadly of two types:
(i) Emission spectra (ii) Absorption spectra.
Difference between Emission and Absorption Spectrum
lA =10-8 em =10-10 m Emission spectrum Absorption spectrum
1 f.,lm= 10-4 em = 1O-ti m 1. It gives bright lines (coloured) It gives dark lines on the brigllt
Inm 10-7 cm= 10-9 m on the dark background. background.
=
1 em 108 A ==10 4 f.,lm==107nm 2. Radiations from emitting It is observed when the white
light is passed through the
source are analysed by the
(ii) Frequency: It is defined as the number of waves which substance and the transmitted
spectroscope.
pass through a point in one second. It is denoted by the symbol radiations are analysed by the
v (nu) and is measured in terms of cycles (or waves) per second spectroscope.
(cps) or hertz (Hz). 3. It may be continuous (if source These are always disconti-
=
AV distance travelled in one second emits white light) and may be nuous.
== velocity == c discontinuous (if the source
c emits coloured light).
or V==-
A Emission spectra: It is obtained from the substances which
(iii) Velocity: It is defined as the distance covered in one second emit light on excitation, i. e. ,either by heating the substances on a
by the wave. It is denoted by the letter 'c'. All electromagnetic flame or by passing electric discharge through gases at low
waves travel with the same velocity, i. e. , 3 x 10 10 cm/sec. pressure or by passing electric current discharge through a thin
filament of a high melting point metal. Emission spectra are of
AV== 3x 1010
two types:
Thus, a wave of higher frequency has a· shorter wavelength (a) Continuous spectra and (b) Discontinuous spectra.
while a wave of lower frequency has a longer wavelength. (a) Continuous spectra: When white light is allowed to pass
(iv) Wave number: This is the reciprocal of wavelength, through a prism, it gets resolved into several colours (Fig. 2.8).
i. e. ,the number of wavelengths per centimetre. It is denoted by The spectrum is a rainbow of colours, i. e., violet merges into blue,
the symbol v (nu bar). blue into green, and so on. This is a continuous spectrum.
_ 1
V==- Ultraviolet
A
l I . Violet 4000 A
It is expressed in cm- or m- .
Indigo
(v) Amplitude: It is defined as the height of the crest or
Blue
tlepth of the trough of a wave. It is denoted by the letter 'a'. It
determines the intensity ofthe radiation. Green
The arrangement of various, types of electromagnetic radiations Yellow

~~~~~~~=======~" Orange
in the order of their increasing or decreasing wavelengths or
frequencies is known as electromagnetic spectrum. ' Red 6500 A
, Frequency
V 3x107 (cycle/sec) )3x10 21 Infrared
).. (cm) = 103 f-(_W_av_el_en_gt_h_ 10-11

'"
,.~
Continuous
"'~ '" !-<
Low energy
~5 ~ High energy Fig. 2.8 Continuous spectrum of white light
ll.l
..l
<,... 0 0
Low frequency i:\:", ~, ;;: High frequency (b) Discontinuous spectra: When gases or vapours of a
~
ll.l
Long wavelength 9~
'">- '">-
~ ~ ~ :i
...l
!!l Short wavelength chemical substance are heated in an electric arc or in a Bunsen
~~ Sl
~ ~ ;;:
'" ..l
;:::l
:><: 'I- a
flame, light is emitted. If ray of this light is passed through a
prism, a line spectrum is produced. This spectrum Consists of a
limited number of lines, each of which corresponds to a different

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74 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

wavelength of Ffgbt. The line spectrum of each element is unique. Red Blue-Green Indigo Violet
Hydrogen spectrum is an example of line emission spectrum or
atomic emission spectrum.
When an electric discharge is passed through hydrogen gas at
low pressure, a bluish light is emitted. When a ray of this light is 6563 4861.4340 4102
passed through a prism, a discontinuous line spectrum of several
. isolated sharp lines is obtained. The wavelengths of various lines Fig. 2.9 (a) Balmer series In the hydrogen spectrum

[~
show that these Jines lie in visible, ultraviolet and infrared
regions. All these lines observed in the hydrogen spectrum can be Lyman series: v 1 = RZ
A
2
12
- ..!..]
n:
claSsified into six series.
Speetral series . Discovered by Appearing in n2 =2, 3,4,5, ...
Lyman series Lyman' Ultraviolet region Obtained in emission as well as in absorption spectrum ratio
Balmer series . Balmer Visible region ofm lh to nth wavelength of Lyman series.

~: =(:::)' ·{::::i:=:}
Paschen. series Paschen Infraled region
Brackett'series Brackett Infrared region
Pfund series Pfund Infrared region

=IlZ 2 [..!.. - ~l
Humphrey series Humphrey Far infrared region
Ritz presented a mathematical formula to find the Balmer ser.ies: v =!
A 22 ni.
wavelengths of various hydrogen lines.
=3,4,5,6, .. ;
V=l=~=R[~-~)
A c nl n2'
. n2
Obtained only in emission spectrum.

Where, R is a universal constant, known as Rydberg constant, ItS

value is 109,678 cm-I, n l and n2 are integers (such that n2 > n l ).
Paschet series: ±
v = =RZ 2 [312 - n1i]
For a given spectral series, n 1 remains constant while n2 varies
n2 =4,5,6,7, ...
from line to line in the same series.
The value of nl =1, 2, 3, 4 and 5 for the Lyman, Balmer,
Paschen, Brackett and Pfund series respectively. n2 is greater
Brackett series: . ±
v == =RZ 2 [ 412 - ::]
than nl by at least 1. .
n2 == 5,6,7,8, ...
Values ofnl and n~ for various series
Spectral series Value ofnl Value ofnz Pfuud series: v= 1 = RZ 2 [~_..!..]
2
A S2n2 .
Lyman series 1 2,3,4,5, ...
Balmer series 2 3,4,5,6, ... n2 = 6; 7, 8, 9, ...
.Paschen series 3 4,5,6,7, ... Note: (i) Atoms give line spectra while molecules give band spectra .
Brackett series 4 5,6,7,8, ... (ii) Balmer, Paschen, Brackett, Pfund series are found in
Pfund series 5 6,7,8,9, ... . emission spectrum.

. Electronic transition Name of line Wave no. Wavelemrth and'colour

n2 = 3 ----? n, = 2 Ha (First line) . v = ..!. =R [ l... _l...2] =5R A = 6563 A (Red)
(M). (L)
. A i 3 . 36
"2 = 4 ----? nl = 2 H,J (Second line) 1] 3R A = 4861 A(Blue-Green)
(N) (L) V=±=R[;2 42 =16
"2 = 5 ----? n1 = 2 H., (Third line) 2lR
:-
A= 4340 A (Indigo).
(0) (L) V=±=R[;2 _ 1
100
"2 = 6 ----? nl= 2 Ha (Fourth fine) 1] 8R A =4102A (Violet)
(P) (L) V=±=R[;2 62 = 36
(Above four lines were viewed in Balmer series by naked eye.)

Absorption Spectrum: Suppose the radiations from a conti- number of wavelengths in the emergent light than in incident
nuous source like a hot body (sun light) containing the quanta of light is called absorption spectrum.
all wavelengths passes through a sample of hydrogen gas, then Let the radiations of wavelengths AI' A2' A3, A4 ,AS are
the wavelengths missing in the emergent light give dark lines.on passed through the sample of hydrogen gas such that AI and A4
the bright background. This type of spectrum that contains lesser are absorbed then the absorption spectrum may be represented as:

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ATOMIC STRUCtURE 75

,Example 2. Calculate the frequency and wave number of

Al
• r-----'---~ Az radiation with wavelength 480 nm.
Az • Sample Solution: Given,
A3 of
• Hydrogen . A ::: 480 nrn.= 480 x 10-9 m
A4 • gas
1 - - - - - - - 1 AS c::: 3 X 108 ml sec '
AS •
c 3 X 108 ms-:I
Frequency, v -:::', ' = 6.25 x 1014 S-I
, A 480x 10-9 m
Fig. 2.9 (b) Absorption Spectrum = 6.25 x 1014 Hz
Example 3. Calculate the energy associated with photon
'2.10' QUANTUM THEORY OF RADIATION
of light having a wavelength 6000 A. [h = 6.624 x I 0-27 erg~s~c.]
The wave theory successfully explains many properties of , 'c
electromagnetic radiations such as reflection, refraction, diff- Solution: We know that, E = hv =h . -
A
raction, interference, polarisation, etc., but fails to explain some
h::: 6.624 X 10-27 erg-sec; c 3 x 1010 cml sec;
phenomena like black body radiation, photoelectric effect, etc. ",
In order to explain black body radiation and photoelectric A 6000A::: 6000 x 10-8 cm
effect, Max Planck in 190 I presented a new theory which is
=(6.624XIO-27)X(3XIOIO)=3312 10-12
known as quantum tbeory of radiation. According to this S0" E 5 • X erg.
theory, a hot body emits radiant energy not continuously but 6x 10-
discontinuously in the form of small packets of energy called Example 4. Which has a higher energy, a photon of violet
quantum (in plural quanta). The energy associated with each light with wavelength 4000 A or a ,photon of red light with
quantum of a given radiation is proportional to the frequency of wavelength 7000 A? [h = 6.62 x 10-:34 Js] ,
the emitted radiation. '
Eocv
Solution: We know that, E ::: hv = h . E
Or E = hv where, h is a constant known as Planck's constant. Its ,A
numerical value is 6.624 x 10-27 erg-sec. The energy emitted or Given, h 6.62 X 10-34 Js, c::: 3 X 108 ms- I
absorbed by a body can be either one quantum or any whole
number multiple of hv,i.e.,2hv,3hv,4hv, ... ,nhv quanta of For a photon of violet light,
energy. A = 4000 A = 4000 x 10-10 m
, 8
Thus, energy emitted or ,absorbed = nhv, where n can have
values I, 2, 3,4, .... Thus, the energy emitted or absorbed is E = 6.62 X 10-34 x 3 x 10 4.96 x 10- 19 J
4x 10-7
quantised.
In 1905; Einstein pointed out that light can be supposed to For a photon of red light,
consist of a stream of particles, called photons. The energy of A::: 7000 A ::: 7000 x 10-10 m
each photon of light depends on the frequency of the light, 8
i. e., E hv. Energy is also related according to Einstein, as E = 6.62x 10-34 x 3 X 10 ::: 2.83 X 10- 19 J
E = mc 2 where m is the mass of photon. Thus, it was pointed out , 7000 X 10- 10
that light has wave as well as particle characteristics (dual Hence, photon of violet light has higher energy than the
nature). photon of red light.

::: :::aW5oME SOLVED ExAMPLES\ a:::"::' " ExampleS. What is the ratio between the energies of two
radiations one with a wavelength of 6000 A and other with
" Example 1. How many protons, electrons and neutrons' 2000 A?
are present in 0.18 g ?~ P? ' ,
Solution: No. of protons in one atom ofP Solution: A) = 6000 A and A2 = 2000A
= No. of electrons in one atom ofP 15 c C
No. of neutrons in one atom of P= (A - Z)= (30-15)= 15 E1 h·-andEz =h·-
Al AZ
30 0.18 '
O.l8g ISP= ::: 0.006 mol E) h· c Az 2000 1
30 Ratio, -=-x-::::; =--=
E z Al h· C Al 6000 3
No. of atoms in 0.006 mol = 0.006 x 6.02 x IOz3
or E z = 3EI
No. of protons in 0:006 mol i~p = 15 x 0.006 x 6.02 x 1023
5.418>< 102z Example 6. Calculate the wavelength, wave number and
frequency of photon having an energy equal to three electron
So, No. of electrons = 5A18 xlOz2 volt. (h =6.62 x 10-27 erg- sec.)
and No. of neutrons 5.418 x lOz2

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76 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Solution: We know that, Number of photons = 1.00 9 =2.01 X 10 18

4.965x10- 1
E=h·v
Example 10. Find the number of quanta of radiations of
v=E (leV = 1.602 x 10- 12 erg)
h frequency 4.67 x 1013 S~I, that must be absorbed in order to melt
12 5 gofice. rhe energy required to melt 19ofice is 333.1.
= 3x (1.602 x 10- )= 7.26x 1014 S-I = 7.26-x 1014 Hz
6.62x 10-27 Solution: Energy required to melt 5 g of ice
10
A=~= 3x10 14 =4.132x10-5 cm
= 5x 333= 1665J
V 7.26x10 Energy associated with one quantum
_ 1 1 4 1 = hv = (6.62 X 10-34 ) X (4.67 X 1013 )
V =- 2.42 x 10 cm-
A = 30.91 X 10-21 J
Example 7. Calculate the energy in kilocalorie per mol of. Number of quanta required to melt 5 g of ice
the photons of an electromagnetic radiation of wavelength 1665 = 53.8; 1021
76001. 30.91 X 10-21
= 538x 1022
Solution: A = 7600A = 7600 x 10-8 cm
.... Example 11. Calculate the wavelength ofthe spectral line,
c 3x10 lO cms-1
when the electron in the hydrogen atom undergoes a transition
c 3 x 1010 from the energy level 4 to energy level 2 ...
Frequency, v = - = = 3.947 X 1014 S-I
Solution: According to Rydberg equation,
8
A 7600 x 10-

Energy of one photon = hv = 6.62 X 10-27 x 3.947 X 10 14 ..!.=R[~2 __l )

A x . y2
= 2.61 X 10- 12 erg
R=109678cm- l ; x 2; y=4
Energy of one mole of photons = 2.61 x 10- 12 x 6.02 X 10 23
= 15.71x 1011 erg 1 = 109678[''!'
A 4
1
16
J
Energy of one mole of photons in kilocalorie
3
=109678x-
= 15.71 x 1011 [1 kcal = 4.185 x 1010 erg] 16
4.185 x 1010 On solving, A 486 nm
= 37.538 kcal per mol Example 12. A bulb emits light of wavelength A = 45001.
Example 8. Electromagnetic radiation of wavelength The bulb is rated as 150 watt and 8% of the energy is emitted as
242 nm is just sufficient to ionise the sodium atom. Calculate the light. How many photons are emitted by the bulb per second?
ionisation energy in kJ mol -I ,h = 6.6256 X 10-34 Js. (lIT 1992) (liT 1995)
Solution: Energy emitted per second by the bulb
Sol~t ': A= 242 nm = 242 x 10-9 m
=150x 8 J
c 3x 108 ms- I 100
8 hc 6.626 x 10-34 x 3x 108
E=hv=h.!: 6.6256 x 10-34 x 3x10 E nergy 0 fl p hoton:::: - = .
A 4500 x 10-1(1
A 242 x 10-9
19
= 4.42 x 10- joule
=' 0.082 X 10- 17 J = 0.082 X 10-20 kJ
Let n photons be evolved per second.
Energy per mole for ionisation = 0.082 x 10-20 x 6.02 X 1023
493.6 kJ mol-I nx4.42xl0- 19 150x~
100
Example 9. How many photons of light having a n = 27.2 x lOiS
wavelength 4000 1 are necessary to provide 1.00 J of energy?
Example 13. A near ultraviolet photon of 300 nm is
'-'0In11',n: Energy of one photon absorbed by a gas and then remitted as two photons. One photon
is red with wavelength of 760 nm. What would be the wave
hv=h·!:
A number of the second photon?
(6.62 X 10-34 )(3.0 X 108 ) Solution: .
Energy absorbed = Sum of energy of two quanta
4000 x hc ·hc hc
---- = + -----
= 4.965 X 10- 19 J 300 x 760 X 10-9 . AX 10-9

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ATOMIC STRUCTURE 77
On solving, we get, AgCI(s) ---7 Ag(s) + Cl
V (wave number) :::: ..!. 2.02 X 10- m- 3 I Escape ofchlorine atoms is prevented by the rigid structure of
A the glass and the reaction therefore, reverses as soon as the light
Example 14. Calculate the wavelength of the radiation is removed. If 310 kJ / mol of energy is required to make the
which would cause the photodissociatio11. of chlorine molecule if reaction proceed, what wavelength of light is necessary?
the Cl-Cl bond energy is 243 kJ mol -I. Solution: Energy per mole Energy of one Einstein
Solution: Energy required to break one CI-CI bond i. e. , energy of one mole quanta
Bond energy per mole
Nhc
:::: A
Avogadro's number
6.023 X 10 23 X 6.626 X 10-34 X 3 X 108
243 kJ 243 x 10
3
J 310x 1000
A
23 23
6.023 X 10 6.023 X 10 . A 3.862 X 10-7 m =3862 X 10-
10
m 3862A
Let the wavelength of the photon to cause rupture of one
CI-Cl bond be A.
fUlimATfONS ·Of OSJ£ClIVE
We know that, 1. The frequency of the radiation having wave number 10m-I is:
34 8 23
A:::: hc :::: 6.6 X 10- X 3 X 10 X 6.023 X 10 (a) tOs- 1 (b) 3 x 107 S-1
E 243 X ( c) 3 X loll S-I (d) 3 X 109
7
:::: 4.90 X 10- m:::: 490 nm [ADs. (d)]
1
Example 15. How many moles of photon would contain [Hint:
sufficient energy to raise the temperature of225 g of water 21°C A
to 96°C? Specific heat ofwater is4.18J g-I K -I andfrequency v cv =- e
= 3 x 108 X 10 =- 3 X 109 S-l]
A
of light radiation used is 2.45 X 109 S -I. 2. The energy of a photon of radiation having wavelength
Solution: Energy associated with one mole of photons 300nmis:
::::No xhxv (a) 6.63 x 10-29 J (b) 6.63 X 10- 19 J
:::: 6.02 X 10 23 X 6.626 X 10-34 X 2.45 X 109 (c) 6.63 X 10-28 J (d) 6.63 x 10- 17 J
2 1
:::: 97.727 X 10- J mol- [Ans. (b).]
Energy required to raise the temperature of 225 g of water by he 6.626 X lO-34 x 3 x 108
75°C =mxsxt [Hint: E A 6.63 X 10-19 J]
300 X 10-9
= 225 X 4.18 X 75 = 70537.5 J
3. The maximum kinetic energy of the photoelectrons is found
Hence, number of moles of photons required
to be 6.63 x 10- 19 J, when the metal is irradiated with a
:::: mst 70537.5:::: 7.22 X 104 mol
radiation of frequency 2 x 1015 Hz. The threshold frequency
Nohv 97.727 x 10-2
of the metal is about:
Example 16. During photosynthesis, chlorophyll-a (a) 1x Hy5 8- 1 (b)2xI015 8- 1
absorbs light ofwavelength 440 nm and emits light ofwavelength
670 nm. What is the energy available for photosynthesis from the (c) 3x 1015 S-I (d) 15 X Id 5 S-I
absorption-emission ofa mole ofphotons? [Ans. (a)]
[Hint: KE =h(v - vo)
Solution: M:::: [NhC] [Nhc] KE
vo = v - -
A absorbed A· evolved h

::::NhC[ 1
Aabsorbed
- .
Aevolved
J = 2X 101~
6.63 X 10- =1 X 1015 s-I]
_
19

6.63 x
23
:::: 6.023 X 10 X 6.626 X 10-34 X 4. The number of photons of light having wavelength 100 nm
which can provide 1 J energy is nearly:
3 X 108 [ 1 _ 1
440 X 10-9 670 X (a) 107 photons (b) 5 xJOIS photons
:::: 0.1197[2.272 x 106 -1.492x 106 ] (c) 5 x 1017 photons (d) 5 x 107 photons
:::: 0.0933 X 106 J I mol:::: 93.3 kJ I mol [Ans. (c)]
[Hint: E= nhe
Example 17. Photochromic sunglasses, which darken
A
when exposed to light, contain a small amount of colourless
AgCI (s) embedded in the glass. When irradiated with Ught, EA
metallic silver atoms are produced and the glass darkens. he

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78 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

5. The atomic transition gives rise to the radiation of frequency (vi) The emission or absorption of energy in the form of
(104 MHz). The change in energy per mole of atoms taking radiation can only occur when an electron jumps from one
place would be: . stationary orbit to another.
(a) 3.99 x 10-6 J (b) 3.99 J f),E:::: E bigh - E low :::: hv
24 30
(c) 6.62 X 10- J (d) 6.62 x 10- J Energy is absorbed when the electron jumps from inner to outer
[Ans. (b)] orbit and is eniitted when it rhoves 'from outer to an inner orbit.
[Hint: E =Nhv .
= 6.023 x loB x 6.626 X 10-34 x 104 X 1(f
= 3.99J]
.2l!~;l~ BOHR'S ATOMIC MODE;L Electrons excited
by absorbing
To overcome the objections of Rutherford's model and to explain
energy
the hydrogen spectrum, Bohr proposed a quantum mechanical (Energy absorbed)
model of the atom. This model w~s based on the quantum theory
of radiation and the classical laws of physics. The important
postulates on which Bohr's model is based are. the following:
(i) The atom has a nucleus where all the protons and neutrons ::::::::;'::::::::::::::::l Energy radia1ed
are present. The size of the nucleus is very small. It is present at when electrons
the centre of the atom. fall back
(ii) Negatively charged electrons are revolving around the n=3 (Energy emitted)
(M) n=4
nucleus in the same way as the planets are revolving around the
(N) n=5
sun. The path of the electron is circular. The force of attraction (0) n=6
between the nucleus and the electron is equal to centrifugal force (P)
. of the moving electron.
Fig. 2.10
Force of attraction towards nudeus centl'ifugal force
(iii) Out of infinite number of possible circular orbits around When the electron moves from inner to outer orbit by
the nucleus, the electron can revolve only on those orbits whose absorbing defini~e amount of energy, the new state of the electron
is said to be excited state (Fig. 2.10).
angular momentur.n * is an integral multiple of !!...., i. e. ,
Using the above postulates, Bohr calculated the radii of.
2n
various stationary orbits, the energy associated with each orbit
mvr:::: n!!.... where m:::: mass of the electron, v:::: velocity of and explained the spectrum of hydrogen atom.
2n .
electron, r = radius of the orbit and n =1; 2, 3, ... number of the Radii of various orbits: Consider an electron of mass' m'
orbit. The angular momentum can have values such as, and charge 't! revolving around the nucleus of charge Ze
(Z =atomic number). Let 'v' be the tangential velocity of the
h , 2h , 3h ; etc., but it cannot have a fractional value. Thus, the revolving electron and ' r' the radius of the orbit (Fig. 2.11). The
21t 2n 21t ,
electrostatic force of attraction between the nucleus and electron
angular momentum is quantized. The specified or circular orbits
(quantized) are called atationary orbits. kZex e kZe 2
(applying Coulomb's law)
(iv) By the time, the electron remains in anyone of the
stationary orbits, it does not lose energy. Such a state is called
v
grou.nd or normal state.
In the ground state, potential energy of electron will be Centrifugal
minimum, hence it will be the most stable state. force
(v) Each stationary orbit is associated with a definite amount
of energy. The greater is the distance of the orbit from the
nucleus, more shall be the energy associated with.it These orbits
are also called energy levels and are numb~red as 1, 2, 3,4, .... or
K, L, M, N, ... from nucleus outwards.
i.e., E, <.E2 < E3 < E4 ...
Fig. 2.11
(E2 - E, » (Ei - E 2 » (E4 - E3 ) ....

*Angwar momentum = lro, where; I b moment of inertia and ro =angular velocity;

O)= !: , where, v =linear velocity and radius; and I ,;" m?-. So angular momentum = mr2 . v = mvr.
r r

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ATOMIC STRUCTURE 79

where, k is a constant. It is equal to 1 ,Eo being absolute of

Energy an electron: Let the total energy of the electron
4TtEo be E. It is the sum of kinetic energy and potential energy.
E =kinetic energy + potential energy
permittivity ofmedi~. In SI units, the numerical value of_l_
4TtEo _ 1 2 kZe 2
--mv - - -
is equal to 9 x 10 9 Nm 2 I C 2 • 2 r
[Note: In CGS units, value of k is equal to 1.] Putting the value of mv 2 from eq. 0),
2 2 2
. As force of attraction = centrifugal force E = kZe _ kZe :::: _ kZe
2 2 2
kZe mv
So, --=--
r2 r
or v2 =kZe
rm
2r r 2r
Putting the value of r from eq. (iii),
Ze 2
2 _ 1 kZe 2 4Tt 2 mkZe 2 21t 2 Z 2k 2 me 4
v - -_.- ... (i) E::::---x· =- ...(iv)
4TtEo rm 2 n2h2 n2h2
According to one of the postulates, For hydrogen atom, Z = 1
h 2Tt 2 k 2 me 4
Angular momentum:::: mvr:::: n - So, E::::
2Tt
nh Putting the values ofTt, k, m, eand h,
or v=-- ... (ii)
21tmr E =_ 2x (3.14)2 x (9 X 109 )2 x (9.1X 10-31 )x (1.6x 10- 19 )4
Putting the value of <v' in eq. (i), n 2 x (6.625 x 10-34 )2 ' '
n 2h 2 kZe 2
-"...-:"-:-. = - - or 21.79 x 10- 19
4Tt 2 m2 r2 mr ----,--- J per atom
n 2 h2
or r=~:---.".. ... (iii)
4Tt 2 mkZe 2 E _RH (where, RH =2.I8x 10-18 J)
Greater is the value of < n', larger is the size of atom. On the n2
other hand, greater is the value of <Z', smaller is the size of the 13.6 V
- - 2 e per atom (1 J:::: 6.2419 X 1018 eV)
atom. Across a period from left to right, atomic number < Z' n
increases with constant value of < n' hence atomic radius
decreases towards right. On moving down the group, both < Z' and -- - - kcall mo1
313.6 (l eV = 23.06 kcallmol)
'n' increase but due to shielding, Z* (effective nuclear charge) n2
remains same. Hence, on moving downwards, atomic radius _ 1312kJ/moI
increases due to increase in' n'. . n2 '

n2h2
For hydrogen atom, Z = I; so r = 2 2 ·' . energy m
Kmetlc . n th she11 = 13.6>< Z 2 ev
:'\7
4Tt mke
Now putting the values of h, It, m, e and k,
"1 energy m
. n th sheII -27.2 2x Z
2 V

r= .•
n 2 X (6.625 X10-34 i Potentla
n
e

4 x (3.l4i x (9.1x 10-31 ) x (1.6x 10- 19 )2 x (9x 109 ) Substituting the values of n :::: 1, 2, 3,4, ... , etc., the energy of
electron in various energy shells in hydrogen atom can be
= 0.529x n 2 x 10-10 m:::: 0.S29 x n 2 A calculated.
8 2
= 0.529 X 10- X n cm Energy sbeII E (Joule per atom) E (eV per atom) E (kcal/moI) .
where h:::: 6.625 X 10-34 J-sec, Tt = 3.14 I' ,..21.79 X 10- 19 .!..13.6 - 313.6

m= 9.1x 10-31 kg, e= 1.6 x 10-19 coulomb 2 -5.44 x 10-19 -3.4 -78.4
3 -2.42 X 10-
19 -1.51 -34.84
k 9xl09 Nm2/C 2
4 -1.36 X lQ-19 0.85 -19.6
Thus, radius of 1st orbit
0.529x 10-8 x 12 = 0.529 x 10-8 cm= 0.529 x 1O-10 m
Radius of 2nd orbit
=0.529x 10-8 x 22= 2.11 X 10-8 cm= 2.l1x 10-10 m o o o
Radius of 3rd orbit (for hydrogen atom)
0.529x 10-8 x 32 4.76x 10-8 cm= 4.76x 10- 10 m
2 .
and so on. and E" =£1 X Z (for hydrogen like species)
r" :::: 11 x n 2 fo~ hydrogen atom
2 where, £1 :::: energy of hydrogen fIrst orbit.
and ... r" =0.529 x -n A (for hydrogen like species)
Z

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80 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Since, n can have only integral values, it follows that total Interpretation of hydrogen spectrum: The only electron
energy of the electron is quantised. The negative sign indicates in the hydrogen atom resides under ordinary conditions on the
that the electron is under attraction towards nucleus, i. e., it is first orbit. When energy is supplied, the electron moves to higher
bound to the nucleus. The electron has minimum energy in the energy shells depending on the amount of energy absorbed.
first orbit and its energy increases as n increases, i. e. , it becomes When this electron returns to any of the lower energy shells, it
less negative. The electron can have a maximum energy value of emits energy. Lyman series is formed when the electron returns to
zero when n = 00. The zero energy means that the electron is no the lowest energy state while Balmer series is formed when the
longer bound to the nucleus, i. e. , it is not under attraction towards electron returns to second energy shell. Similarly, Paschen,
nucleus. Brackett and Pfund series are formed when electron returns to the
For hydrogen h'ke species such as He . 2+ , etc."
+ ,L· third, fourth and fifth energy shells from higher energy shells
En =Z2 xEn for hydrogen atom. respectively (Fig. 2.l2).
Velocity of an electron: We know that,
Centrifugal force on electron
== force of attraction between nucleus and electron
mv Ze
2 2
(in CGS units) ... (i)
r
The angular momentum of an electron is given as:
mvr= 2n
nhf .. ·(i0
From eqs. (i) and (ii), we have Balmer 1.0--1---1 Brackett

v(;~)= Ze 2
series series

v= ~ [2n;2)
v = ~ x 2.l88 X 108 cmf sec ... (iii)
n
2.188 x 108 Lymanserles
v cm! sec (For hydrogen, Z = 1)
n Fig. 2.12
VI = 2.188x 108 cmlsec Maximum number of lines produced when an electron jumps
1 . .. n(n --, 1)
v2 = x 2.l88x 108 cm!sec from n th level to ground level IS equal t o ' . For example,
2 2
in the case of n = 4; number of lines -produced is 6. (4 -t 3,4 -t 2,
I x 2.188x 108 cm!sec 4 -t 1,3 -t 2, 3 -t 1,2 -t n When an electron returns from n2 .to
3 nl state, the number oflines in the spectrum will be equal to.
Here, VI ,V2 and V3 are the velocities of electron in first, (nz - nl )(nz nl + 1) ,
second and third Bohr orbits in hydrogen. . 2
From equation (iii), If the electron comes back from energy level having energy E 2
VI _ 2 V3 1 to energy level having energy E[, then the difference may be
and - == - and so on.
V2 1 . VI 3 expressed in terms of energy of photon as:
Orbital frequency: Number of revolutions 'pet"'second by E2 - E[ =!lE == hv
an electron in a shell is called orbital frequency; it may be or the frequency of the emitted radiation is given by
calculated as, !lE
v -
Number of revolutions per second by an electron in a shell h
Vel~city 2)V l( Since, !lE can have only definite values depending on the
Circur:.ference = = 2nr h 7 definite energies ofE2 and E 1 , v will have only fixed values in an
atom,
= Z2 x 6.66 x rol5 or v
c !lE
n3 A h
where, E. = Energy offrrst shell. he
Time period of revolution of electron in nth orbit (Tn) : or A
!lE
3 .
Since, hand e are constants, !lE corresponds to definite
~ x 1.5 x 1O-[? sec energy; thus, each transition from one energy level to another will
Z2
produce a light of definite wavelength. This is actually observed
as a line in the spectrum of hydrogen atom.

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ATOMIC STRUCTURE 81

Thus, the different spectral Hnes in the spectra of atoms

cOiTespond to ditferent transitIOns of electrons from higher
Afirst
=R[_1
n[
_ 1 '1
+ J (nl 1)2
energy levels to lower energy levels.
Similarly for second, third and fourth lines,
Derivation of Rydberg Equation
n2 nl + 2; n z = nl + 3 and n z n, + 4 respectively
Let an excited electron from n2 sheli come to the n[ shell with the
release of radiant energy. The wave' number v of the corres- .. Rydberg equation may be written as,
ponding spectral line may be calculated in the foIIowingmanner: v= 1 RZz,nt
[ I 1]
21t 2 mZ 2 e 4 21t 2 mZ2 e 4 A (nl +X)2
M=E z -E[ (-) 2 2 -(-) 2' 2 '
n2 h n[ h
• where, x number of line in the spectrum.
he 2
21t mZ e
2 4 [_1__ 1 e.g., x = 1,2,3,4, ... for first, second, third and fourth lines in the
h2 nt ' n~ spectrum respectively. '
he
(ii) Series limit or last line of a series: It is the line of
where, M = hv shortest wavelength or line of highest energy.
A For last line, nz 00
i = J.. =21t2mZ2e l'_I___1_) , 4 R
=
'A 2
eh 3 , n I2 n2 )~last
· R' ; R
L yman 1·
orv=RZ [_1n __n1) 2
2 2
Imlt= Balmer limit

I 2 R ;
Pasch"en, 1"Imll =? B raek ett '1Imlt
" = R
21t 2me 4
, 3- 42
where, R= Rydbergconstant= 109743 cm- I
ch 3
This value of R is in agreement' with ,experimentally
Pfund limit R,
52
H~mphrey limit = 6'~
determined value 109677,76 cm- 1 • Rydberg equation for (iii) Intensities of spectral 'lines: The intensities of spectral
hydrogen may be given as, lines in a particular series decrease with increase in the value of
, l
v 1 = R [ -\- -
A nl
-\-J
n2
n2' i. e., higher state.
e.g.,
, Lyman series (2 ---; 1) > (3 ---; 1) > (4 ---; 1) > (5 ---; 1)
("2 ~nl)
ModificatIon of Rydber,gEquation
Balmer series (3 4 2) > (4 ---; 2» (5 ~ 2) > (6---;2) .' .
According to the Rydberg ~quation:

wave~umber [n~2 n~]

2 4 DeGfeasing intensity of the spectral lines
21t mZ 2e
lontzationEnergy and 'EXCitation 'E.nergy
It can be considered that the electron and the nucleus revolve
around their common. centre of mass. Therefore" instead of .the, Excitation potential for nl' ---;,112
Electronic charge
mass of the electron, the.,reduced mass of the system was
introduced and the equatiOilbecomes: E
Ionization potential for n 1 ---; 00 == 111

v = 21t J.1Z 2e
.
2
1: 3
4
[_1 __1]
2
, Electronic charge
Cl nl n22 , , The energy required to remove an electron from the groul\d
. , state to form cation, i.~. , to take the electrop to infinity, is called
Reduced mass '11' can be calculated as, ionization energy.
1 1 1
-=-+- IE E"" - Eground
11 m M
where, mass of electron
m IE:::: 0- El (H)= 13.6eV atom-I
and mass of nucleus
M :::: 2.17 X 10- 18 Jatom- I
mM
.. J.1 Z2
m+M IE=- x 13.6eV
(i) First line of a series: It is called 'line of longest , n2
wavelength' or 'line of shortest energy'. Z2
I1 _I_,X
For first line, •
10'
'"
n 12 Z2
2

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82 I G. R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

(IE)H X Z 2 radial velocity and the other perpendicular to the radius vector
(IEh = 2 called transverse or angular velocity. These two velocities give
.. n
rise to radial momentum and angular or azimuthal momentum.
If an electron is already present in the excited state, then the
Sommerfeld proposed that both the. momenta must be integral
energy required to remove that electron is called· separation
energy. multiples of!!...... [Fig. 2.13 (b)].
21t
E separation E ~ - E excited
h
The following points support Bohr theory: Radial momentum n -
r 21t
(i) The frequencies of the spectral lines calculated from . h
Bohr equation are in close agreement with the Azimuthal momentum =n<l>-
frequencies' observed experimentally in hydrogen 21t
spectrum. nr and n<l> are related to the main orbit 'n' as:
(ii) The value of Rydberg constant for hydrogen calculated n nr + n<l>'
from Bohr equation tallies with that determined n nr + n<l> Length of major axis
or
experimentally. . n<l> n<l> Length of minor axis
(iii) The emission and absorption spectra of hydrogen like (i) 'n<l>'. cannot be zero becau~e un,der this condition, the
species such as He + , Li 2+ and Be 3 + can be explained ellipse shall take the shape of a straIght hne.
with the help of Bohr theory. (ii) 'n<l>' cannot be more than' n'because minor axis is always
Limitations of Bohr Theory smaller than major axis.
" • 't
(iii) 'n<l>' can be equa~ to ' n'.. Under this ~onq.ition, the major
(i) It does not explain the spectra ofmulti-electron atoms.
axis becomes equal to mmor aXIS and the elhpse 41kes the shape
(Ii) When a high resolving power spectroscope is used, it is of a circle. Thus, n<l> can have all integral values up to ' n' but not
observed that a spectral line in the hydrogen spectrum is zero. When the values are less than' n',orbits are elliptical and
not a simple line but a collection of several lines which when. it becomes equal to 'n', the orbit is circular in nature.
are very close to one another. This is known as fine
For n == I, n4J can have only one value, i. e., 1. Therefore, the
spectrum. Bohr theory does not explain the fine spectra
first orbit is circular in nature.
of even the -hydrogen atom.
(tii) It does not explain t~e splitting of spectral lines into a
For n =2, n. can have tWo values 1 and 2, i. e., ~e ~econd
orbit has tWo sub-orbits, one is elliptical and the other IS CIrcular
group of finer lines under the influence of magnetic field in nature.
(Zeeman effect) and electric field (Stark effect).
For n == 3, n4J can have three va~ue.s I, 2 and 3, i..e., t~ird or~it
(iv) Bohr theory is not in agreement with Heisenberg'S has three sub-orbits, two are elhphcal and one IS CIrcular m
uncertainty principle. ' nature.
For n = 4, n4J can have four values I, 2, 3 and 4, i. e., fou~h
III SOMMERFELD'S EXTENSION OF orbit has four sub-orbits, three are elliptical and fourth one IS
BOHR THEORY circular in nature (Fig. 2.14).
To accouht for the fine spectrum of hydrogen atom, Sommerfeld,
in 1915, proposed that the moving electron might descri~e
elliptical orbits in addition to circular orbits and the pucleus IS
situated at one of the foci. During motion on a circle, only the
angle of revolution changes while the distance from the nucleus
remains the same but in elliptical motion both the angles of
revolution and the distance of the electron from the nucleus
change. The distance from the nucleus is termed as~adius v,ector
and the angle of revol\ltion is known as :azimutluda~gle. The
tangential velocity of the electron at a particular instant can be
resolved into two components: one along the radius vector called
V Radial velocity Fig. 2.14
. V (Tangential velocity)
Sommerfeld thus introduced the concept of subenergy shells.
In a main energy shell, the energies of subshells differ slightly
from one another. Hence, the jumping of an electron from one
r-..1----I---..!---1 Major axis
energy shell to another energy shell will involve slightly different
amount of energy as it will depend on subshell also. This explains
to some extent the fine spectrum,Qf hydrogen atom. However,
Minor axis Sommerfeld extension fails to explain the spectra of
(a) (b)
multielectron atoms.
Fig. 2.13

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ATOMIC STRUCTURE 'I 83

2i.7xlO- 12
: :: ::::_SOME SOLVED ExAMPLES\ =:::: : .
---'--.,,---'-'- erg = 5.425 x 10-12 erg,
22
~~ii"E~ample 18. Calculate the wavelength arid energy of
radiation emitted for the electronic transition from infinite to and. E"", = 0
stationary state of hydrogen atom, '(Given, R 1.09678 M == Change in energy E"",...., 5.425 X 10- 12 erg ,
xl0 7 m-l,h=6.6256xlO~34J-sandc 2.9979xl0 8 m:s.,-I) Thus, energy required to remove an electron from 2nd orbit
, = 5.425 X 10- 12 erg
Solution: l
A
R[~
nf ·~l
ni According to quantum equation,
c

r~nn~:2, 1
M=h·
A
R or A ~
M
or A I 1 9.11x 10-8 m (h = 6.625 x 10-27 erg - sec; c 3 x 1010 cml sec)
R 1.09678 x 107
and M = 5.425 X 10- 12 erg
We know that,
A (6.625 X 10-27 ) X (3 X 1010 )
E := hv h. c == 6.6256 x 10-34 X 2.9979 x So,
A 9.11 X 10-8 5.425 X 10- 12
=3.7xl0-5 cm
2.17 X 10- 18 J
Thus, the longest wavelength of light that can cause this
Example 19. Calculate the velocity (cm /sec) of an transit40n is 3.7 x 10-5 cm. '
electron placed in the third orbit of the hydrogen atom. Also
calculate the number of revolutions per second that this electron Example 21. Calculate the shortest and longest
makes around the nucleus. wavelengths in hydrogen spectrum ofLyman series,
Solution: Radius of 3rd orbit Or
32 x 0.529 x 10-8 4.761x 10-8 cm Calculate the wavelengths of the first line and the series limit
We know that, for the Lyman series for hydrogen. (R H == 109678 cm- I )
nh nh Solution: For Lyman series, n l I
mvr or v = - -
2n 2nmr For shortest wavelength in Lyman series (i. e. ,series limit), the
3 x 6.624 x 10-27 energy difference in two states showing transition should be
2 x 3.14 x (9.108 x 10-28 )x (4.761 X 10-8 ) maximum, i. e. , n2 00.,

= 0.729 x 108 cmlsec So,

T ·ime taken &lor one revolutton
' -2nr
v
A=_l_, =9.117x 10-6 cm
Number of revolutions per second 109678
1 v
'2nr 2nr = 911.7A
For longest wavelength in Lyman series (i.e., first line), the
v
energy difference~"two states showing transitiqR should be
0.729 x 108 minimum, i. e. , n2 ±it< .~,
2 x 3.14 x 4.761 x
= 2.4 X 1014 revolutions I sec So,., ±= L;?-RH 2; J ! RH
, Example 20. The electron energy in hydrogen atom is
4 I 4
. bE
12
Cl or A - x - = - - - - 1215.7x 10-8 c~
given y = - 21.7xlO- a cu Iate the energy reqUlre
. d to
3 RH 3 x 109678
remove an electron completely fromn= 2 orbit. What is the = 1215.7 A
longest wavelength (in cm) of light that be used to cause this
transition?
,Example 22. Show that the Balmer series occurs between
3647 A and 65631. (R 1.0968 x 10 7 m- I )
21.7x 10-12 Solution: For Balmer series,
Solution: E=- , .' 2 erg
n
Electron energy in the 2nd orbit, i. e. , n 2,
1Rr~-
A L22 n2
1.

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G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

where, n = 3,4,5, ... 00 ·l~~ample.2~.:. If the. ' energy differenge,be/ween two

. . . , ,.. . .' .• . I' . . . .
To obtain the limits for Balmer series n = 3 and n = 00 elegtronic state.~ ..is 46:12 kcal mor. " what wjll be. lhe:freQllency
respectively. of tile light emitted when the electrons drop from hi.~he,. tolower
36
Amax (n 3) - - - - - = - states? (Nh == 9.52 x 10-14 kcal sec mor l ,where, N is the
R[_l _~]
22
5R
3 2
Avogadro s number a~d h is the Planck s constant)
Solution: l::iE = 46.12 kcal mol- 1
36 According to Bohr theory, l::iE == Nhv
= m
5 x 1.0968x 107 l::iE 46.12
or v =- :: -----:-:-
= 6563A Nh 14
9.52 X 10-
1 4
Amin (n=oo)=-----
R[~
22
__l ]
00
2
R = 4.84 X 1014 cycle sec-I

Example 26. According' to Bohr theory, the electronic

energy of the hydrogen atom in the nth Bohr orbit is given by
-~---,,-m
19
1.0968 x 10 7 E = -:. 21.76x 10- J
= 3647A . n . n2

f?L:~_,~~ample 23. Light of wavelength 128181 is emitted when Calculate the longest wavelength of light that will be needed
the electron ofa hydrogen atom drops from 5th to 3rd orbit. Find to remove an electron from the 3rd orbit of the He + ion.
the wavelength ofthe photon emitted when the electron falls from (liT 1990)
3rd to 2nd orbit. Solution: The electronic energy of He + ion in the nth Bohr
St)lution~ We know that, . orbit

.!A = R [-\- - -.12]

nl n2
where, Z= 2
When, III ='3 and n2 = 5,
Thus, energy of He + in the 3i:d Bohr orbit
1 [1 1] 16R 19
12818=R 9- 25 = 9x25 = 21.76 x 10- x4J
9
or 12818 9x 25 ... (i)
16xR l::iE = E~ - E3

When, n l == 2 and n2 3, == 0- [_ 21.76~ ~0-19 x 4]

.! R [.! _~] = 5R
A 4 9 . 36 21.76 x 10- 19 x 4
A 36 ... (ii) 9
5R he 6.625 x 10-34 x 3 X 108 X 9
Dividing eqn. (ii) by eqn. (i),
We know that, A ::
l::iE 21.76 x x4
A 36 16R 64 = 2.055 x 10-7 m
-- -x--
12818 5R 9x 25 125
Example 27. Calculate the ratio l?f the velocity of light
and the velocity of electron in the first orbit of a hydrogen
A == ~ x 12818= 6562.8 A atom. (Given, h = 6.624 X 10-27 erg-sec; m = 9.108 X 10-28 g,
125
r= 0.529 x 10-8 cm)
Example 24. The ionisation energy o/hydrogen atom is
1l6eV. What will be the iOliisafion energyofHe+ and Li 2+ ions? h
V=--
~~ Ionisation energy:: - (energy of the 1st orbit) 21tmr
6.624 x 10-27
Energy of the 1st orbit of hydrogen = 13.6 yV
Energy of the 1st orbit of He+ = -13.6 x Z2 (Z for He+ = 2) 2 x 3.14 x 9.108 x 10-28 x 0.529 X 10-8
= -13.6x 4 eV= - 54.4 eV == 2.189 X 108 cml sec
So, Ionisation energy ofHe+ ;. -(-54.4) =54.4 eV
2
. Energy of 1st orbit of Li + = -13.6 x 9 (Z for Li 2:. =: 3)
137

== -122.4 eV
Ionisation energy of Li 2+ == -(-122.4) = 122.4 eV

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ATOMIC STRUCTURE I· 85
Example 28. The wavelength of a certain line in Balmer 3R
series'is observed to be 4341 A. To what value of 'n' does this 4
correspond? (R H =109678cm- l ) For hydrogen atom,

Solution: .!. =RH

A
[~-1.:]
22 n 2
.!., =R [_1 I2
A nt n2, ]
1 1 1
So, 3R = R
4
[l.: -.1-]
nt ni
= - - --'-----:-"----
4 4341x x 109678 1 3.
or
0.04 n; ni 4

=25 i.e., nl = 1 and n 2 = 2

0.04
Example 31. Calculate the energy emitted when electrons
or n·= 5 of 1.0 g atom ofbydrogen undergo transition giving the spectral
Example 29. Estimate the difference in energy between line of lowest energy in the visible region of its atQmic spectrum.
the first and second Bohr orbit for hydrogen atom. At what (RH = 1.1 x 107 m-I;c= 3x 108 ms-I;h = 6.62x 10-34 J-s)
a
minimum atomic number would. transition from n = 2 to n = 1 (liT 1993)
energy level result in the emission of X-rays with Solution: The transition occurs like Balmer series as
A = 3.0 x 10-8 m ? Which hydrogen-like species does this atomic spectral line is observed in visible region.
number correspond to? (liT 199~) Thus, the line of lowest energy will be observed when
Solution: AE hv = -
h·c transition occurs from 3rd orbit to 2nd orbit, i. e., nl =2 and
A n 2 = 3.
and .!.= R
A
[_In~ __ 1]
ni ±=R[2; - 3~] :6 R
AE , [1-2-2"1]
R·h·c E=hv h·~=6.62xJO-34 x3xlO8 x2 x 1.1 x lO'
,nl n2 A 36
3 = 3.03 x 1O~19 J per atom
AE=h.c.
, 4R
Energy corresponding to ).0 g atom of hydrogen
{j.625 X lQ-34 x 3 X 108 x 1.09678 X 10 7 x 3
="-----~------------- 3.03 x ) O-I~ x Avogadro' Ii number
4
~ 3.03 x 10- 19 x 6023 X 1023 J
=1.635 X 10- 18 J
= 18.25 X 104 J
For hYdro:n~I~:;::[iesi __1_] '" Example 32. How many times does the electron go around
ihe first Bohr s orbit ofhydrogen in one second? .
, nt ni 'v
SoJutif}o: Nmrtber of revolutions per second ... (i)
1]
.!.=Z2R[_1 __
2nr
8
A nl
2' 2 2.188x 10 /
n2 v= ' em see
n
1 " = Z2 x 1.09678 X 107 x
3.0x 10-8 12
[..!. -~]
22 ,..',X 10~;,,'
v = 2',18.8 ':~,' / see
'" ,=; 2'' ."'I'S'8"-x '1'08-, em
- 1'--'
Z2= 4 z4 Il
Z
3 x 10-8 x 1.09678 X 10' x 3 r=-xO.529A
Z
or Z=2 12 ,
The species is He + . = - x 0.529x 10-8 cm
1
, "Example 30. What transition in the hydrogen spectrum
0.529 x 10-8 em
have the same wavelength as Balmer transition n = 4 to n = 2 of
He+ spectrum? (liT 1993)
· 2.l88xl08
SolutiQD: For He + ion, .'. N urnber 0 f revo IutlOUS per sec = - - - - - - - - - - , -
2 x 3.14 x 0.529 x
I =Z2R[_1 _~]
A, nf n~, = 6.59x 1015

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86 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

;;)<:> Example 33. Calculate the wavelength of radiations [Hint: . 1iHe+ ) = Z 211 (H)
emitted, produced in a line in Lyman series, when an electron 22 x E = 4E]
falls from fourth stationary state in hydrogen atom. 8. The number of spectral lines that are possible when electrons
(R H L! X 107 1
111- ) (liT 1995) in 7th shell in different hydrogen atoms return to the 2nd
shell is:
Solution: 1 - R [-I __1_'j\
A- n; ni (a) 12
" [Ans. (b)]
(b) 15 (c) 14 (d) lQ

Llx 10 7 (~, _ _
11 (fl:! - nl )(llz n t + 1)
, '4 ) e 2 [Hint: Number of spectral lines
2
969;6 x IO-10metre (7 - 2)(7 2 + 1) = 15] <
A= 969.6A 2
Example 34. What is the degeneracy of the level of the 9. The ratio of radii of first orbits ofH, He + and Li 2 + is:
(a)I:2:3 (b)6:3:2 (c)I:4:9 (d)9:4:1
. hydrogen atom that has the energy (- RH
, . 9
J, ? [Ans. (b)]
n2
Solution: E =_ RH =_ RH [Hint: r=~ x 0.529 A
n, 'n2 ,9 Z
n 3 rH : rHe + : rLi2+

, Thus, 1= 0 and m ::::: 0 (one 3s--orbital) 1: 1 .!

1=1 and m = 1,0, + I (three 3 p-orbitals) ; ,~ , 2 3
1=2 and m =-
2, -1, 0, + 1,+ 2 (five 3d-orbitals) 6 : 3 : 2]
Xhus, degeneracy is nine (1 + 3 + 5 , 9 states). 10. The energy of second orbit of hydrogen is equal to the energy
i>:;:i?U'r1:~" ..." . of:
~;~;'?Jl:xample 35. Calculate the angular frequency of an
(a) fourth orbit of He + (b) fourth orbit ofLi 2+
electron occupying the second Bohr orbit ofHe + ion.
(c) second orbit ofHe+ (d) second orbit ofLi 2+
. ' 2nZe 2
Solution: Velocity of electron (v) = - - ... (i) [Ans. (a)]
nh Z2
2 2 [Hint: E= -2 x 13.6eV
.
RadIUS . .m an orb'It (In) =
0 fHe +Ion n h 2 ... (ii) n
4n 2 mZe E2 -- - 4 ~,or 'H'.
13,6
Angular frequency or angular velocity (00) ,
z2
v '21tZe 2 4n 2 mZe 2 Sn 3 mZ 2 e 4 .E = - x 13.6 eV
-=--x =-- --
rn nh n 2h2 n 3h 3 13.6 Z2
28 10 -2 x 13.6
Given, n == 2,m"" 9.1 x 10- g, Z== 2,e= 4.Sx 10- esu 4 n
Z2 1
h =:6.626 x 10-27 erg -sec (Z =1, n =. 2)]
4

00=
8x (2;r '
x 22 x 9.1 X 10~28 x (4.Sx 10-10 )4
'r-'- - -
11. What is the energy in eV required to excite the electron from
n =' lto n = 2 state in hydrogen atom? (n = principal quantum
number) teET (J&K) 20061
(2)3 x (6.626x 10-27 )3 (a) 13.6 (b) 3.4 (c) 17 (d) 10.2 '
[Ans. (d)]
[Hint: b.E = E2 EI

, 6. If the speed of electron in first Bohr orbit of hydrogen be 'x',

( - 13.6 J-( - 1~~6 J
then speed of the electron in second orbit ofHe+ is:
(a)x 12 (b) 2x (c) x (d) 4x 13.6 ( 1- ~J ! x 13.6 = 10.2 eVl
[Ans. (c)] 12. An electron in an atom undergoes transition in such a way that
[Hint: v =VI X Z = x x 2 = x] its kinetic energy changes from x to x , the change' in
II n ,2 4
7. If first'ionisation energy of hydrogen is E, then the ionisation potential energy will be :
energy of He+ would be: 3' 3 3 3
(a) + -x (b) -x (c) +-x (d)-x
(a) E (b) 2E (c) 0.5E (d) 4E 2 8 4 4
[Ans. (d)] [Ans. (a)]

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ATOMIC STRUCTURE 87

[Hint: PE = - 2KE value of mass (m) increases, i. e. • in the case of heavier particles,
the wavelength is too small to be measured. de Broglie equation .
:. PE will change from - 2x to _ 2x is applicable in the case of smaller particles like electron and has
4
no significance for larger particles.
Change in potential energy (- ~ ) - (-2x)
(A) de Broglie wavelength associated with charged
particles
=_::+2x=3x]
2 2 (i) For electron:

,#,~,i~} PARTICLE AND WAVE NATURE OF

ELECTRON (ii) For proton:
In 1924, de Broglie proposed that an electron, like light, behaves
both as a material particle and as a wave. This proposal gave birth
I\=-rvA'.
~ O.286

to a new theory known as wave mechanical theory of matter. (iii) For a-particles:
According to this theory, the electrons, protons and even atoms,' A=0.10I A
when in motion, possess wave properties.
de Broglie derived an expression for calculating the
rv
wavelength ofthe wave associated with the electron. where, V =accelerating potential oftQe~~particles.
Accoram15to Planck's equation,
'-. '., · · · e (B) de Broglie wavelength associated with
E=hv h·,- ... (i) uncharged particles
:A
The energy of a photon on the basis of Einstein's mass- energy (i) For neutrons:
relationship is h 6.62 X 10-34
A=-- =-======
E = me2 .,. (ii) .J2Em ~2 x 1.67x 10-27 x E
where, e is the velocity of the electron.
Equllting both the equations, we get . = 0.286 A
e
h-=me
2' JE (eV) .
A ...
(ii) For gas molecules:
A =!!-. = h h
me p A=---
mxvnns
Momentum of the moving electron is inversely proportional to
h
its wavelength.
Let kinetic energy of the particle of mass 'm' is E. .J3mkT
I 2 where, k = Boltzmann constant
E=-mv
2 Bohr theory versus de Broglie equation: One· of the
2Em=m 2v
2 postulates of Bohr theory is that angular momentum of an electron
42Em = mv = p(momentum) is an integral multiple of~. This postulate can be derived with the
2n >

A=!!..= h help of de Broglie concept of wave nature of electron.

P :J2Em Consider an electron moving in a circular orbit around
Davisson and Germer made the following modification in de nucleus. The wave train would be associated with the circular
Broglie equation: orbit as shown in Fig. 2.15. If the two ends ofan electron wave
Let a charged particle, sayan electron be accelerated with a meet to give a regular series of crests and troughs, the' electron
potential of V; then the kinetic energy may be given as: wave is said to be in phase, i. e. ,the circumference of Bohr orbit is
1 2 equal to whole number multiple of the wavelength ofthe electron
- mv =eV wave.
2
m2v 2 = 2eVm
mv .J2eVm = p
h
A=--
. .J2eVm
A= ~ for charged particles of charge q
,,2qVm
de Broglie waves are not radiated into space, i. e., they are
always associated with electron. The wavelength decreases if the Fig. 2.15

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88 G.A.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

So, 21tr= n/.. In the case of bigger particles (having considerable mass), the
/.. 21tr value of uncertainty product is negligible. If the position is
or ... (i)
n known quite accurately, i. e. , Ilx is very small, Av becomes large
from de Broglie equation, and vice-versa. Thus, uncertainty principle is important only' in '
/..= h, the case of smaller moving particles like electrons.
... (ii) • for other canonical conjugates of motion, tIie equation for
mv.'
IT 21tr Heisenberg uncertainty principle may ~e given as:
Thus, -'=---=-- momentum = mass x velocity
mv n
. velocity .
h ==massx xbme
mvr = n . - (v =;: velocity of electron time
, 21t
and r=;: radii of the orbit) == force' x time
momentum x distance == force x distance x time
i. e., . '\.ngu1ar momentum = n . - h . .. ( III
... )
==energy J< time
21t
Ap Ilx == AE At
This proves tlut the de Broglie and Bohr concepts are in .' ·h·
perfect agreement with each other.· ' . AE At Co - (for energy and time)
41t
Similarly, A Ij) AS Co.!!:.... (for angular motion)
iil~ll;. ",HEISENBERG UNCj:RTAINTY 4~
PRINCIPLE . On the basis of this principle, therefore, Bohr picture of an
electron in an atom, which gives a fixed position in a fixed orbit
Bohr theory considers an electron asa material particle. Its
and definite velocity to an electron, is no longer tenable. The best
position and momentum can be determined with accuracy. But,
we can think of in terms of probability of locating an electron
when an electron is considered in the form of wave as suggested
with a probable velocity in a given region of space at a given
by de Broglie. it is not possible to asc~rtain simultaneously the
time. The space or it. three dimensional region round the nucleus
exact position and velocity of the electron more precisely at a
where there is maximum probability of finding an electron of a
given instant since the wave is extending throughout a region of
specific energy is called an atolUic orbital.'
space. To locate the electren, radiation with extremely short
wavelength is required. Radiation that has short wavelength is
very energetic in nature. When it strikes the electron, the Impact
: : ::::a_SOME SOLVED EXAMPLES\a::::::
causes a change in the velocity of the electron. Thus, the attempt Example 36. Calculate the wavelength associated with an
to locate the electron' changes ultimately the momentum of the electron moving with a velocity of10 10 cm per sec.
, electron. Photons, with longer wavelengths are less energetic and 28
Solution,: Mass of the electron = 9.10 x 10- g
cause less effect on the momentum of the electron. Because of
Velocity of electron == 1010 cm per sec
larger wavelength, such photons are not able to locate the
position of an electron precisely. h == 6.62 X 10-27 erg- sec
In 1927, Werner Heisenberg presented a principle known as According to de Btoglie equation,
Heisenberg uncertainty principle which states: "It is ,. h 6.62 X 10-27
/ . . = - = - - 28
----
impossible to measure simultaneously the exact position and . mv 9.l0x 10- x 1010 .
exact momentum of a body as smaU as an electron."
== 7.72x 10- 10 cm
The uncertainty of measurement of position, Ax and the
uncertainty of momentum, IIp or llmv are related by Heisenberg's 0.0772 A
relationship as: Example 37. Calculate. the uncertainty in.the position ofa
1lx·1:!.p Co hi 41t partieie when the uncertainty in momentum is:
or Ilx . llmv Co h I 41t (a) Ix 10-:-3 gcmsec- 1 (b) zero.
where, IT is Planck's constant. Solution: (a) Given, ..
28
for an electron of mass m(9.l0x 10- g), the product of AP= lxlO-3 gcmsec- 1
uncertainty is quite large~ h 6.62 x 1.0-27 erg- sec
6.626 x 10-27 1t == 3.142
Ilx ·llv Co - - ' - - - - -
.41tm According to uncertainty principle,
. h
6.626x 1O-2~ A,;r: .I:!.p Co -
Co - - - - - - - - - 41t
4x 3.14x9JOx . '27
So, Ilx Co .!!:.... • _1_ Co 6.62 x 10- x _I_
'" 0.57 erg- sec per gram approximately
. 41t Ap ··4 x 3.142 10-3 .
Ilx' Av = uncertainty product
=
When Ilx = 0, Av 00 and vice-versa. ' == 0.527 x 10-24 em
(b) When the viuue of Ap 0, the value of Ilxwill be infinity.

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I
. ATOMIC STRUCTURE 89

'>', J1x.a~ple 38. .Calculate the momentum ofa p(1rtide which, Example 43. The kinetic energy of an electron is
has a de Brogiie wavelength of2.5 x 10- 10 m. 4.55 x 10-25 J. Calculate the wavelength, (h = 6.6 x 10-34 J-sec,
(h = 6.6 x lQ~34 kgm 2 s -I ) mass of electron = 9.1 x 10-31 kg).

Solution: MOnlentuhl =:! (using de Broglie equation) Solution: KE = ~ mv 2 = 4.55 x 10-25

A 2
6.6 x 10-34
or ~ x 9.1 X 10-31 x v 2 = 4.55x 10-25
2.5 X 10- 10 2
= 2.64 x 10- 24
kg m se'c-\ 2 2 x 4.55 X 10-25
or V =-----::-:--
9.1 X 10-31
Example 39. What is the mass of a photon ofsodium light
with a wavelength of5890"A'? v = 103 ms- I
(h = 6.63 x 10-27 erg - se~, c = 3 X 1010 cml sec) Applying de Broglie equation,
A= ~ ~
34
h 6.6x 10- = 0.72 x 10-6 m
Solution: A=-
mc mv 9.1 x 10-31 x 10 3
I h
or m=- Example 44. The speeds of the Fiat and Fe;rari racing
Ac cars are recorded to ± 4.5 x 10-4 m sec -I. Assuming the track
6.63 X 10-27
So, m = -------::------,-:- distance to be known within ± 16 m, is the uncertainty principle
5890 X 10-8 x 3 X 10 10 violatedfor a 3500kg car?
·=3.752xl0-33 g Solution: llx ft!.v = 4.5 X 10-4 X 16
' .. Example 40. The uncertainty in position and velocity of a =7.2 X 10-3 m 2 sec-I. . .. (i)
particle are 10- 10 m and 5.27x 10-24 ms- I respectively.
34
h 6.626 x 10-
Calculate the mass of the particle. (h = 6.625 x 10-34 J - s) ... (ii)
41tm4 x 3.14 x 3500
Solution: According to Heisenberg'S uncertainty principle,
h = 1.507 x 10-38
llx·mft!.v=-
41t Since, llx ft!.v ::2: hi 41tm
h Hence, Heisenberg uncertainty principle is not violated.
or m=----
41t llx· ft!.v Example 45. Alveoli are tiny sacs in the lungs whose
6.625 X 10-34 average diameter is 5 x 10-5 m. Consider an oxygen molecule
4 x 3.143 X 10- 10 x 5.27x 10-24 (5.3 x 10-26 kg) trapped within a sac. Calculate uncertainty in
= 0.099 kg the velocity of oxygen molecule.
Example 41. Calculate the uncertainty in velocity of a Solution: Uncertainty in position llx = Diameter of Alveoli
cricket ball ofmass 150 g ifthe uncertainty in its position is ofthe = 5x 10- 10 m
order ~fl A (h =6.6 x 10-34 kg m 2 s -I ). . h
llxft!.v::2:--
h 41tm
Solution: ft!.x·mft!.v=-
41t 6.626 X 10-34
ft!.v::2:-------:-----~
ft!.v=~_h__ 26 10
4x 3.14x 5.3 x 10- x 5x 10-
41t llx· m ft!.v"" 1.99 ml sec
6.6xlO-34
.6f-OBJEcTIVt· QUESTIONS'
4 x3.143 X W- IO x 0.150
= 3.499 X 10-24 ms- I 13. If the kinetic energy of an electron is increased 4 times, the
wavelength of the de Broglie wave associated with it would
. Example 42. Find the number of waves made by a Bohr
become:
electron in one complete revolution in the 3rdorbit. (liT 1994)
(a) 4 times (b) 2 times
Solution: VeloCity of the electron in 3rd orbit = ~
. 21tmr .) 1 .
(c - tImes (d) J.. times
where, m = mass of electron and r = radius of 3rd orbit. 2 4
_Applying de Broglie equation,
[Ans. (c)]
A = ~ = !!... x 21tmr = 21tr
mv m' 3h 3 . h
[Hint: A = - - where, E = kinetic energy
.' . 21tr 21tr- .J2Em'
No. of Waves == - := ~ x 3 = 3
A 21tr

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90 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

When, the kinetic energy of electron becomes 4 times, the ~.~ WAVE MECHANICAL MODEL OF ATOM
de Broglie wavelength will become half]
The mass of photon having wavelength 1 nm is: The atomic model which is based on the particle and wave nature
14.
of the electron is known as wave mechanical model of the
(a) 2.21 x 10-35 kg (b) 2.21 x 10-33 g
atom. This was developed by Erwin Schrodinger in 1926. This
(c) 2.21 X 10-33 kg (d) 2.21 x 10-26 kg model describes the electron as a three-dimensional wave in the
[ADS. (c)] electronic. field of positively charged nucleus. SchrOdinger
derived an equation which describes wave motion of an electron.
[Hint: A= h
me The differential equation is:
h 6.626 x 10-34 d2W d 2 W d 2W 8n 2m
m - = ---,:-----;:- +--+ +--(E V)W=O.
Ae 1 x 10-9 X 3 X 108 dy2 h2 ;
= 2.21 X 10-33 kg] where, x, y and z are cartesian coordinates of the electron;'
15. The de Broglie wavelength of 1 mg grain of sand blown by.a m = mass of the electron; E = total energy of the electron;
20 ms- I wind is: . V = potential energy of the electron; h = Planck's constant and
(a) 3.3 x 10-29 m (b) 3.3 x 10-21 m W(psi) = wave function of the. electron.
Significance of W: The wave function may be regarded as
(~) 3.3 x 10-49 m (d) 3.3 x 10--42 m
the amplitude function expressed in tenns of coordinates
[Ans. (a)]
34
x, yandz. The wave function may have positive or negative
[Hint: A =.!!..- = 6.626 x 10- = 3.313 x 10-29 m] values depending upon the values of coordinates.
mv 1 X 10-6 X 20 The main aim ofSchrOdinger equation is to give a solution for
16. In an atom, an electron is moving with a speed of 600 m sec-I the probability approach. When the equation is solved, it is
with an accuracy of 0.005%. Certainty with which the observed that for some regions of space the value of Wis positive
position of the electron can be located is: and for other regions the value of W is negative. But the
(h = 6. 6x 10-34 kgm2 s-I ,mas~ofelectron= 9.1 x 10-31 kg) probability must be always positive and cannot be negative. It is,
(AIEEE 2009) thUs, proper to use W2 in favour ofW.
(a) 1. 52x 10-4 m (b) 5.1x 10-3 m Significance of W1. : W2 is a probability factor. It describes
(c) 1.92x 10-3 m (d) 3.84 x 10-3 m the probability of finding an electron within a small space. The
[Ans. (c)] space in which' there is maximum probability of finding an
electron is termed as orbital.
[Hint: ACcuracy in velocity =0.005% .
The solution of the wave equation is beyond the scope of this
Av = 600 x 0.005= 0.03 book. The important point ofthe solution of this equation is that it
100 provides a set of numbers, called quantum numbers, which
According to Heisenberg's uncertainty principle, describe energies ofthe electrons in atoms, information about the
h shapes and orientations of the most probable distribution of
AxmAv;::'- electrons around the nucleus. .
411:
Wave function \If can be plotted against distance Or' from
6.6 X 10-34
Ax = -------:;::;--~- nucleus as,
4x3.14x9.1x x 0.03
1.92xI0-3 m] '"
<.:
17. Velocity of de Broglie wave is given by:
'If
e2 h 2
(a) -
V
(b) ~
me
(c) me
h
(d) VA t
[Ans. (b)]
.h h Node
[Hint: A
mv p
h
p=- 'If
A
mv=
hv t
e
v hv ]
me

For hydrogen wave function, number of nodes can be

calculated as, #

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ATOMIC STRUCTURE 91.

(i) Number of radialnodes = (n -/-1)

(ii) Number of angular nodes = I 2s 2 (~J312 (2~ :: )e-ZrI2QO
(iii) Total number of nodes (n I)
(iv) Number of nodaLplanes =I
If the node at r = is also considered then no. of nodes will be
2p 2 _1 (zr )312 (zr) e-Zrl2ao
Note:
'n' (not n 1).
00
.J3 2a o ao
Examples: (i) For Js..orbital n =1, I 0, it will have no where, Z atomic number, ao = radius of flTSt Bohr orbit of
hydrogen.
radial or angular node.
(ii) For 2s-orbital, n 2, I O,it will have only one radial Plot of Radial Wave Function 'R' :
node.
18 28
(iii) For3s-orbital, n = 3, 1= 0, it will have two radial nodes. 2p

1~
(iv) For 2p-orbital, n 2, 1=1, it will have no radial node
but it has only one angular node.
(v ) For 3p-orbital, n 3, 1= l, it will have one radial and one
angular node.
For s-orbitals: ----+ r ----+

For p-orbitals:
°
(n - l) radial nodes + angular' node (n - 1) total nodes.
r
Number of radial nodes = (n -/- 1).
At node, the value of'R 'changes from positive to negative.
(n '2)radial nodes + I angular node = (n I) total nodes.
Plot of Radial Probability Density 'R 2,:
For d-orbitals:
(n - 3) radial nodes + 2 angular nodes (n - I) total nodes. 28 2p
d 2 like all d-orbitals has two angular nodes; The difference is
z .
t\ 1s . .,.
thanhe angular nodes are cones in a d 2 orbital, not planes.
z R2~
Operator form Schrodinger Wave Equation
H'P=E'P (Operatorfonn) r ----+ r ----+ r ----+
~ [ - -h- V ' 2 +v~]
2
The plots of probability, i.e., R2 or '1'2 are more meaningful
whereH= Hamiltonian operator
.
2
81.t m than the plots of functions themselves. It can be seen that for both
Is and2s orbitats, the probability has a maximum value at r= 0,
=1'+ V i.e., in the nucleus. In case of 2s orbital, one more maximum in
Here, t = Kinetic energy operator the probability plot is observed.
V Potential· energy operator
Plot of Radial Probability Function (41tr2R2):
Complete wave function can be given as
In order to visualize the electron cloud within a spherical shell
'I' (r,9,(j)= R(r) e(9)«I>«(j) is placed at radii 'r' and 'r + dY from the nucleus. Thus radial
'---v-' ~
Radial' part Angular part ,probability function describes the total probability qffinding the
Dependence of the wave function on quantum number can be electron in a spherical shell of thickness 'dr' located at the
given as, ' distance r from the nucleus.
'Pn1m (r, 9, (j) Rnl (r) elm (9)«1> m«(j) R.P.F.= (Volume of spherical shell) x Probability density
The function R depend only on 1'; therefore they describe the , . = (41tr 2 dr) x R2
distribution of the electron as a function of r from the nucleus.
These functions depend upon two quantum numbers, n and I. The 18 2s 2p
two functions e and <l> taken together . give the angular
distribution of the electron. N

The radial part ofthe wave function for some orbitals may be
given as, 1
n
312 In the plot of radial probability against 'r', number peaks, i.e.,
o 2 ~
Is
( ) e-Zrlao
region of maximum probability n - I.

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92 G.R'.B. PHYSiCAL CHEMISTRY FOR COMPETITIONS

[Hint: For Is orbital, probability of finding the electron at the

fUlJmATfONf OF:OIl)ECTIVEQ.UESTIONS nucleus is zero.]
22. The radial wave equation for hydrogen atom is :
18.
,\f = , Ir:; (~}312 [(x l)(f 8x + 12)] e:'" .t12
t
Radial
~ov4 ao
where;, x = 2r / ao; ao = radius of first Bohr orbit.
probability . The minimum and maximum position of radial nodes from
density: :
I . Y:
,
Distance from n~cleus -+
nucleus are :
ao
(c)-,a
a
(d)~,4ao
.
o
If the above radial probability curve indicates '2s' orbital, the 2 2'
distance between the peak points X, Y is : [Ans.(b)]
(a) 2.07A '(b) l.59A (c) 0.53A (d) 2.12A [Hint: At radial node, \f 0
[Ans. (a)] ... From given equation,
[Hint: X = 0.53A, Y = 2.6A , x-l=Oandx2 -8x+12=0
Y-X=,2.6-0.53 2.07Al x 1=0 =) x 1
19. Plots for 2s orbital are :
i.e. , ~ =1; r = i' (Minimum)

x 2_ 8x + 12 = 0
(x 6)(x - 2) =0
when x-2=0
x=2
r--+ 2r = 2, i.~.,r =~ (Middle value)
llo
X, Y and Z are respectively
(a)R,R 2 and4n?R2 (b) R2, Rand 41t?R2, when x-6=O
2 2 x ='6'
(c)4n?R ,R andR (d) R2, 4n?R2 and R
2r =6
[Ans. (b)]
llo
[Hint: Y will be definitely 'R' because value of R ,cannot be
r = 31lo (Maximum)]
negative, thus X will be R2 and Z will be 41t?R. Z represents
radial probability function; its value will be zero at origin]
2~16 QUANTUM NUMBERS
20. The wave function CI') of 2s is given by :

\f2s 2lrt(:J f

2 {2~ ~} e~rl2ao "

As we know, to search a particular person in this world, four
things are needed:
(i) The country to which the person belongs
At r =ro, radial node is formed. Thus for 2s, 'iJ in terms of ao (ii) The city in that country to which the person belongs
is : (iii) The street in that city where the person is residing
, (a) 'iJ = ao (b) 'iJ = 2ao (c) 'iJ = ao /2 (d) 'iJ = 4ao (iv) The house number
[Ans. (b)]
Similarly, four identification numbers are required to locate a
[Hint: When r = ro, 'liz. = 0, then from the given equation: particular electron in an atom. These identification numbers are
2- r =0 called quantum numbers. The four quantum numbers are
Go discussed below..
r= 21lo]
21. The wave func!ion for Is orbital of hydrogen atom is given Principal Quantum Number
by: It was given by Bohr; it is denoted by 'n'. It represents the
1t -rfao name, size and energy ofthe shell to which the electron belongs.
tTl
Tis -
....
.J2 e The value of 'n' lies between I to "". '
where, ao = Radius of first Bohr orbit n = 1,2,3,4, ... 00
r =Distance from the nucleus (probability of finding Value of n 1• 2 3 4 5 6 7
,the electron varies with respect to it) Designation of shell K, L M N 0 P Q
What will be the ratio of probabilities offindiilg the electron~
at the nucleus to first Bohr's orbit ao ? (i) Higher is the vahie of' n', greater is the distance·, of the
shell from the nucleus.
(a) e (b) e 2 (c)1/ e 2 (d) zero
[Ans.(d)] Ii <"2 < r3< r4 < rs < ...

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ATOMIC STRUCTURE 93
n2 : s- subshell -t 2 electrons d-subsheH -t 10 electrons
r=-xO.529A
Z . p-subshell-t 6 electrons f-subshell -t 14 electrons
(it) Higher is the value of 'n', greater is the magnitude of g-subshell -t 18 electrons
energy. .
Magnetic Quantum Number
El <E2 <E3 <E4 <E5 ...
Z2 This quantum number is designated by the symbol 'm'. To
E x.21.69x 10-19 JI atom explain splitting of a~gle spectral line into a number of closely
spaced lines in the presence of magnetic field (Zeeman effect),
Z2 Linde proposed that eleotron producing a single line has several
= - -2 X 313.3 kcal per mol
n possible space orientations for th~ same angular momentum
Energy separation between two shells decreases on moving . vector in a magnetic field, i. e. , under tbe influence of magnetic
away from nucleus. field each subshell,is further sub-divided into orbitals. Magnetic
(E2 -Ed>(E3 -£2»(E4 -E 3 »(E5 -E4) quantum number describes the orientation or distribution of
electron cloud. For each value of 'I', the magnetic quantum
(iii) Maximum number of electrons in a shell* = 2n 2 number 'm' may assume all integral values from -1 to + I
(iv) Angular momentum can also be calculated using including zero, i. e. , total (21 + 1) values.
principal quantum number
nh
°
Thus, when I 0, m = (only one value)
when I = I, m = -1, 0, + I (three values)
mvr=-
2n i. e. , three orientations.
One orientation corresponds to one orbital. Three orientations
Azimuthal Quantum Number
(orbitals) are designated as Px' pyand pz.
It was given by Sommerfeld; It IS also called angular When I 2,m -.2,-1,0,+ 1,+2 (five values), i.e., five
quantum number, subsidbuy quantum number orseeondary orientations.
r;
quantum number. It is denoted by' its value lies between 0, I, The five orbitals are designated as:
2. ' .. (n -I),
dxy' dyz , dzx, dx 2 - Y 2 and d z 2'
It describes the spatial distribution of electron cloud and
CL'1gular momentum. It gives the name of the subshell associated' When J= 3, m 3, +
1,0, I, + 2, + 3 (seven values),i.e. ,
with the main shell seven orientations.
I", 0 s-subshell; / I p-subshell; Different values of 'm ' for a given value of'/' provide the total
I ~ 2 d-subshell; l == 3 f-subshell; number of ways in which a given s, p, d and f subshells in
presence of magnetic field can be arranged in space along
I 4 g-subshell. ..
s',P.' d, f and g are spectral terms and signify sharp. principal.
x, yand z axes or total number of orbitals into which a given
diffused. fUndamental and generalised respectively. subshell can be divided.
The energies of the various subshells in the same shell are in When I :: 0, m = 0, i. e. , one value implies that's' subshell has
the order of s < P < d < f < g (increasing order), Subsh~lls only one space orientation and hencYl> it can be arranged in space
having equal I values but with different n values have similar only in one way along x, y or z axes. Thus, 's' orbital has a
shapes but their sizes increase as the value of 'n' increases. symmetrical spherical shap~,and is usually represented as in Fig.
2s-subshell is greater in size than Is-subshelL Similarly 2.16.
2p, 3p,4p-subshells have similar shapes but their sizes increase In case' of Is-orbital, the electron cloud is maximum at the
in the order 2p < 3p < 4p. . nucleus and decreases with the distance. The electron density at ~
Orbital angular momentum of an electron is calculated using particular distance is uniform in all directions. The region of
the expression maximum electron density is called antinode. In case of
'2s'-orbital, the electron density is again maximum at the nucleus
I-tl ~/(l+1)~=~/(/+1)1i and decreases with increase in distance. The '2s'-orbital differs in
2n
detail from a Is-orbital. The electron in a '2s'-orbital is likely to be
h
here, 1'1 " ' - found in tWo regions, one near the nucleus and other in a
2n spherical shell about the nucleus. Electron density is zero in
The magnitude of magnetic moment I-tL may be given as: nodal region.
. I-tL ~/(l+ I) BM .
where, BM == Bohr Magneton

1 BM=~=9.2732x 10-14 J
41tmc
Maximum electrons present in a subshell = 2(21 + I)
1s 2s-orbital

*No energy' shell in atoms of known elements possesses mor" than 32 electrons.

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94 I G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

z z

Fig. 2.16
When 1= 1, 'm' has three values -1,0, +1. It implies that 'p' dxy dyz d;x .dx2 d;
Orbital i
subshell of any energy shell has three space orientations, i. e. , m ±2 ±1 ±1 ±2 0
three orbitals. Each p-orbital has du-mb-bell shape. Each one is
disposed symmetrically along one of the three axes as shown in Nodal planes:
Fig. 2.16. p-orbitals have directional character. Orbital Nodal planes
Orbital P z Px Py d X)' xz,yz
m
Nodal plane xy
° ±l
yz
±1
zx
d·yz
dzx
XY,zx
xy,yz
When 1= 2,' m' has five values -2, ...:.1, 0, + 1, + 2. It implies d x2 :.. y2 x-y O,x+ y=O
that d-subshell of any energy shell has five orientations, i. e., five dz2 No nodal plane, it has
orbitals. All the five orbitals are not identical in shape. Four. of a ring around the lobe
the d-orbitals dX)'· ,dyz, d zx ,dx 2 - y 2 contain four lobes while
There are seven I-orbitals designated as I 2 2' I y(x 2 y2)'
fifth orbital d z2 consists of only two lobes. The lobes of d X)' • x(x - y)

orbital lie between x and y-axes. Similar is the case for I z(x 2 - y 2) ' Ixyz ,Iz 3, f yz 2 and 1._2'
x< Their shapes are complicated
d yz and d zx' Four lobes of d x2 _ 2 orbital are lying along x and y-
axes while the two lobes of d 2 ~rbital are lying along z-axis and ones.
z
contain a ring of negative charge surrouliQing the nucleus in Positive values of ml describes the orbital angular momentum
xy-plane (fig. 2.16). component in the direction of applied magnetic field while the
negative values of mlare for the components in opposite
direction to the applied magnetic field.

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ATOMIC STRUCTURE 95
Degenerate Orbitals
Orbitals which are located at the same energy level on the energy
level diagram are called degenerate orbitals. Thus, electrons have
equal probability to occupy any of the degenerate orbitals.
Px' Py and pz--,------t 3-fold degenerate
d-orbitals --,------t 5-fold degenerate
f-orbitals --,------t 7-fold degenerate
Degeneracy of p-orbitals remains unaffected in presence of
external uniform magnetic field but degeneracy of d and
-3 f-orbitals is affected by external magnetic field.
Fig. 21.6 (a) Space quantization in magnetic field Spin Quantum Number
It is denoted by's' and it was given by Goldschmidt.
Characteristics of Orbitals
Spin quantum number represents the direction of electron spin
(i) All orbitals of the same shell in the absence of magnetic around its own axis_
field possess same energy, i. e. , they are degenerate. (i) For clockwis~ spin,s = + Ji (i arrow representation).
(ii) All orbitals of the same subshell differ in the direction of (ii) For anticlockwise spin, s = - Ji (J- arrow representation).
their space orientation. Spin electron produces angular momentum equal to I-l s given by
, h
(iii) Total number of orbitals in amain energy shell is equal to I-l s = ~s( s + 1) - , where, s = + Ji
n 2 (but not more than 16 in any of the main shells of the known 2n
elements). Total spin of an atom = n x Ji. (n = number of unpaired
n = 1 No. oJ orbitals = (1)2 = 1(ls) electrons)
n = 2 No. of orbitals = (2)2 = 4 (2<>, 2px, 2py, 2pz ) Spin magnetic moment (!-.I. s ) is given by

n = 3 No. of orbitals = (3)2 == 9 (3s, 3Px' 3p y' 3p z , 3d XY' I-ls =~s(s+ 1)~
2nmc
3d yz , 3d zx , 3dx2 _ y2 , 3d z2 )
Each orbital can accommodate two electrons with opposite
n=4 No. of orbitals = (4)2 =16 spin or spin paired; paired electrons cancel the magnetic moment
and develop mutual magnetic attraction as shown in the
The division of main shells into subshells and that of subshell
following Fig. 2.17.
into orbitals has been shown below:
N S
Note: Magnetic quantwn nwnber also represents quantized value of
.. . . MagnetiC field .....,.. .
z-component of angular momeritwn of the electron in an orbital
through the expression
/-
....,,~.- ....,
--, /
... ",--_ .....
--
,/./.................
.
.... "-- .... , .... ,,--~,
............... -- -- , \ /
.

Lz=mC:) /
['
I
. /":'

[
..... :\ I "'. . ','
II
\

I \
/
['
I
/"

[
, ~ I .,.. " "
II
I \
I I \. I I I \ 1
If El is the angle between z-axis and angular momentum vector, 1 1 1 1+1/2 -1121 1 1 1
\ 1 I I \ 1 I [
Lz=L cos El I I
II
,~ // " II,
[
'[
[ I I
II
" ...,"
[
'[
[

jI,
m (~) = ~l(l + I) ~ ." .... _- / , --" jI ~ .... -- / , --" ,t
cos El ..... _-,,., '---~ ' ..... _-,,., '---;/
2n 21t
or m =~l(l + I) cos El
Main shell Subs hells Orbitals S N
1st·shell ___ _
(K-shell) 'n = 1
Fig. 2.17
15 (/= 0) 15(m=0)

2nd-shell -------< "",,- 25 (/ = 0) 25(m=0) Electrons having same spin are called spin parallel and those
(L-shell) n=2 ...............
'---
2p(/=1)
..,,""~O)
-- ------
............ ,2py(m=±1)
having opposite spin are called spin paired.
Spin paired ~ II]]
2Px(m=± 1)
---------------
35(m= 0)
Spin parallel ~ [fJ]
3rd-shell Spin multipliCity: .
(M-shell)
Spin multiplicity = [2:U + 1]
where; s = spin quantum number
e.g., carbon ls2 2<>2 2p2

Normal state [!IJ [!IJ IL--i----'--I_i-----'-_---'

Excited state '[!IJ [IJ IL--i----JIL--i----JL--i----J

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96 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

. )

Spin multiplicity == 2[ ( +~) 5+(-~) J+ I . ~ PriDeipal Azimuthal Magnetie

Q.No. Q.No. Q.No.
Spin
Q.No.
No. of No. of
electrons electrons
On' om' Os' ona ona
'I'
subshell mainsheQ
2~it7 PAULI'S EXCLUSION PRINCIPLE
Each electron in an atom is designated by a set of four quantum .4 O(s) 0 +~, - Yz 2
numbers. In 1925, Pauli proposed that no two electrons in an. 1(P) 6
atom r;an have same values ofall the four quantum numbers. 2(d) 10
An orbital accommodates two electrons with opposite spin; :"'3 +Yz, -;j
these two electrons have same values of principal, azimuthal· and
magnetic quantum number but the fourth, i. e., spin quantum
-2 +Yz2' -Yz2
number will be different. -I +Yz,- Yz 32
This principle, can be illustrated by taking example of . 3 (f) 0 +Yz,-Yz 14
nitrogen. +1 + Yz2' - /2
1/

N7 1s2' 2s2 2i +2 +Yz,-Yz

+3 +Yz,-Yz
1s2 2s2 2p; 2p.:. 2p~
Conclusions:
[ill;@]; (i) The maximum capacity ofa main energy shell is equal to
Principal quantum number (n) 2 2 2 2 .2n 2 electrons.
.I (ii) The maximum capacity Of a subshell is equal to 2(2/ + 1) .
Azimuthal quantum number (l) 0 0
electrons.
Magnetic quantum number·(m) 0 0 +1 -I 0 Azimuthal Maximum eapadty of
Subenergy sbell .
Spin quantum number (s) +Yz - Yz;+ Yz - Yz; + Yz +Yz + Yz Q. 'I' eleetrons 2(U + 1)
.s 0 2(2 x 0 + 1)= 2
Out of seven electrons no two have same values of all four
quantum numbers. With the help of this principle, it is possible to p I 2(2 x I + I) =6
calculate the maximUm number of electrons which can be d 2 2(2 x 2 + l) = lO
accommodated on main energy shells and subshells. [ 3 2(2x3+ 1)=14

No. of No. of (iii) Number ofsubshells in a main energy shell is equal to the
Prindplll Aziinutfud Magn~1!' SpiD value ofn. .
eleetrons eleetrons
Q.No. Q. 1\19. Q. No. Q.No.
ona ODa . No. of subenergy
on'· .'1' 'mf '!I' Valueofn Desiguted as
.ubsbeR maiD .beR .beIII
O(s) 0 +Yz,-Yz 2 2 I I Is
2 2 2s,2p
2 O(s) 0 +Yz,-Yz 2 3 3 3s; 3p, 3d
1(P) -I 8· 4 4 4s, 4p, 4d, 4[
+)-;'-)1, \
0 +Yz,-Yz 6 (iv) Number of orbitals in a,main energy shell is equal to n 2 .
+1 +Yz,-Yz n No. of orbitalt
== I s
3 O(s) . 0 +Yz,-Yz 2
-I
2 (2i =4 s, Px; Py' p=
+)1,,-)1, )
+~"-Yz ..
3
1(P) 0 6
(v) . One orbital cannot have more than two electrons. If two
+1 '+Yz,-Yz electrons are present, their spins should be in opposite
-2 +Yz,-Yz 18 directions. .
-I +Yz,-Yz IUlI$IMfONS ·01 iOBJlCTN£QUUOOJt5
2(d) 0 .+ Yz, - Yz lO
23. The orbital angular momentum of an electron in a d-orbital is:
+1 +Yz,-Yz (DeE 2007)
+2 +Yz,-Yz (a).J6~ (b).fi h (c) h . (d) 2h
2n 2n 2n 2n
[Ans. (a)]

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ATOMIC STRUCTURE 97

[Hint: Orbital angular momentwn =,//(1 + I) -

·~h ~L.LI AUFBAU PRINCIPLE
21C
~h h Aufbau is a German word meaning 'building up'. This gives us a
== ,/2(2 + 1)- "".J6 sequence in which various subshells are filled up depending on
21C 21C
(Here, 1 2, for d-orbitals)] the relative order of the energy of the subshells. The subshell
24. Which of the following sets of quantum numbers is correct for with minimum energy is filled up first and when this obtains
an electron in 4 f-orbital? maximum quota of electrons, then the next suhshell of higher
(Jamia Millia lslamia Engg. Ent. 2007) energy starts filling.
(a) n =4, 1= 3, m =+ 4, s = + ~ The sequen~e in which. the various subshells are filled is the.
(b)n=4,1 4,m -4,s=- ~ following: .:
(c) n::: 4, I = 3, m = + 1, s =+ ~.
1== 0 1= 1 1== 2 (= 3
(d) n =3, I = 2, m::; - 2, s::; + ~
[Ans. (c)]
[Hiot:For4f, n = 4,1 = 3, m::; - 2, 1, 0, + 1, + 2, + 3 n=1
s=- Xor+ Xl
n=2
25. Match the List-I with List~II and select the correct set from
the following sets given below: .
List-I· List-U
(A) The number of sub-energy levels in an (I) n2
n=4
energy level . .
(B) The number of orbitals in a sub-energy (2) 3d n=5
level
(C) The number of orbitals in an energy I~vel (3) 21 + I .n==6
(D) n:;; 3, I::; 2, m = 0 (4) n
n=7
[PET (Raj.) 2005)
Sets (A) (B) (C) (0)
(a) 4 3 1 2 n::8
(b) 3 I 2 4
(c) 1 2 3 4
(d) 3 4 I 2
[Ans. (a)]
[Hiot: Number of orbitals in a shell = n2 Fig.2.18 Order of filling of various subshells

Number of subshells in a .shell = n Is,2s,2p,3s,3p,48,3d, 4p,5s,4d, 5p,6s,4f,5d ,6p, 78, 5f,6d, 7p.
Number of orbitals in a subshell = (21 + 1) The sequence in which various subshells are filled up can also
n = 3, 1 = 2, m = orepresents 3d] be determined with the help of(n + I )value for a given subshell.
The subsheU with lowest (n + I ) value is filled up tirs~.WheD
26. Which of the following is not possible?
[BCECE (Medical) 2007) two or more subshells have same (n + I) value, the.subshell
(a) n = 2, I = I, m = 0 (b) n ::; 2, I::; 0, m::; -I with lowest value of 'n' is fdled up first.
(c) n = 3, I ::; 0, m = 0 (d) n 3, l:::i: I, m:::-I
[Aos. (b)] SubsheU If 1 (If + I)
[Hint: When I = 0, 'm' will also be equal to zero.] Is I 0 1
2s 2 0 2
27. What is the maximum number of electrons in an atom that can
have the quantum numbers n ::; 4, me + 1?
[PMT (Kerala) 2007)
2p
3s
2
3
.I
0
3
3
} Lowest value of n

(a) 4
[ADs. (e)l
(b) 15 (c) 3 (d) 1 (e) 6 3p
48
3
4
'~J'
," .
I
0
4
4
} Lowest value of n

[Hint: n = 4; I 0; me =0 3d 3 2 5
1= 1; me = 1,0, + 1 '"',
1=2; m" = - 2, -1, 0, + I, + 2
4p
58
4
5
1
0
5
5 J Lowest value of n

./ 3; me ::; - 3, - 2, -1, 0, + I, + 2, + 3
There are three orbitals having me = + I, thus maximum numbe~
of electrons in them will be 6.]

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98 G.R.B. PHYSIC,AL G~.E;MIl:?TRYFOR COMPETITIONS

Subshell n· ::ud; p x , Py and p z • All the three have sarnt: energy. The electron can
be accomniodated on anyone of thP. 2p-orbitals.fu the case of
~a1u~"of n
4d 4 2
carbon, sixth electron is also accommodated on 2p subshell and
5p 5 : ) . Lowest
its electronic configuration is represented as 1s 2 2s 2 2p2 but three
6s 6 0 ·6
orbital diagrams can be expected ..

~.
4f 4 3
, iJ,ii' .
~
5d
6p
5
6
2
7 1.' Lowest value of n . (I) [He] -
.
- - - ' Electrons are present on two
2s .L 2p .J different orbitals with parallel spins.
7s 7 0 7
, 5f 5 3 8 (ii) [He]i J, 1. -±-- Electrons are.present on two
2s .L 2p.J d'Ifferent
. orb'Ita. Is WI'th opposite
' spms.
.
6d. 6 2 8 ) Low", "lu, of n ,
7 8 (iii) [He] i t i J, _ _ Both the electrons are present on
. The energy of electron in a hydrogen atom and other single L2p.J . one orb'ItaI Wit. h' OPP9SI~
. spms.
. .
electron species like He + , Li 2+ and Be 3+ is determined solely by Experiments show that (i) orbital diagram is correct while (ii)'
the principal quantum number. The energy of orbitals in and (iii) are not correct. This has given birth to a new rule known
hydrogen and hydrogen like species increases as follows: as Hund's rule of maximum multiplicity. It states that electrons
1s< 2s= 2p < 3s= 3p = 3d < 4s= 4p= 4d = 4/ < ... are distributed among the orbitals of a subshell in such..a way·
The complete electronic' configuration of all the known as to give the maximum nuniber of unpaired electrons witll
elements have been given in the table on next page. It is observed parallel spins. Thus, the orbitals available in a subsheH are flrst
that few of the elements possess slightly different electronic filled singly before they begin to pair. This means that pairing of
configurations than expected on the basis of Autbau Prlnelple. electrons occurs with the introduction of .second electron in
These elements have been marked with asterisk (*) sign. s-orbitals, the fourth electron in p-orbitals, sixth electron in
:?2~1:.·; HUND'S RULE OF MAXIMUM d-orbitals and eighth electron in / -orbitals. The orbital diagrams of
nitrogen, oxygen, fluorine and neon are as given below:
. MULTIPLICITY (Orbital Diagrams)
Nitrogen (7) [He] 1t 1 i i
There is one more method of representing the electronic. config-
,. uration which is usually called as orbital diagra.~. In this method,
the' electron is shown by an arrow: upward direction i (clockwise . Oxygen (8) [He] i J, iti"i
spin) and downward direction t (anti-clockwise spin). ~L2p.J
To indicate the distribution of electrons among the orbitals of Fluorine (9) [He] itiJ,iJ,i
an atom, arrows are placed over bars that symbolise orbitals. 2s

Hydrogen, for exam~le, is represented as f. The next element

Neon. (10) [He] iJ,iJ,i,J,lt
~L2[JT
. atomtc
With " number 2'IS· he I'lum. It 'IS represente d as i J, , I,, e" . The orbital diagrams of elements from atOinic number 21 to 30
. . . 1s can be represented on similar lines as below:
both the electrons are present on the same orbital1s and are paired
(spins are in opposite directions). The next two. elements are Li Sc [Ar] 3d 1 4s2 [Ar] I IJ,
an:d Be with three and four electron~, respectively. These are
. represented by orbital diagrams as: Ti [ArJ3d 2 4s 2 1 1 1'-l-
i J, i V [Ar] 3d 3 4s 2 1 1 1 it
Li
1s2s
i J, i t Cr [Ar] 3d 5 4S1 i i i i i 1
Be --
1s2s Mn [Ar] 3d 5 4s2 I 1 1 1 'I tJ,
>:
These can also be written as: -
, i Fe . [Ar] 3d 6 4i IJ, 1 i 1 1 'tJ,
Li [He]-
2s Co [At-] 3.d 7 4s 2 IJ, tJ, 1 i 1 tJ,
Be [He] i J,
2s Ni [Ar) 3d 8 4s 2 IJ, IJ, IJ, 1 1 iJ,
-
In Be, 2Y-orbital has been completed. The fifth electron in the 1 J," . H
Cu· [Ar] 3d 10 4s1 IJ, tJ, H t
case of boron enters the next available subshell whiCh is 2p. Thus,
the electronic configuration of boron is 1s 2 2s 2 2pl . In the orbital Zn [Ar) 3d 10 4s2 It H i1 H tJ, tJ,
diagram [He] i J, - ~- , the 2p subshell has three orbitals 4s
. 2s 2p

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ATOMIC STRUCTURE 99
" . ELEC1lRONIC,C,ONFIGURAnONOFELEMENTS
!
, Element At,·No. ls 2s 2p; .3s 3p ,3d ~: 4d ,
"'4p 4f ·S.s5pStlSJ. - - -
H
I
I I !
He 2 2 (Is completed)
Li
Be
B
3
4
5
2
2
2
I 1
2
2 I
1 . (2scompleted)

C 6 2 2 2
N 7 2 2 3 i
i
0 8 2 .2 4
F 9 2 2 5
i i
Ne 10 2 2 6 (2 p completed)
!
Na U 2 2 6 I
Mg i 12 2
i
2 .6 2
i
. (38 completed)
Al 13 2 2 6 2 1
Si 14 2
P 15 2
2
2
6
6
2
2
2
3
I ,-

i I I
S 16 2 2 6 2 4
CI 17 2 2 6 2 5
AT i i
18 l 2 6 2 6 (3 P completed)
K 19 2 2 6 2 6 1
Ca 20· 2 2 i 6 2 6 2 (48 completed)
Sc 21 2 2 6 2 6 1 2
Ti 22 2 2 6 2 6 2 2
V 23 2 2 6 2 6 3 2
*Cr 24 2 2 6 2 6 5 1
Mn 25 2 2 6 2 6 5 2
Fe 26 2 2 6 2 6 6 2
Co 27 2 2 6 2 6 7 2
Ni 28 2 2 6 2 6 8 2
*Cu 29 2 2 6 2 6 10 1
Zn £10 2 2 6 2 6 10 2 :(3d completed)
Ga 31 2 2 6 2 6 10 2 1
Ge 32 2 2 6 2 6 10 2 2
.As 33 2 2 6 2 6 10 2 3
Se 34 2 2 6 2 6 10 2 4
Br 35 2 2 6 2 6 10 2 5
Kr 36 I 2 2. 6 2 6 10, 2 6 ! (4p completed)
Rb 37
i
2 2 6 2 6 10 2 6 I t
Sr 38 2 2 6 2' 6 10 2 6 2 (5scompleted)
y 39 2 2 6 2 6 10 2 6 1 2
Zr 40 2 2 6 .2 .6 10 2 6 2 2
*Nb 41 2 2 .6 2 6 10 2 (; 4 I
·*Mo 42 2 '2 6 2 6 10 :2 6 5 ' 1
Tc 43 2 2 6 2 6 10 . 2 6 5 2
*Ru 44 2 2 6 2 6 10 2 6 7 1
*Rh 45 2 2 6 '2 6 10 2 6 8 1
i !
·Pd i 46 2 2 .6 2 6 10 2 6 10
*Ag 47 2 2 6 2 .6 10 2 6 10 1
Cd 48 2 i
i 2 6 2 6 10 2 6 10 2 (4d completed)
In 49 2 2 6 2 6 10 2 6 10 2 1
Sn 50 2 2 6 2 6 10 2 6 10 2 2
Sb 51 2 .2 6 2 6 10 2 6 10 2 3
Te 52 .2 2 6 2 6 10 2 6 I 10 2 4
I 53 2 :.t. 2 2
I 6 6 10 2 6 10 2 5
I ,
- ,I I I
Xe 2 Cf
!
54 I ""'""~~
2 6 2 6
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100 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

ELECTRONIC CONFIGURATION OF ELEMENTS

m_.tl·~N~ I I K
L M 46 4.P- 4d ,,4/ 56 5p 5d Sf I 6.¥ 6p16d16f 16
Cs 8 18 2 6 6

ffi-f1
I
Ba 8 18 2 6 6 2 (68 completed)
'"La 57 8 18 2 6 6 I 2
'"ee 58 2 8 18 2 6 6 I 2
Pr 59 2 8 18 2 6 10 3 2 6 2
Nd 60 2 8 18 2 6 10 4 2 6 2
'"Pm 61 2 8 18 2 6 10 5 2 6, ' 2
8m 62 2 8 18 2 6 10 6 2 6 2
Eu 63 2 8 18 2 6 10 7 2 6 2
'"Gd 64 2 ,8 18 2 6 10 7 2 6 I 2
Tb 65 2 8 18 2 6 10 9 2, 6 2
Dy 66 2 8 18 2 6 10 10 2 6 2
Ho ,67 2 11 18 2 6 , 10 '11 2 6 2
Er 68 2 8 18 2 6 10 12 2 6 2
Tm 69 2 8 18 2 6 10 ' 13 2 6 :2
Vb 70 2" 8 18 2 6, t6;' 14 2 6 2
Lu 71 2 8 18 2 ,.,6 10 14 2 6 I 2 (4/ completed)
Hf 72 2 8 18 32 6 2 . 2
Ta 73 2 8 18 32 6 3 2
W 74 2 8 /
'i8 32 6 4 2
Re 75 2 $ 18 32 2 6 5 2
Os 76 2 " 8 18 32 2 6 6 2
Ir 77 2 8 18 32 2 6 7 2
'"Pt 78 2.. 8 18 32 2 6 9 1
'"Au 79 2 8 18 32 2 6 , 10, J
Hg 80 2 8 18 32 2 6 10 2 (5d completed)
Tl 81 2 8 18 32 2 6 10 2 1"
Pb 82 2, 8 18 32 2 6 10 2 2
Bi 83 2 8 18 32 2 6 10 2 3
Po 84 2 8 18 32 2 6 10 2 4
At 85 2 8 18 32 2 6 10 2 5
Rn 86 2 8 18 32 2 6 10 2 6(6p completed)
Fr
Ra
'"Ac
·Th
·Pa
PH+90
91
2
2
8
8
8
8
8
18
18
18
18
18
32
32
32
32
32
2
2

;
2
6
6
6
6
6
10
10
10
10
10
0
2
2

;
2
6

I
2
I
1
2 (7scompleted)
2
2
2
'"U 92 2 ~ 18 32 2 6 10 3 2 6 I 2
·Np 93 2 8 18 32 2 6 10 4 2 6 1 2
Pu 94 2 8 18 32 2 6 10 6 2 6 2
Am 95 2 8 18 32 2 6 10 7 2 6 2
·Cm 96 2 8 18 32 2 6 10 7 2 6 I 2
*Bk 97 2 8 18 32 2 6 10 8 2 6 I 2
Cf 98 2 8 18 32 2 6 10 10 2 6 2
Es 99 2 8 18 32 2, 6 10 11' 2 '6 2
Fm 100 2 8 18 32 2 6 10 12 2 6 2
Md 101 2 8 18 32 2 6 10 B ,2 6 2
No 102 2 8 18 32 2 6 10 14 2 6 2
*Lr 103 2 8 18 32 2 6 10 14 2 6 I 2 (5/ completed)
KuorRf 104 2 8 18 32 2 6 10 14 2 6 2 2
HaorDb. 105' 2 8 18 32 2 6 10 14 2 6 3 2
Sg 106 2 8 18 32 2 6 10 14 2 6 4 2 I!!
Dh 107 2 ,8 18 i,~·g
32 2 6 10 14 2 6 5 2 ,ggi
Hs 108 2 8 18 32 2 6 10 14 2 6 6 2 !Jtj
Mt 109 2 8 18 32 2 6 10 14 2 6 7 2 "8
*UunorDs 110 2 8 18 32 2 6 10 14 2 6 9 I
*Uuuor Rg III 2 8 18 32 2 6 10 14 2 6 10 1
Uub 112 2 8 18 32 2 6 10 14 2 6 10 2 (6d completed)
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I
.\

ATOMIC STRUCTURE 101

I

All those atoms which consist 'of at least one of the orbitals
singly occupied behave as paramagnetic materials because these
•

:::=
::[II- I]
hi
o
.. A
are weakly attracted to a magnetic field, while aU those atoms in
which all the orbitals are .doubly occupied behave as . where, Vo and Ao are threshold frequency and threshold'
diamagnetic materials because they have no attraction for wavelength respectively.
magnetic field. However, these are slightly repelled by magnetic Stopping potential: The minimum potential at which the
field due to induction. plate photoelectric. current becomes zero is called stopping
Magnetic moment may be calculated as, potential.
IfVo is the stopping potential, then
11=~n(n+2)BM eVo = h(v- v o )
eh
I BM (Bohr Magneton) -- Laws of Photoelectric Effect
.•. 4nmc·
where, n =no. of unpaired electron (i) Rate of emission of photoelectrons from a metal surface
is directly proportional to the intensity of incident light.
Exceptions to Aufbau Principle (ii) The maximum kinetic energy of photoelectrons is
In some cases, it is seen that actual electronic arrangement is directly proportional to the frequency of incident
slightly different from arrangement given by aufbau principle. A radiation; moreover, it is independent of the intensity of
simple reason behind fhis is that half-filled and fuU:'filled light used.
subshells have got extra stability. (iii) There is no time lag between incidence of {ight and
Cr24 -----+ ls2,2!i2l,3i3l3d 4 ,4s2 (wrong) emission of photoelectrons.
2 2 s 1 (iv) For emission ofphotoeJectrons, the frequency ofincident
-----+. lsL,2s 2l,3s 3l3d ,4s (right)
light must be equal to or greater than the threshold
Cu 29 -----+ . ls2, 2s 22l, 3s 23l3d. 9,4s2 (wrong) frequency.
-----+ ls2, 2s22p6, 3s23l3dlO, 4S1 (right)

Similarly the following elements have slightly different

configurations than expected: 28. The maximum kinetic energy of photoelectrons ejected from
Nb 41 ~[Kr]4d4581
a metal, when it is irradiated with radiation of frequency
· 2 X 1014 s-I is 6.63 x 10- 201 The threshold frequency of the
M042~[Kr]4dsSsl metal is: (PMT (Kerala) 2008)
Ru 44 ~ [Kr]4d 7 58 1 (a)2x Id4 s- 1 (b)3x Id4 s- 1
Rh 4s ~[Kr]4d8 S~I (c) 2 X 10- 14 S-I (d).l X 10- 14 s-I
Pd 46 ~[Kr]4dlO58° (e) I x 101 4 s- I
[Ans.(e)]
Ag 47 ~[Kr]4dlo581
[Hint: Absorbed energy Threshold energy + Kinetic_energy
Pt78.~[Xe]4f 145d9 fu'l of photoelectrons
Au 79 ~[Xe]4f 14 Sd lO fu'l . hv=hvo+KE
hvo =hv -' KE
La S7 ~ [Kr] 4d lO 58 2 Sp6 5ti 1fu'2 6.626 x 10- 34 X Vo = 6.626 x 10-.34 X 2X 1014 - 6.63 x 10- 20
CeS8 ~[Kr] 4d lo 4f2 58 2 Sp6 Sd°fu'2 1.3252 X 10- 19 _ {i63 x 10- 20
Yo""
6.6~6 x 10- 34
Gd64 ~[Kr]4dlo 4f7 58 2Sp 6 Sd I fu'2
Vo =9.99 X 1013 = 1014 S-l ]
220 PHOTOELECTRIC EFFECT 29. 1fAo and Abe- the threshold wavelength and the wavelength
of inCident light, the velocity of photoelectrons ejected will
Emission Qf electrons from a metal surface when exposed to light be:
radiations'f,fappropriate wavelength is called photoelectric
effect. The emitted electrons are called photoelectrons. · (a) ~;;(AO - A) (b) ~~(AO ~ A)
Work function or threshold energy may be defined as the
minimum amount of energy required to eject electrons from a
metal surface.
(c) 2hC(AO - A) .
m A.A.o
2h(1 I)
(d) _ . .....,..--
m A A o
According to Einstein, [Ans.(c)]
Maximum kinetic energy of the ejected electron · [Blnt: AbsOrbed energy = Threshold energy + Kinetic energy
=absorbed energy - work function . . of photoelectrons
I 2 he he 1 2
-mv
2
=hv-hv o -=-+-mv
max A Ao· 2

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102 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS .,.

\
, h
v = 2hc (1..0 A)] (Slope}z/(Slope)l -=e]
~ m AAo I.
Me. ,.
j • " \. 1
32. Ground state energy of H-atom is (- E) ),the velocity of
30. A radiation of wavelength A illuminates a· metal iJnd ejtlCts
photoelectrons of maximum kinetic energy of leV. Another photoelectrons emitted when photon of energy E z strikes
~, photoelectron~ of
stationary Liz+ ion in ground state will be:
radi!)tion of wavelength ejects
_ ~2(Ez ..:. E\ ) (b) _ 1'-2-(E-2-+-9E-1-) ,
maximum kinetic energy of 4 eV. What will be the work
()
av-
m
v-" ,m
function of metal?
(a) leV (b) 2eV (3) 0.5eV (d) 3eV (c) v =~2(E2: 9E1 ) : (d)v ~2(E2 ~3Ed
(Ans. (c)]
[Hint: Absorbed energy = Threshold energy + Kinetic energy
[Ans. (c)]
" 'ofphotodectrons [Hint: Threshold energy ofLi 2+ =9E) . .
c Absorbed energy = Threshold energy + Kinetic energy of
h- Eo+ leV ... (i) photoelectrons
,A
c I 2
3h-=Eo + 4eV ... (ii) E2 = 9E) + -Zmv .
A
3(Eo + leV) =Eo + 4eV
mv 2 =·2(E'z 9E1)
=0.5eV] I2
Eo . _. _v== V (E: :,!EI? 1
31. The ratio of siopes of maximum· kinetic 'energ)pversus
frequency and stopping potential (Vo) versus frequency, in
photoelectric effect gives: ' 2.21 SOME OTHER FUNDAMENTAL
(a) charge of electron (b) planck's constant PARTICLES
(c) work function (4) threshold frequency Besides protons, neutrons and electrons, many more elementary
[Ans. (a)] particles have been discovered. These particles are also called
[Hint: hv = hvo +ef/o Fundamental particles. Some of these particles are stable while
eVo=hv":'hvo the others are unstable. Out of stable particles, the electron, the·
h h proton, the antiproton and the positron are four mass particles
Vo =-v --Yo ... (i) ,
·e e while neutrino,photon and graviton are three energy particles.
(Slopelt=h / e Among these, unstable particles are neutron, meson anq
(KE)max =hv - hv 0 ... (ii)
v-particles. The main' chara~teristics of the particles are given in
(SI()pe)2 = h table 2.1 below.
Table 1.1
Particle Symbol Nature ebara' au x 10= 10 MIll (amu) DIteovered. by
Positron ,e+, leO, + + 4.8029 0.0005486 Anders()n (1932)
Neutrino v 0 0 <0.00002 Pauli
Antiproton _ p- -4.8029 1.00787 Chamberlain Sugri and Weigh land (1955)
Photon hv 0 0 0 Planck
Graviton G 0 0 0
Positive mu meson Il+ + + 4.8029 0.1152 Yukawa (1935)
, Negative mu meson .. Il - 4.8029 0.1152 Anderson (1937)
Positiye pi meson 1t+ + + 4.8029 0.1514
Negative pi meson
NeUtral pi meson
1t

1t
-
0
0
-4.8029
0
0.1514
0.1454
} Powell (1947)

2.22 ISOTOPES
Isotopes are the atoms of the same element having differen l on atomic mass. Isotopes were first identified in radioactive
atomic masses (see determination of isotopic mass). The term elements by Soddy. In 1919, Thomson established the existence
'isotope' was introduced by Soddy. This is a Greek word of isotopes in a non-radioactive element, neon. Until now, more
=
meaning same position (lsos same, topes = position), since all than 1000 isotopes .have been identified (natural as well as
the isotopes of an e1ement occupy the same position in the artifichd). Out of these about 320 occur in nature, approximately
periodic table. Isotopes of an element possess identical chemical 280 of these are stable and the remaining 40 are radioactive.
properties but differ slightly in physical properties which depend

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I
ATOMIC STRUCTURE 103

Conclusions ',' same two particles (i. e., protons) are at least forty times greater
i . .. "
than the repulsive forces; Thus, two major forces exist in the
(i) Number of n~trons present in the nuclei of various
nucleus. These rrre electrostatic and nuclear. The nuclear forces
isotopes of an ~lement is alw~ys different. The number of '
are stronger and the range of these forces is extremely small. The
neutrons is detennined by applying the formula N =A - Z
forces which operate between nucleons are referred to as
where A is mas~ number and Z is atomic number.
exchange forces. In order to account for the stability of the
Hydrogen has tHree isotopes, ~ H, fH and H. i nucleus, a theory known as meson theory was put forward by
A (MIll ~umber) Z No. of neutrons Yukawa, in 1935. Yukawa pointed out that neutrons .and protons
o are held together by very rapid exchange of nuclear particles
called pi mesons. These mesons may be electrically neutral,
2
positive ornegative (designated as nO , n + and n - ) and possess a
'tH 3 2 mass 275 times the mass of an electron. Nuclear forces arise from
Oxygen has tliree isotopes, 16 0 , 170and 18 O. a constant exchange of mesons between nucleons with very high
velocity (practically the velocity of light).
A Z No. of neutrons Let a neutron be converted into a proton by the emission of a
1:0 16 8 8 negative meson. The emitted meson .is accepted by another
proton and converted into a neutron.
llo 17 8 9
--
"J8 . _'"_TT __ "
."
n A ~ p~ +n-
gO 18- 8 '10'
n- + p; ~nB
(ii) In a neutral atom, the number of protons and the number
Similarly, a proton after emitting a positive meson is
of electrons are aht'ays the same, i. e., the electronic converted into a neutron and vice-versa,
configuration of all the isotopes of an element is the
p~ ~nA +n+
same. Thus, all the isotopes of an element show the same
chemical properties. However, the rates of reactions may n+ +nB ~p;
be different for different isotopes of an element.
or simply
(iii) All the isotopes of an element occupy the same position
in the periodic table. n ~ n- + p
(iv) The isotopes of an element differ slightly in physical There may be two more types of exchange, i. e., between
properties. The compounds formed by these isotopes will neutron-neutron and proton-proton, involving neutral pi mesons.
also have different physical properties. p n. 11.0 nO
"'-no '\no orsimply p"--; p and n~n
Determination of Isotopic Mass p It" nit" .n nO
37
Chlorine has two isotopes 17 Cl 3S and 17 Cl ; these are found
in nature in 3 : 1 ratio or 75%: 25% respectively. Isotopic mass'
Mass Defect-Binding Energy
may be calculated as: ". . , . It is observed that the atomic mass of all nuclei (except
Isotopic mass of chlorine hydrogen) is different from the sum of the masses of protons and
neutrons. For example, the helium nucleus consists of2 protons
3s 7
%of
- -- . 0 f Cl35 + %of ct3
Cl - x mass X mass 0 f Cl 37
and 2 neutrons. The combined mass of 2 protons and 2 neutrons
100 100 should be
75 25 = 2 x 1.00758 + 2 x 1.00893
=- x 35 + x 37 =35.5 4.03302 amu
100 100
OR The actual observed mass of helium nuclei is 4.0028 amu. A
Isotopic mass of chlorine . difference of 0.0302 amu is observed between these two values,
This difference is termed as mass defed.
Ratio of Cl 35 x mass of Cl 3s + Ratio of Cl 37 x mass of Cl 37
Mass defect = Total mass of nucleons Observed atomic mass
Sum of ratio
This decrease in mass (i. e., mass defect) is converted into
= 3 x 35 + 1 x 37 = 35.5
energy according to Einstein equation E mc 2 • The energy
4
released when a nucleus is formed from protons and neutrons. is
2.23 THEORIES OF NUCLEAR STABILITY called the binding energy. This is the forfe which holds all the',
Since, a nucleus contains positively charged protons, there must nucleons together in the nucleus. Binding eI1ergy can be d~fined in
exist a strong repulsive force between them. It has been other ways also, i. e. , the energy required tohreak the nucleus· into
calculated that there exists an electrostatic repulsion of constituent protons and neutrons. Binding energy is measured in
approximately six tons between two protons situated at a nuclear ' MeV (Million Electron-Volts), i. e. , 1 aml.l 931 MeV.
distance but at the same time the forces which bind the nucleus Binding energy =Mass defect x 931 MeV
are very high. It has been found that. nuclear forces attracting the

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,
104 ' G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Bjnding energy can also be calculated in erg. This is ':t«i THE WHOLE NUMBER RULE AND
Mass defect (amu) x J .66 X 10- 24 x (3 X 1010 )2 erg PACKING FRACTION
6
(l MeV = 1.60 x 10 erg) Aston believed that mass number values (sum of protons and,
The binding energy increases with the increase in atomic neutrons) of isotopes should be whole numbers on the scale of
number of the element. This indicates that heavier nuclei should oxygen ( 16 0 = 16) but actually it was observed that these were
, be' more stable than lighter nuclei. But, it is not so because not integers. The difference in the atomic mass of an isotope and
heavier nuclei above atomic number 82 are unstable. It is thus mass number was expressed by Aston (1927) as packing
clear that total binding energy of a nucleus does not explain the"" " fraction bythe following expression:
stability of the IUlcleus.
The total binding energy of a nucleus when divided -by the P k' fra' Isotopic atomic mass'- Mass number 104
, ac 109 ctlOn = " x
number of nucleons gives the average or mean binding energy Mass number
per nucleon. The binding energy per, nucleon is actually the
measure of the stability of the nucleus. The greater the binding ' the pack~109 fra'
energy per nucleon, more stable is_the nucleus. Thus, ctlOn 0 f IH = l.OO78~ I x 104 78 and
I
Binding energy per nucleon
Total binding energy
the pacmgk ~ fra'ctlon
. '0,f 3sCI
,=
34.980 35.0 x 104 =-5.7. The
Total nwnber of nucleons 35.0
packing fraction of oxygen is zero.
When binding energy per nucleon of a number of nuclei is
It is clear that the value of packing fractlon varies from one---~~­
plotted agairlst the corresponding 'maSs number, a graph is
atom to other. This is sometinie positive or zero but more often
, obtained (Fig. 2.19) whose characteristics are as follows:
negative.
Region of A negative packing fraction means that atomic mass is less
greatest stability than nearest whole number and this suggests that some mass has

f:
':> 8
I---r----,--
been converted into energy when the particular isotope has been
constituted. This energy is r~sponsible for nuclear stability. All
those having negative values of packing fraction are stable
Q) nuclei.
67 Apositive packing fraction generally indicates instability of
ju 6 1-1'---+- the nucleus. However, this statement is not correct for lighter
nuclei.
E5 In general, lower the value of packing fraction, the gr~ter is
~ 4H---+- the stability of the nucleus. The lowest values of packing
>.
fractions are observed for transition elements or iron family
~3
c:: indicating thereby maximum stability of their nuclei.
~ 21i---+-
c::
'g ,~~ ,THE MAGIC NUMBERS
iii
It has been observed that atoms with an even number of nucleons
o 20 40 60 80 100 120 140 160180200220 240 in their nuclei are more plentiful than those with odd number.
, Mass number, A - This indicates that a nucleus made up of even number of nucleons
Fig. 2.19 is more stable than a nuclei which consists of odd number of
(i) Binding energy per nucleon increases from l.l to 8.0 nucleons. It has also been observed that a stable nuclei results
MeV from mass number 2to 20. when either the number of neutrons or that of protons is equal to
(ii) Binding energy per nucleon increases from 8 to 8.6 MeV one of the numbers 2, 8, 20, 50, 82, 126. These numbers are
from mass number 20 to 40. ' called magic numbers. It is thought that the magic numbers form
closed nuclear shells in the same way as the atomic numbers of
(iii) Binding energy per nucleon remains 8.6 -8.7 MeV from
mass number 40 to 90. Iron (56) has the maximum value inert gases form stable electronic configuration. In general,
of8.7 MeV per nucleon. elements that have nuclei with magic number of !:fotons as well
(iv) The value of binding energy per nucleon decreases from a.-, magic number of neutrons such as jHe, I~O, 20 Ca, 2~~Pbare
8.6 to 7.5 MeV from masS number 90 to 240. highly stable and found in abundance in nature.
(v) Points for helium, carbon; oxygen lie quite high in the A survey of stable nuclei found in nature shows the following
graph showing that these nuclei are highly stable. trend:
The binding energy per nucleon can be increased in two ways: Protons Even Even Odd Odd
(i) Either by breaking heavy nucleus to those of NeutronS Even Odd Even Odd
intermediate mass numbers (process of fission) or No. of stable nuclei 157 52 50 5
(ii) By fusing lighter nuclei to form heavier nuclei (process Only five stable odd-odd nuclides are known; these nuclides
of fusion). 2H' 6 · 10 14N' d 180 '
are I ',3 LI, sB, 7 an 73 Ta.

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ATOMIC STRUCTURE 105
.
: : :::n_SOME SOLVED EXAMPLES\ I:::: : : Example 50. Write down the values of quantum numbers
of all the electrons present in the outermost orbit of argon
Example 46. The minimum energy required to overcome (At. No. 18).
the attractive forces between an electron and the suifaceof Ag Solution: The electronic configuration of argon is
metal is 5.52 x 10- 19 J. What will be the maximum kinetic energy 1s 2 , 2s 22p 6 , 3s 2 3p2x 3p2y 3p2z
of electrons ejected out from Ag which is being exposed to UV Values of quantum numbers are:
light ofA = 360A'? n / m s
Solution: Energy ofthe'photon absorbed 3i 3 0 0 +~.-~
h . c 6.625 X 10-27 X 3 X 1010 3p; 3 ±l +~.-~
=- =--------:::---
8
A 360 x 10- 3p.~ 3 ±1 +~,-~
= 5.52 X IO- It erg 3p; 3 0 +~,-~
= 5.52 x 10- 18 J Example 51. (a) An electron is in 5f-orbital. What
E(photon) work function + KE possible values ofquantum numbers n, I, m and s can it have?
KE=5.52xlO- 18 -'7.52xIO- 19 , (b) What designation is given to an orbital having
(i) n = 2, I =I and (ii) n = 3, 1= O? .
=47.68 X 10- 19
J
Solution: (a) For an electron in 5f-orbital, quantum numbers
"''''E,lample 41~ Let a light ()f wavelength' Aand intensity ']' are: .----,-~--~-

strikes a metal su1face to emit x electrons per second. Average n = 5; , 1= 3 ; m =- 3,- 2, - I, 0, + 1, + 2, + 3

energy ofeach electron is y' unit. What will happen to 'x' and y'
when (a) A is halved (b) intensity] is doubled? and s =elt'hel' + -I or - -I
Solution: (a) Rate of emission of electron is independent of 2 2
wavelength. Hence, 'x' will be unaffected.' (b)(i) 2p, (ii) 3s
Kinetic energy of photoelectron = Absorbed- Threshold ,.Iit. ~1i6J..altlple 52. Atomic number of sodium is II. Write down
energy energy 'the four quantum numbers of the electronhavillg highest energy.
Solution: The electronic configuration of sodium is:
he
Y"'J:-w o 1s2 ,2s2 2p6 ,3s 1

when, A is halved, average energy will increase but it will not 3s-e1ectron has the highest energy. Its quantum numbers are:
become double. 1 1
n=3, 1=0, m=O" s=+ ' 2 or--2
(b) Rate of emission of electron per second 'x' will become
double when intensity ] is doubled. Average energy of ejected Example 53. An element has 8 electrons in 4d-subshell.
electron, i. e., 'y' will be unaffected by increase in the intensity of Show the distribution of 8 electrons in the d-orbitals of the
light. . ' element with~n small rectangles.
'" Example 48. How many orbits, orbitals and electrons are Solution: 4d-subshell has five d-orbitals. These are first
there in an atom having atomic mass 24 and atomic number 12? occupied singly and then pairing occurs. The distribution can be
Solution: shown in the following manner:
Atomic number = No. of protons = No. of electrons = 12 4d
Electronic configuration 2, 8, 2 [iJ,!iJ,!H! i ! i J
No. of orbits = (K. Land M ) Example 54. How many elements would be in the third
No. of orbitals on which electrons are present period ofthe periodic table if the spin quantum number ms could
=(one 1s + one 2s + three 2p + OI~e 3s) I
have the value - - ,0 and + - ?
I
Example 49. A' neutral atom has 2K electrons, 8L 2 2
electrons and 6 M electrons. Predict from this: Solution:
(a) its atomic number, (b) total number of s-electrons. (c) total n=3,/=O,m=0
number ofp:electrons, (d) total number ofd-electrons.
Solution: (a) Total number of electrons ' . ms = -1/ 0 +1/

~
/2,,72

(2+ 8+ 6)= 16 / = I; m -I, 0, + I ms = -~, 0, +~

So, Atomic number 16 . m - -1/ 0 +1/
s - 72' , 72

Electronic configuration = tS'2 ,2s 2 2 p 6 , 3$23 p 4 72' 0, +1/

m,t -- -1/ 72 }
(b) Total number of s -electrons =(1s 2 + 2s 2 + 3s 2 ) = 6 for each value of
/ =2; m =~ 2, -.1, 0, + I, + 2 magnetic '
(c) Total number of p-electrons = (2 p 6 + 3p4)= 10 {
quantum no ..
(d) Total number of d -electrons 0

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106 G.A. B. PHYSICAL ~HEMISTRY FOR COMPETITIONS

i . • c

Number of elements 3s (3e) .' =I6.~ 8 x (lOOQ5486:= lS.99S~ aniu

3p (ge) Mass of the nucleus of l~O
3d (15e)
:= mass of 8 protons + mass of 8 neutrons
.. 27 elements will be there in third period pfperiodic table.
Example 55. The binding energy of He is 28.57 Me V. i 8x 1.0,0,757+ 8x 1.0,0,893 16.l32amu
What shall be the binding energy per nucleon of this element? Mass defect =(16.132-15.9956) Q.1364amu
Solution: The nucleus of j He consists of 4 nucleons. Binding energy =0,.1364 x 931=127MeV
. . Total binding energy Example 57. There are four atoms which have mass
So, 13mdmg energy per nucleon := - - - . --=---=- numbers 9,10,11 and 12 respectively. Their binding energies are
No. of nucleons
54,70,,66 and 78 MeV respectively. Which one of the atoms is
= 28.57 = 7.14 MeV most stable?
4 Solution: Stability depends on the "alue of binding energy
per nucleon.
Example 56. . Calculate the binding energy of the oxygen
isotope 1:0. The mass of the isotope is 16.0, amu. (Given A B C D

e= Q.QQQ5486amu, p = 1.0,0,757 amu and n l.OO893amu.) Binding energy (MeV) 54 70 66 78

Solution: . The isotope. I~O contains 8 protons, No. of nucleons 9 10 11 12
8 neutrons and 8 electrons. Binding energy per nucleon (MeV) 6 7 6 6.5
Actual mass of the nucleus.of I: 0 Thus, B is most stable.
:= 16- mass of8 electrons

MISCELLANEOUS NUMERICAL EXAMPLES~'/L/L<i

'. ... ~ .
..... .

These examples will give the sharp edge to the aspirants for Solution:
lIT and various other entrance examinations. Coulombic force of attraction =Centrifugal force
Exa.mple 1. The SchrOdinger wave equation for hydrogen 1 Zexe mv 2
atom is
4nEo T=~
_ 1 (1 [2
'112s - - - - - -
)3/2 ro ] -rlao
-- e . where, v:= velocity of electron
4~n ao ao
ao = distance between electron and nucleus
where ao is Bohr radius. If the radial node in 2s be at· ro ' then
find r in terms ofao. (liT 2004) 1 Ze 2 _ 2
-----mv
Solution: Given, 4nEo ao .

_ 1 (1 [2
'112s - - - - - -
)3/2 ro ] e -rlao
-- KE
1 2 1 Ze 2
-mv = - - - ; r -
2 . ' ~nEQ·.Lao
4~n ao ao
PE=-2xKE .
'II~s = Qat node 1 I, Ze 2
-2x~x-=
4ntb· 2ao
Example 3. Hydrogen atoms are excited from ground
state. Its spectrum contains wqvelength 486 nm. Find, what
ro = 2a o transition does the line corresponds to. Also find from this
information what other wavelengths will be present in the
Example 2. Consider the hydrogen atom to be a proton
spectrum?
embedded in a cavity of radius ao (1!ohr radiUS) whose charge is
neutralized by the addition ofan electron to the cavity in vacuum Solution: Wavelength 486 nm, i. e., 4860, A indicates that
infinitely slowly. Estimate the average total energy ofan electron the spectrum is in visible region, i. e. , Balmer series.
in its ground state in a hydrogen atom as the work done in the
above neutralization process. Also, if the magnitude of average 1. = Rz.2 [_1 ....:_11
KE is half the magnitude of average potential energy, find the A nr ni
average potential energy. (lIT 1996)

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ATOMIC STRUCTURE 107

•
_ _1_ _ g ::::: 109677.76 x 12 [~- ~l . [1 1J
=109678 - - -
32 4 2
4860x 10- 2 n2

On solving, we get A = 1.8 X 10-4 cm

ni == 16 Example 7. A negatively charged particle called
Negatron was discovered. In the Millikan oil-drop experiment, s
n2 == 4 the charges of th£ oil-drops in five experiments are reported as
Thus, transition is from, 4 ~ 2 12 x 10- 19 coulomb; 4.8 x 10- 19 coulomb; 6.4 x 10-19 coulomb;
Other transitions in the spectrum are 8 x 10-:-19 coulomb and 9.6 x 10- 19 coulomb. Calculate the charge
4~3~2 on the negatron.
Solution: In Millikan's oil-drop experiment; the charges on
1 . [1 IJ
-::::: 109677.76 x ex - 2 - - 2.
A 3 4
the oil-drops are integral multiples of the charge of the particle.
" Dividing the charges of droplets bytbe lowest charge:
3.2~ 10- 19 __ 1 4.8 X 10- 19
A == 1875 X 10-7 cm ro 12x 10- 19
00
12x 10-19
15
Example 4. lfuncertainties in the measurement of
position and momentum of an electron are equal; calculate 6.4 x 10- 19 ." 8x 10- 19
uncertainty in the measurement of velocity. .
(iii)
12x 10- 19
=2 (iv)
12x
=2.5
Solution: According to Heisenberg's uncertainty principle, 19
9.6x 10- ::::: 3"
. h (v)
Ax.~p~- 12x 10- 19
41t
All the values are not integral; they can be converted to

. .~ = V~
Given, Ax ::::: [""h : : : 0.726 X 1O-I7 integers on rriultiplyingby 2.
:.Charge of the negatron will be
~:::::m~V 19
3.2 x 10- =.1.6 X 10- 19 C
. ~ 17 2
or AV= P ::::: 0.726 x 10- =7.98 x 1012 ms- I '. . .
m 9.1 x "Example 8. When a certain metal was irradiated .with
light offrequency 12x 1016 Hz, the photoelectrons emitted had
Example 5. How much· energy will be released when a
sodium ion and a chloride ion, originally at infinite distance are twice the kinetic energy as did photoelectrons emitted when the "
brought together to a distan.ce of2. 761 (the shortest distance of same metal was irradiated with light offrequency 2.0x 1016 Hz.
approach in a sodium chloride c rys tal)? Assume that ions act as Calculate v 0 for the metal.
point charges, .each with a magnitude of 1.6 x 10- 19 C. Solution: Applying photoelectric equation,
Permittivity constant of the medium is 9x 109 Nm 2C -2.
KE hv-hvo
Solution: Energy released
" KE
9 19 or (v-vo)==
== _ K = _ 9 X 10 x (16 X 10- )2 = -8.35 X 10- 19 J h
r . 2.76 x 10- 10 Given,
. Example 6. The angular momentum of an electron in a
... (i)
Bohr orbit ofH-atom is 4.2178 X 10-34 kg m 2/sec. Calculate the
spectral line emitted when an electron fails from this level to the
next lower level. and ... (ii)
. "h
Solution: We know, mvr= n Dividing equation (i) by equation (ii),
21t
34 v 2 -vo == = 2KEI =2
4.2178x 10-34 = n x 6.626 x 10-
2x 114 VI - Vo KEI KEI

n=4 or v 2 -vo =2v 1 -2v o

16 16
or v O =2v I - V2 2(2.0xI0 ) (3.2 x 10 )

=8.0 X 1015 Hz

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108 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Example 9. An electron moves in an electric field with a ... (iii)

kinetic energy of 2.5 eV. What is the associated de Broglie
wavelength?
Solution: Kinetic energy

::::~mv2 (v:::: :1.)

::::~m(:J2 Thus, AE3~1 =fj£3~2 + AE2~1
::::
I h
--
2
.. Example n. If an electron is moving with velocity
2mA2 500 ms- I, which is accurate up to 0.005% then calculate
2 uncertainty in its position. [h == 6.63 x 10- 341s, mass of electron
or 1.2 =! h
== 9.1 x 10- 31 kg] [AIPMT (Mains)2008)
2mxKE

A==
.
h
28
[m=9.108x 10- g
. h 6.626 X 10-27 erg ~sec
1 Solution: Uncertainty in velocity
A
uv= 600 x 0.005 3
::::.x 10-2
. .
ms -I
100 .
~2mx KE leV= 1.602 x 10- 12 erg According to Heisenberg's
.. ---- •. - - -- uncertainty
-
principle
--c- -- - .,---------
c- 7 -

6.626 x 10:"27 h
AtAv~-
4rcm
~2 x 9.108 x 10-28 x 2.5 x 1.602 x 10- 12
At ~ h
== 7.7 x 10-8 cm 4nmAv
c.· Example 10.· Consider the following two electronic
transition possibilities in a hydrogen atom as pictured below: 6.63 x 10- 34
~----------------------
·4x3.l4x9.1x

=tt
n=3
. n=2 == 1.9 x to- 3 m

. n== I Example 12. Applying Bohrs model when Hatom comes

from n == 4 to n == 2, calculate its wavelength. In this process,
(a) The electron drops from third Bohr orbit to second Bohr write whether energy is released or absorbed? Also write the
orbit followed with the next transition from. second to first Bohr range ofradiatian. RH =2.18 X 10- 18 1, h = 6.63 x 10- 34 Js.
orbit.
(AIPMT 2008)
(b) The electron drops from third Bohr orbit to first Bohr
orbit directly. Show that the sum of energies for the transitions
Solution: Energy is released in this process; and the radiation
n = 3 to n =2 and n :::: 2 to n =·1 is equal to the energy oftransition
will belong to visible region (Balmer series)
forn=3ton=1. .. E= hc == RHZ2 [-\-_ -\-] -
. A nl n2
Solution: Applying, AE Riel [_I __ 1 ]. 2
n 12 n. 22 . X== R H'Z [-\- _ -\-]

For n == 3to n:::: 2;

.1. hc nl ni
2 18
2.18xlO- xI [.1 I]
AE3
~
i =RH [~-~]
22 32
=RH X 5
36
... (i) =6.63XIO-34 X3XI0 8 4-16
18
. 2.18xlO':' xtz [3]
For n = 2 to n = I; == 6.63 x lO-j4 x 3 x-108 16
8
... (ii) A= 6.63 x 10- 34 X 3 X 10 x .16 = 4866 x 10- 10 m
3x2.l8xto-
For n == 3to n 1; = 4866 A
' .....,.

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ATOMIC STRUCTURE 109

SUMMARY AND IMPORTANT POINTS TO REMEMBER

1"", I " , "11, My • ~ $JIIItI ... ' us ,. .. ~

1. Atom is the smallest indivisible particle of matter 8. Isotones: These are atoms of different elements having
(proposed by John Dalton in 1808). same number of neutrons in the nucleus, e.g.,
2. All atoms except hydrogen atom are composed of three
ItC: l~, 19O
fundamental particles, namely, electron, proton and neutron.
Hydrogen atom has one electron and one proton but n~ neutron.
9. Electromagnetic radiations are energy waves containing
(a) Electron: The nature and existence of electron was both electric and magnetic vector perpendicular to each other.
established by experiments on conduction of electricity through
(i) These are transverse waves.
gases, i.e., discovery of cathode rays. In 1897, J.J. Thomson
(ii) They do not need any medium for their propagation. They·
determined e/ m value (-1.7588 x 108 coulomb/g) and proved travel with the velocity of light. '
that whatever gas be taken in the discharge tube and whatever be
the material ofthe electrodes, the value of e/ mis always the same. (iii) v:::: C ,V =frequency, c =veiocity oflight,
A '
Electrons are, thus, common universal constituents of all atoms. A = wavelength
Electron is a' subatomic pru.:ticle which carries charge
V ='= .~1. T
I wave'Ilwiwer, .:.;,;I = tIme
. per1
' 'od.
-1.60 x 10-19 coulomb, i.e., one unit negative charge and has A," v
maSs 9.1xlO-28 g (or 9.1 x 10-31 kg), i.e., _l_th mass of (iv) According to Planck's quantum theory, the energy is
1837 emitted or absorbed in the fOlTIl of energy packets called quanta.
hydrogen atom (0.000549 amu). The name electron was given by Quantum of visible light is called photon.
Stoney. ' Energy of one quantum == hv
(b) Proton: The nature and existence of proton was
established by the discovery of positive rays (Goldstein 1886). ==h5..
A
Proton is a subatomic particle which carries +1.6x 10- 19
coulomb or one unit positive charge and has mass 1.672 x 10-24 g h = Planck's constant
27
(or 1.672 x 10- kg), i. e., 1.0072 amu. The el m was determined 6.626 x 10-34 J sec
by Thomson in 1906 and the value is +9.579 x 104 coulomb/g. It
"': 10. Hydrogen spectrum: Hydrogen spectrum is a line
was named as proton by Rutherford. spectrum. The lines lie in visible, ultraviolet and infrared regions~
(c) Neutron: It is a subatomic particle which carries no All the lines can be classified into five series. Ritz presented a
charge. Its mass is 1.675 x 10-24 g (1.675 X 10-27 kg) or 1.0086 mathematical formula to find the wavelengths of various lines,
amu. It is slightly heavier than proton. It was discovered by
Chadwick in 1932 by bombarding beryllium with a.-particles. -.!.=V=R[~-~l
A. n n l 2

where, R is Rydberg constant (R == 10 9678cm- I ).

The e/ rh value of neutron is zero.
HI H2
3. According to the Rlith~rfotd's model of atom, (i) it
consists of nucleus of very small size and high density Lyman series (UVregion) 1 2,3,4,5, ...
(ii) electrons revolve round the nucleus in a circular path. BaImer series (Visible region) 2 3,4,5,6, ...
15
Radius of nucleus = 10- m Paschen series 3 4,5,6,7, ...

4.
Density of nucleUs = 108 tonnes/cc
Atomic number (Z) =Number of protons in the nucleus
Brackett series
Pfund series
} (IRregion) 4
5
5,6,7,8, ...
6,7,8,9, ...

S. Mass number (A) =Number of protons + Number of Balmer series consists offour prominent lines Ha., IIp, Ry and
neutrons H3 having wavelength 6563 A, 4861 A, 4340 A and 4102 A
6. Isotopes: These are atoms of same element having respectively. . .
same atomic number but different mass numbers, e. g., Balmer equation is,
<l H, ;H, fH): mCI, r;Cl) -.!.=R[~ __
l2 ]
A 22 n
7. Isobars: These are atoms of different elements having
same mass number but different atomic numbers, e.g., where, H == 3,4,5,6, ...
The Rydberg formula is used to calculate the wavelength of
:~Ar, teK, ~Ca any line of the spectrum

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I
110 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

V::V Ke2 ' '-

-,.~. ' . mr
where, x :::;: number of lines in the spectrum; x = 00 for series limit where,K _1_ = 9x 109 Nm2/C2
or last fine. Let, transition of electrons takes place from ni to n l 4nEo
shell; then the number oflines can be calculated as:
v ~ x 2.188 X 108 emf sec
· (n2 nl )(n2 - nl + 1) n
Number 0 f . Imes:::;: ...:.....:=---...!...:...:..-=---'----'-
2 (vii) Potential energy of electrons in a particular shell:
2
11. Bohr's atomic model: It is based on Planck's PE:::;: -KZe =- 27.2 xZ 2 eV
quantum theory. Its main postulates are summarised as: r n2
(i) Electrons revolve round the nucleus in circular path of
(viii) Kinetic energy of electrons in a particular shell:
fixed energy called stationary states.
(ii) Angular momentum of electrons are quantised, i. e. , KE=~KZe2 =+13.6 Z2eV
2 r n2
mvr=n(:n) 2
Total Energy, TE =- -1 -
KZe
-
2 r
(iii) The energy as weU as angular momentum both are 1
quantised· for electrons. It means they can have only certain TE=-PE"
'2
. yalues of energy and angular momenta. TE -KE
12. Importantf(JrJllulatioQs .. obtained from Bohr's (ix) Number of revolutions per second by an electron in a
atomic model which are valid for single electron species like shell:' .
H, He, + L':1+
I , Be3+ ,et c..•
Velocity v
(i) EI <E2 <E3<E4 = CircUJl?ference =-=-
21tr . h
(ii) (E2 -Ed>(E3 E 2 »(E4 -E3 ) ...
where, Ei ,E2 ,E3 , ... are energies of cOrresponding shells. (~) Frequency of electroJ,ls in nth orbit:
2 2 v' ..
n h '=-'-
. (iii) r.:::;: --:c--.....,,-- ,2nr
n 4n2 K~2 mZ
:::;: 662 XlO15Z 2
'K =_1_ =9x 109 Nm 2fC;
4nEo n3 •
2 . (xi) Period of revolution of electrons in nth orbit (Tn)'
r =~Z x 0.529 A (where, r is the radius of Bohr orbit of
2nr 1.5xlO- 16 n 3
T.=-= . sec
electrons.) n Vn Z2
(iv) Energy of.electrons in a partiCular sheli can he calculated
as:

E=--
Z2 2n 2 mK.2'e4
2 (xii) Ionization energy =E~ - En'
n h2

E ='-~ x 21.79 x 10- 19 J/atom

=o-(-~: x 13.6eV}
2
n Z2
. Z2 =-x116eV
2
= - -2 x13.6eV n
n . Z2
Z2 :;:: - 2 x 21.79 X 10-'9 J / atom
=- 2 x l312 kJ/mol n
n . I Z2' 2
Z2 RE (xiii) .-l. =_1- X ~
I 2 Z2 2.
n 12
n2 . .

RE = - l3.6 eV (Rydbetg energy) I, and 12 are ionization energies of two elements 1 and 2.
2
(v) En :::;:E\ln 2 ;En =E1 X~forhydrOgen-likespecies.
n . (xiv) AE'{Energy of transition) := RE
. , . ..
[-i-.n, -~)n2
(vi) Velocity of electrons in a particular shell or orbit can be
calculated as: . . }{f::::-13.6 eV

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ATOMIC STRUCTURE 111

RH = Rydberg constant = RE:::: 109677 cm~-J .

he he 1 2
--'-=-+- mv
he . A Ao 2
Defects of Bohr theory: (i) It fails to explain the spectra of Vo and Ao are called threshold frequency and threshold
multi-el~ctron atoms. (ii) It fails to explain fine spectrum of
wavelength respectively.
hydrogen. (iii) It does not provide an .explanation why angular
18. Quantum numbers: The set of four integers required
momentum should always be an integral multiple of hl2n. (iv) It
to define an electron completely in an atom are called quantum
does not explain splitting of spectral lines under the influence of
numbers. The first three have been derived from Schrodinger's
magnetic field (Zeeman effect) and electric,. field (Stark effect). wave equation. . . ..
13. Sommerfeld's extension: Sommetreld (1915) introduced
(i) Principal quantum number: It describes the name,
the. idea of elliptical orbits. Except first orbit which is only
size and energy of the shell to which the electron belongs.
circular, the other orbits are elliptical. The second orbit has One
eIliptical and one circular suborbit. The third orbit has two n = 1,2,3,4, ... represent K, L, M, N, ... shells respectively.
elliptical and one circular suborbit.
14. Dual nature: Light has dual character, i. e. ,it behaves Formulae for radius, energy and angular momentum of
electrons are given earlier.
s~metimes like particles and sometimes like waves. de Broglie
(1924) predicted that small particles such as electrons should (ii) Azimuthal quantum number: It is denoted by'/'o It
describes the shape of electron cloud and number of subshells in
show wave-like properties along with particle charact~r. The
a shell.
wavelength (A) associated -with a particle of-mass m and moving ----------

with velocity v is given by the relationship A = ...!!:.- ; where, h is 1= 0, 1,2,3, ... ,(n -1)
mv 1= 0 (s-subshell); 1= 1(p-subshell); I = 2 (d-subshell);
Planck's constant. 1=3 (f-subshell). .
The wave nature was cQnfl11lled by Davisson and Germer's Orbital angular momentum of electron
experiment.
Davisson and Germer gave some modified equations for = ~/(l + l)..!!...- = ~/(l + 1)/i
calculation of de Broglie wavelength: 2n

A= h ; where, E = kinetic energy of the particle. when I = 0, electrons revolve in a circular orbit and when I 0, *
·hEm the electrons revolve round the nucleus in an elliptical path.
(iii) Magnetic quantum number: It is denoted by 'm'. It
A= h ; where, q = charge ofthe particle accelerated by describes the orientations of the subshells. It can have values
~2qVm from -I to +1 including zero, i. e., total (21 + 1) values. Each valQe
corresponds to an orbital. s-subshell has one orbital, p-subshell
the potential of V volt.
has three orbitals (p x , P y and p z), d-subshell has five orbitals
15. Heisenberg uncertainty principle: It is imp~ssible to
. (d xy " d}'2' d zx , d> _ y2 and d z2) and f-subshell has seven
measure simultaneously,both the position and moment"umof any
microscopic particle with accuracy. Mathematically, orbitals. One orbital can .accommodate either one or two
h electrons but not more than two. s-orbital is spherically
llx I:l.p::: 4n; where, llx = uncertainty 10 position and
symmetrical and non-directionaL p-orbitals have dumb-bell
I:l.p = uncertainty in momentum. It introduces the concept of shape and are directional in nature. Four d-orbitals have double
probability of locating the electron in space around the nucleus. dumb-bell shape but· d z2 has a baby soother shape. The total
16. de Broglie concept as well as uncertainty principle have numl?er of orbitals present in a main energy level is 'n 2 '.
no significance in everyday life because they are significant for (iv) Spin quantum number (s): It describes the spin orthe
only microscopic systems. electron. It has values + II2 and '-'1/2. (+) signifies clockwise
17. When radiations of a certain minimum frequency (vo ), spinning and (-) signifies anticlockwise spinning.
called threshold frequency, strike the surface of a metal, electrons
called photoelectrons are ejected from the surface. The ininimum
Spin angular.momentum = ~s(s +. l)..!!...-
2n
energy required to eject the electrons from the metal surface is
called threshold energy or work function.
( where, s = ~J
Absorbed energy = Threshold energy + Kinetic energy of
photoelectrons Total spin of an atom or an ion = n x.!.; where, 'n' is the
E=Eo +KE 2
number of unpaired electrons.
1 2
hv=hvo +-mv Spin !Jlultiplicity of an atom = (2rs + 1)
2

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112 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Singlet state (Nonnal) Singlet excited 22. Autbau principle: Aufbau is a Gennan word meaning
i.! ± building up. The electrons are filled in various orbitals in an order
of their increasing energies. An orbital of lowest energy is filled
Spin multiplicity 1 first. The sequence of orbitals in the order of their increasing
;: : n:s+ 1 Spin multiplicity 1 energy is:
]S, 2s, 2p,3s, 3p,4s, 3d, 4p,5s,4d, 5p,6Y, 4.f, 5d, 6p,7s, 5f,6d, ...
=2xO+l=1 The energy of the orbitals is governed by (n +/) rule.
Triplet excited state (i) Subshell with lower of (n + I) has lower energy, hence
1 filled fIrSt, e.g., .
i 3p(n + 1 = 4) will be filled before 3d(n + I 5).

Spin multiplicity (ii) When (n + I) values are same, then the subshell with
lower value of 'n' is filled fIrst, e. g. ,
2XG+~)+1=3 3p(n+I=4) will be filled before 4s(n+1 4.)
because 3p has lower value of n..
19. (i) Number of subshells in a shell =n 23. Hund's rule: No electron pairing takes place in the
orbitals in a subenergy shell until each orbital is occupied by one
(ii) Number of maximum orbitals in ashell =n 2
.. ~le.ctro.n~ ~it1!~p..!U:~!Ie.Lspi!1' Ex:a(;.t1YJ!1l1f:fill~ar~~~llYiJi.lle.~
(iii) Number ofmaxiInum orbitals in a subshell =21 + 1 orbitals make the atoms more stable, i. e., p ,p ,d , d ,.f
(iv) Maximum number of electrons in a shell ;:::: 2n 2
(v) MaximUm number of electrons in a subshell
and /4 configurations are most stable.
All those atoms which consist of at least one orbital singly
=
2(21+ 1)
occupied behave as paramagnetic while all those atoms in
(vi) Z-component of the angular momentum depends
which all the orbitals are doubly occupied are diamagnetic in
upon magnetic quantum number and is given as:
nature.
, h) . Magneticmoment;::::~n(n+2)BM .
Lz =ml21t n = number of unpaired electrons
(vii) Number ofradiaVspherical nodes in any orbital 24. Half-filled and fully-filled suhshells have extra stability
due to greater exchange' energy and spherical symmetry
=(n~/:"'l)
around the nucleus.
]s orbital has no node; 2s orbital has one spherical
25. It is only dz2 orbitals which do not have four lobes like
node; 2porbitalhas no spherical node; 3porbital has
other d-orbitals.
one spherical node.
26. The d-orbital whose lobes lie along the axes is d 2 2 •
(viii) Schrodinger wave equation does not give spin x -)'
quantum number. . 27. Wave mechanical model of atom: It was
(ix) A plane passing through the nucleus at which the SchrOdinger who developed a new model known as wave
probability of finding the electron is zero, is called mechanical model of atom by incorporating the conclusions of de
nodal plane. Broglie and Heisenberg uncertainty principles. He derived an
The number of nodal plane in an orbital 1 = ... equation, known as Schrodinger equation.
.\'-Orbitals have no nodal plane; p-orbitals have one d 2", d 2", d 2", 8x2m
nodal pJane, d-orbitals have two nodal planes and so - -2+ - - + - - + (E-V)", 0
on. dx dy2 dz2 .

20. Pauli's exclusion principle: No two electrons in an The solution of the equation provides data which enables us to
atom can have the same set of all the four quantum numbers, i. e., calculate the probability of finding an electron of specific energy.
an orbital canuot have more than 2 electrons because three It is pOssible to determine the regions of space around the nucleus
quantum numbers (principal, azimuthal and magnetic) at the where there is maximum probability of locating an electron of
most may be same but the fourth must be different, i. e., spins specific energy. This region of space is tenned orbital.
must be in opposite directions. It is possible to calculate the
'" is the amplitude of the wave at a point with coordinates
maximum number of electrons which can be acconunodated on a
x, yand z. 'E' is total energy called eigen value and V denotes the
main energy sheD or slibenergy shell on the basis of this potential energy of the electron;
principle.
",2 gives the probability offmding the electron at (x, yand z).
21. Electronic configuration: The arrangement of
electrons in various shells, subshells and orbitals in an atgm is Operator form of the equation can be given as:
tenned electronic configuration. It is wrltten in tenns of nl x HW = E",
where n indicates the order of shell, 1indicates the subshell and x
the number of electrons present in the subshell.

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ATOMIC STRUCTURE 113.

if ;: [- 4
8n m
1
A? :.- V called Hamiltonian operator·
.
30. Plot of radial probability density 'R 1 ,:

=T+V
T= Kinetic energy operator t \. 1ISS. ." . . .

V = Potential energy operator R2~

28. Complete wave function can be given as.: r --II>
",(r,O,<I»= R(r) ; e(o) 11>(<1»
'--v--' '---v---"
Radial part Angular part

Dependence of the wave function on quantum number can be

given as:
'" nlm (r,O, <1»= R~, I (r) e1m(0) ~m (<I» ,1s 2s 2p
29. Graph of radial wave fUnction 'R': At node" the
value of' R' changes from positive to negative.

2s 2p

In the plot of radial probability against 'r', number peaks; i. e.,

. region of maximum probability n - l.
r--ll>·

Number of radial nodes = (1'1 -I-I),

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~"~"."... --_._,.""'_......--------

114 I G.R. B. PHYSIC~L GHI=MISTRy' FOR COMPETITIONS

1. Match the following: (c) Paschen series (r) Absorptionspectrum

[A] ( d) Brackett series (s) Ultraviolet region
(i) Aufbau principle (a)' Line spectrum in visible region
(ii) de Broglie
" ' '

(b) Orientation of an eI(:ctron in an

[D] Match the List-lwith List~II:'
, 'orbital List~: List-ll
(iii) Angular momentum (c) Photon'
(a) K-shelL (P) Electrons in elliptical
, (iv) Hund's rule (d) A ,hl(,,!v) " orbit
: (v) Barmers~ries' ,:n . (i)'tiedrofu6 conflgm.~ii~n
(b) L-shell (q) Electrons in circular orbit
(vi) Planck's law (f) mvr
(c) Hydrogen atom, (r) Shell of lowes,! energy
[B] ",
(i) , Thomson (d) Boron atom in groun,d (s) B04(s atomic model
(a)' Exclusion principle
state
(ii) Pauli (b) Radioactivity
(iii) Becquerel' , ' , (c) "Atomiemodel [E] Match the ions of List-I with the properties ofList~I1:
(iv) ,S~ddy (d) Cathode rays Liit-I List-ll
(v) 'Bohr (e) Neutron
(a) Mn 2 + (p) Diamagnetic
(vi) Chadwick (f) Isotopes
[C] (b) V 2+ (q) Paramagnetic
(i) Cathode rays (a) Helium nuclei (c) Zn 2+ (r) Coloured compounds , .
(ii) Dumb-bell ,(b) Uncertainty principle (d) Ti4+ (s) Magnetic moment = 2.82 BM
(iii) Alpha particles (c) Electromagnetic radiation
[F] Match the List-I with List-II:
(iv) Moseley (d) p -orbital
(v) Heisenberg (e) Atomic number List.;.I List-ll
(vi) X~rays (f) Electrons (a) Mg2+ (P) Zero spin multiplicity
2. Matrix Matching Problems (For lIT Aspirants):
(b) Fe 2+ (q) Spin multiplicity = 3
[A] Match the Colurnn-I and Column-II:
(c) C0 3 + (r) Total spin = 0 ,
Column-I ColulIID-D
(d) Ca 2+ (s) Total spin :::: 2
(a) X-rays (P) Davisson and Germer
, experiment , [G] Match the prope.rties of List-I with the formulae in List-II:
(b)' Atomic number (q) Crystal structure List-I ' . List'-ll
determination , determination
(c) Dual nature of matter (r) Moseley's law
(a) Angular momentum of ' (P) JI(l+I)~
electron 2n
(d) Dual nature of radiation (s) Bragg's law (b) Orbital, angular (q) lro
[B] Match the Columil-I and Column-II: momentum
(c) Wavelength of matter (r) nhl2n
Column'" Co)uDID-ll wave
(a) Lyman series (P) Visible region (d) Quantised. value( s) (s) hlp
(b) Balmer series (q) 'UVregion [H] Match the orbitals of List-I with the nodal properties of
"'0·
(c) Pfund ,series (r) IR region List-II:. '
(d) Light emitted by, sodium (s) Line emission spectrum List-I·, List-n
lamp
(a) 2s (P) , Angular node = 1
[C] Match the List-I with List-II in hydrogen atom spectrum:
(b) Is //I" (q) Radial node 0
List-I List-II (c) 2p (r) Radial node = 1
t.:
(a) Lyman series (P) Visible region , (d) 3p (s) Angular node = 0
, (6) Balmer series (q) Infrared region

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ATOMIC STRUCTURE
"r 115
[1] Match the dectronic transitions of List-I with spectral [M] Match the List-I with List-II:
properties o~ List-1I: ~:;:T)Hl?f' ~
, List-I List-ll
List-I List-ll
(a) Radius of nth orbital (P) Inversely pj-opOltional to Z
(a) n=6~n 3 (P) 10 lines in the spectrum
(b) Energy of nth shell (q) Integral mul!ip!e' of h/21t
(b) n=7~n=3 (q) Spectral lines in visible
region
(c) Angular momentum of (r) Proportional to n 2
electron
(c) n= 5.~,n=2 (r) 6 lines in the spectrum
(d) Velo<.;ity of electron in nth (s) Inversely proportional to
(d) n=6~n 2 (s) Spectral lines in infrared orbit 'n'
region
[N] Match the entries in Column-! with the correctly related
[1] Match the List-1 with List-II: ~·quantumni.1mbet(s) inColulnn-II:
List:.t List·ll Column.l
,"

(a) Radius of eleGtron orbit (P) Principal quantum (a) Orbital angular momentum (P) Principal quantum
number of the electron in a hydrogen- number
(b) Energy of electron (q)-Azimutbalquantum. like at.Qmi,G orl:>ita,l
number (b) A hydrogen-like one electron (q) Azimuthal quarttum
(c) Energy of subshell (r) Magnetic quantum wave function obeying Pauli number
number principle

(d) , Orientation of the atomic (s) Spin quantum number (c) Shape, size and ori,entation (r) Magnetic quantum
of hydrogen-like atomic , number
orbitals
orbitals
[K] M.atch the List-I with List-II: (d) Probability density of (s) Electron spin quantum
List-I List-ll electron at the nucleus in number
hydrogen-like atom
(a) ElectrolJ. cannot exist in (P) de Broglie wave
[0] Match the List-lWith List-II: . ,(lIT 2006)
the nucleus :~!:-.

(b) Microscopic particles in (q) Electromagnetic wave List-I List-ll

, motion are associated with' (a) Wave nature of radiation (1') Photoelectric'
(c) No medium is required for (r) Uncertainty principle effect
propagation (b) Photon nature of radiation , (q) Compton effect
(d) Concept of orbit was (s) Transverse wave (c) , I,nteraction of a photon with an (r) Diffraction
replaced by orbital electron, such that quantum energy
is slightly equal to or greater than
[L] According to Bohr theory: (lIT 2006)
the binding ,energy of electron, is,
En == Total energy more likely to result in:
K /I = Kinetic energy
VII = Potential energy l.:.d) Interaction of a photon with an (s) Interference
,
electron, such that photon energy is
r" Radius of nth orbit
much greater than the binding
Match the following:
energy of electron, is more likely to
Column-I 'Column-ll result in:

(a) VnlKn ?' (p) 0 [P] Match the Column-! with Column-II:
(b) lfradius of nth orbit oc Enx; x = ? (q) Column-I Column-ll
(c) Angular momentum in lowest orbital (r) - 2 (a) Orbital angular (P) ~s (s + l)h/2n
momentum of an electron
(d) 1 ocZY ; y=? (s) 1
r" (b) Angular momentwn of (q) ~n (n + 2) BM
electron
(c) Spin angular mQ,mentum (r) nh/2n
of electron
(d) Magnetic 'moment of atom (s) h/2n

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116 G.R.B. PHYSICAL CHEMI,STRY FOR COMPETITIONS·

[Q] Match the Coh,unn-l with Column-II: [R] Match the Column-I with Column-II:
Column-l (;vlumD~lI .- Column-I Column-II
(a) Scintillation (P) , Wave nature (a) Radial function R (P) Principal quantum
(b) Photoelectric effect (q) P~\ltiCle nature' number 'n'
(c) Diffraction (r) Particle nature dominates (b) Angular function e (q) Azimuthal quantum
over wave nature number 'I'
(d) Prin~iple of electron " (s) Wave nature dominates (c) Angular function CP (r) Magnetic quanfum
microscope over particle nature number'm'
(d) Quantized angular (s) Spin quantum number's'
momentum

1. [A] (i-e); (ii-d); (iii-f); (iv-b); (v-a); (vi-'c) [I] (a-,r, s) (b-p, s) (c~, r) (d-p, q)
[B] (i-d); (ii-a); (iii-b); (iv-'---f); (v-c); (vi-e) [J] (a-p) (b-p) (c--p, r) (d-r)
[C] (i-f); (ii-d); (iii-:-:a); (iv-e); (v-,b);(vi-c) [Kl{a-rHb--p)(c~;s) (d-r)-
2. [A] (a-q, r, s) (I~r) (c-p) (d-dpes not match) [L] (a-r) (~)(c-p) (d-s)
[B] (a-q, s) (b-p, s) (c-r, s) (d-fP, s) [M] (a-r, p) (b-r) (c-q) (d-s)
[C] (a-r, s) (b-p) (c-q) (d-q)~ [N] (a-p) (b-s) (c-p, q, r) (d-p, q)
[D] (a~, r) (b-p, q) (c-s) (d-p, q) [0] (a-r, s) (b-p, q) (c-p) (d-q)
[E] (a-q, r) (~, r, s) (c~p) (d-p) [P] (a-s) (b-r) (c-p) (d-q)
[F] (a-p, r) (~, s) (c~, s) (d-p, r) [Q] (a-q) (b-r) (c-p) (d-p, s)
[G] (a~, r)~) (c-s) (d~, r)
[R] (a-p, q) (~, r) (c-r) (d~, s)
[H] (a-r, s) (~, s) (c~, p) (d-p, r)

RACTICEPROBLE • '
1. An atom of an element contains 13 electrons. Its nucleus has 14 4. Calculate the number of neutrons in 18 mL of water. (Density
neutrons. Find out its atomic number and approximate atomic of water = 1)
mass. An isotope has atomic mass 2 units higher. What will be [Ans. 48.16 x 1023 ]
the number of protons, neutron~ and electrons in the isotope?
[Hint: One molecule of water contains = 8 neutrons J
[Ans. At. No. =13, atomic mass =27; the isotope will have
same number of protons and electrons 13 but neutrons will be 5. Find (i) the total number of neutrons and (ii) the total mass of
14 + 2 = 16] neutrons in 7 mg of 14C (assuming that mass of neutron
2. From the following find out groups of isotopes, isobars and mass of hydrogen atom). '
isotones: [Ans.(i) 24.08 x 1020 and (ii) 4 mg]
16 0 391(, 14C 239U 14N 40C 238U 77G'
8 '19 6 ' 92 , 7 , 20 a, 92 '-32 e, 6. Calculate the wavelength of a photon in Angstroms having an
17As 18 0 76G 78S energy of 1 electron volt.
33 ' 8 '32 e, 34 e
[Ans. Isotopes--same at. no. but different at. masses. [Hint: 1 eV = i.602 x 10- 19 joule;
16 0 18 0' 239 238 ' 77Ge 76 Ge h 6.62 x 10-34 J-s, c='3 X 108 ms- I
8 ' 8 '92 U ',92 U , 32 '32
h· c . 7 3
Isobars-same atomic masses but different at. numbers A= - :::: 12.42 x 10- m = 12.42 x 10 A]
E
14C 14N_ 77Ge 771>c
6 ',7 '32 '33'~ 7. ' A photon of light with wavelength 6000 A has an energy E.
Calculate the wavelength of photon of a light which
Isotones-same number of neutrons,
corresponds to an energy equal to 2E.
I~O, l~C; :gK, jgca;T:As, nSe] [Ans. 3000 A]
3. An element has atomic' number 30. Its cation has 2 units 8. Calculate the energy in kilocalorie per mol of the photons of
positive charge. How many protons and electrons are present an electromagnetic radiation of wavel~ngth 5700 A.
in the cation? [Ans. 56.3 kca1 per mol]
[Ans. Protons = 30, Electrons =28]

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ATOMIC STRUCTURE I 117

9. Light of what frequency ·and wavelength is needed to ionise 6.626 X 10-34

sodium atom. The ionisation potential of sodium is
[Hint: A=----=j=======~========~
.J2Em
8.2 x 10-19 I.
[Ans. v = 1.238 X 1015 Hz; A = 242 nm]
=2.2 x 10- 12 m]
19. Determine the number of revolutions made by an electron in
10. Determine the energy of 1 mole photons of radiations whose
one second in the 2nd Bohr orbit of H-atom.
frequency is 5 x 1010 S-I. (h == 6.62 x 10-34 1- s) 2nr
[Ans. n= ]
. [Ans. 19.91] v
2
11. Find e / m for He + ion and compare with that for electron, 20. What is the speed of an electron whose de Broglie wavelength
[Ans. 4.87 x 107 coulomb kg-I] is 0.1 nm? By what potential difference, must have such an
electron accelerated from an initial speed zero?
12. A ball of mass 100 g is moving with a velocity ofl 00 m sec -1.
[Ans. 7.28 x 106 m/sec; 150 V]
Find its wavelength.
21. A green ball weighs 75 g; it is travelling towards observer at a
~ = 6.626 x 10-
34
[Hint: A == 6.626 x 10-35 m] speed of 400 crn/sec. The ball emits light of wavelength
mv 0.1 x 100
5 x 10-5 cm. Assuming that the error in the position of ball is
13. Calculate the wavelength of radiation and energy per mol
the same as wavelength pfitself, calculate error in the
necessary to ionize a hydrogen atom in the ground state.
momentum of the green ball.
[Ans. A == 9.12 X 10-8 m; 1313 kJ 1 mol] h
<[Hint:I:!x·t1p?-<-<
14. Bond energy of F2 is 150 kJ mol-I. Calculate the minimum 4n
frequency of photon to break this bond. h
t1p?:.--
[Ans. 3.759 x 10 14 8- 1] 41t I:!x
15. If an Einstein (E ) is the total energy absorbed by J mole of a t1 > h
.p - 4nA
substance and each molecule absorbs one quantum ofenergy,
then calculate the value of'E' in terms of A in cm. 6.626 x 10-27 23
t1p "" '" 1. 055x 10- ]
8 4 x 3.14 x 5 x
1.198 x 10 . 1-1 ]
[A ns. A < erg mo
22. What is the relationship between the e V and the wavelength in
16. How many chlorine atoms can you ionize hi. the process? metre of the energetically equivalent photons?
Cl ~ Cl+ + e by the energy liberated from the following [Ans. A == 12.4237 X 10-7 metre]
process: 23. What is the velocity ofanelectron(m= 9.1 Ix 10-31 kg) in the
CI + e ~ cr for 6 x 1023 atoms innermost orbit of the hydrogen atom?
given that electron affinity of chlorine is 3.61 eV and (Bohr radius 0.529 x 10- 10 m)
ionization energy of chlorine is 17.422 e V.
. [Ans. 2.187 x 106 m/sec]
[Ans. L24 x 1023 atoms]
24. In a hydrogen atom, an electron jumps from the thud orbit to
17. Find the velocity (ms-'I) of electron in first Bohr orbit of the first orbit. Find out the frequency and wavelength of the·
radius ao. Also find the de Broglie wavetength (in 'm'). Find spectral line. (RH L09678 x 107 m:- I )
the orbital angular momentum of 2 p orbital of hydrogen atom
[Ans. 2.925 x 1015 Hz, 1025.6A]
in units of hI 2Ti.
2.188 x 10 6 . _I
. 25. The energy of the electron in the second and third Bohr orbits
[Hint: v m sec . of hydrogen atom is -5.42 x 10- 12 erg and -2.41 x 10- 12 erg
. n
respectively~ Calculate the wavelength of the emitted radiation
2188 x 10 6
v= 2188 X 10 6 m sec-I when the electron drops from third to second orbit.
1
[Ans. 6.6 x 103 A]
h 6.626 X 10-34
A=-
mv 9.1 X 10-31 x 2.188 X 106
26. Calcuhi.te the wavelength m
angstroms of the photon that is
emitted when an electron in Bohr orbit n ;, 2 rCtums to the orbit
;:: 3.3 x lO- lo m 11 1 in the hydrogen atom. The ionisation potential of the
ground state oi"hydrogen atom is 2.17 x 10- 11 erg per atom.
Orbital angular momentum = ~ l(l + 1) h <

. . 2n [Hint: Energy of the electron in the I st orbit =- (ionisation

~1(1 + 1) h (": I = 1 for 2p) 11
potential), t1E = (31 4) x 217 x 10- erg per atom]
21t
[ADS. A 1220 A]
=..fi h ]
21t 27. Calculate the wave number for the shortest wavelength
18. The energy of an a-particle is 6.8 x 10-18 J. What will be the transition in BalrDer series of atomic hydrogen. (liT 1996)
wavelength associated with it? ICBSE-PMT (Mains) 2005) [ADS. 27419.25 cm- I ]

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1181 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

28. The wavelength of the first member of the Balmer series of 33. The velocity of an electron in certain Bohr orbit of H-atom
hydrogen is 6563 x IO-IO m. Calculate the wavelength of its bears the ratio I : 275 to the velocity of light. (a) What is the'
second member. quantum number 'n' of the orbit? (b) Calculate the wave
number of the radiation when the electron jumps from (n + I)
[Hint: :1 [;2 -),
== RH and :2 RH [;2 12 J
4 state to ground state.
[Ans. v 9.75 x 104 em-I]
5 16 20
==-x v I 3 X 1010 8
1..1 36 3 27 [Hint: (a) - = orv == == 1.09 X 10 em
c 275 275
nh nh
1..2 == 20 x 6563 X 10- 10,:" 4861 X 10.- 10 m] . v == - - == ------_=_~
27 21tmr 21tm X 0.529 x I
29. According to Bohr theory, the electronic energy of hydrogen h
or 11= -------:~-
atom in the nth Bohr orbit is given by, 21tm X 0.529 x XV
2176 x 10- 19 6.625 X 10-27
Ell == J
2 X 3.14 x 9.1 x
Calculate theiongest wavelength of light that will be needed
=2
to remove an electron from the 2nd orbit of Li 2 + ion. .
(b) Thus, II + 1 = 2 + 1 3. The electron jumps from 3rd orbit to
[Ans. 4.059 x 10-8 m] 1st orbit.]
30. Calculate the frequency, energy.. and .wavelengili<of the
radiation corresponding to spectral. line of lowest frequency in
34. Find out the wavelength of the next line in the series having
Lyman series in the spectra of hydrogen atom. Also c'alculate lines of spectrum of H-atom of wavelengths 6565 A, 4863 A,
4342 A and 4103 A.
. the energy of the corresponding line in the spectra of Li 2+.
(lIT 1991) [Ans. 3972 A]
7
[Ans. I.. == 121'; X 10- m;v = 2.47 x lOIS cycle , [Hint: All these lines are in visible region and thus, belong to
Balmer series, Next line is, therefore, from 7th orbit.]
E 16.36 X 10- 19 j, ELi 2+ ~ EH 9 x i6.36 X 10- 19 J
147.27 X 10-19 J]
35. Which jump is responsible for the wave number of emitted
radiations equal to Q.7490 x 106 m- I in Lyman series of
31. Calculate the ratio of the veIocity oflight and the velocity of
electron in the 2nd orbit. of a hydrogen atom. (Given hydrogen spectrum? (R 1.09678 x 107 m- 1 )
h = 6.624 X 10-27 erg-sec;m . 9.108 x 10- 28 g; [ADs. 3]
r 2.11 X 10-8 cm) 36. Calculate the ionisation energy of the hydrogen atom. How
[Ans. 273.2] . much energy will be required to ionise 1 mole of hydrogen
32. What hydrogen-like ion has the wavelength difference atoms? Given, that the Rydberg constant is 10974 X 107 m- I .
betWeen the first lines of Baf~er and Lyman series equal to [Ans. IE per hydrogen atom = 2.182 x 10-1.8 J
59.3 nm (RH = 109678 em-I)? IE per mole = 1314 kJ mol-I]
[Hint: Wavelength df 1st line in Balmer series, 37. Calculate the ionisation energy of (a) one Li 2 + ion and (b) one
r =Z 2RH
I..B
[l.. _l..J
22 32
2. R Z2
36 H
mqJe ofLi2+ ion. (Given, R == 10974 X 10-7 m- I )
[Ans. (a) 19.638 x 10- 18 J (b) 1.118 X 104 kJ mol-I]
t.. _ 36 38. A series of lines in the spectrum of atomic hydrogen lies at
or
B - 5R Z 656.46 nm, 4i6.27 nm, 439.17 nm and 410.29 nm. What is the
. H
wavelength of the next line in this series? What is the
Wavelength of 1st line in Lyman series is, ionisation energy of the atom when it is in the lower state of
J....
I..L Z2R
B [~
12 l..l
22.
transition?
[Ans. I..nexl 397.15 nm; IE = 3.40 e':l
4 39. A certain line of the Lyman series of hydrogen and a certain
or AL == ----;:-2
. 3xRH Z line of the Balmer series of He + ion have nearly the same
wavelength. To what transition do they belong? Small
Differencel..B - I..L = 59.3 X 10-7 == ~ - ~ differences between their Rydberg constant may be neglected.
5RHZ 3RH Z
[Ans. Hydrogen Helium
1
RHZ2
[36
5" '3
4J 2~1
3 __ 1
4 __ 2
6 __ 2
88 4 __ 1 8 __ 2]
z 2 == 7 . 9.0
59.3 x 10- x 109678 x 15 40.. What element has a hydrogen-like spectrum whose lines have
.. , or Z='3 wavelengths four times shorter than those of atomic hydrogen?
[ADS. He+]
Hydrogenclike species is Li 2+.]

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ATOMIC STRUCTURE 119

41. What lines of atomic hydrogen absorption spectrum fall 53. CalCulate the momentum of the particle which has de Broglie
within the wavelength ranges from 945 to 130 nm? . wavelength I A (10- 10 m)and h = 6;6 x 10-34 J- se<;.
[Ans. 97.3; 102.6; 121.6 nm] [Ans. 6.6 x 10-24 kgm sec-I]
42. The binding energy of an electron in the ground state of an
54. The uncertainty of a particle in momentum is 3.3 x 10-2
atom is eq:ualto 24.6 eV Find the energy required to remove
both the electrons from the atom. . ,kg ms-I. Calculate the uncertainty in its positiQ'Q.
[Ans. 79 eV] (h = 6.6 x 10-34 J- sec)
43. What is the ratio of the speeds of an electron in the first and [Ans. 3.1 x 10- 14 m]
second orbits of a hydrogen atom?
55. Calculate the product of uncertainties of displacement and
[Ans. 2; 1] .
velocity ofa moving electron having a mass 9,1 x 10-28 g.
44. Find out the number of waves made by a Bohr electron in one
complete revolution in its third orbit. (lIT 1994) [Ans. 5.77 x 10-5 m2 S-I]
[Ans. 3] [Hint: ~ ·Ilv ~ ~]
45. The wave number of first line in Balmer series of hydrogen is 4~ .
15200 cm -I. What is the wave number of first line in Balmer 56. (a) A transition metal cation x 3 + has magnetic morrient .,[35
series ofBe 3+? BM. What is the atomic number of x 3+?
[Ans. 2.43 x 105 cm- I ] (b) Select the coloured ion and the ion having maximum
46. Calculate the speed of an electron in thegroJln(istate of
hydrogen atom. What ·fraction of the speed of light is this
1111 1111 11 1 D
value? How long does it take for the electron to complete one magnetic moment (i) Fe +, (ii) Cu +, (iU) Sc 3+ and
2

revolution around the nucleus? How many times does the

electron travel around the nucleus in one second? 2
(b)Fe + ~ 11111 11 11 11 I·
[Ans. 2.186 x Hf ms-J ; 7.29 X 10-3 ]
47. An electron, in a hydrogen atom, in its ground state absorbs
1.5 times as much energy as the minimum required for its
escape (i.e~ 13.6 eV) from the atom. Calculate the value of A
for the emitted electron. (iy) Mn 2 +

[Ans. 4.69 A]
[Hint: (a) 26, 26Fe~ 3d 6 4s 2

48. The radius of the fourth orbit of hydrogen is 0.85 nm. Fe3i; ~ 3d 5 4sO
Calculate the velocity of an electron in this orbit
(me 9.1 x 10-31 kg).
Il =In(n + 2) =..J5X7 = 55
Both these ions will be coloured and magnetic moment of Fe2 +
[Ans. 5.44 x 105 m sec-I]
will be greater.]
49. A beam of electrons accelerated with 4.64 V was passed 57. A photon of wavelength 4000 A strikes a metal surface, the
through a tube having mercury vapours. As a result of work function of the metal being 2.13 eV Calculate (i) energy
absorption, electronic changes occurred with mercury atoms 9f the photon in eV; (ii) kinetic energy of. the emitted
and light was emitted. If the full energy of single electron was photoelectron and (iii) velocity of the photoelectron.
converted into light, what was the wave number of emitted
[Ans.E = 3.10 eV; KE= 0.97 eV; Velocity = 5.85 x 105 Ins-I]
light?
[Hint: I eV = 1602 X 10-19 J]
[Ans. : 3.75 x 104 em- I I
so. An electron jumps from an outer orbit to an inner orbit with
58. Calculate the ratio between the wavelengths of an electron and
the energy difference of 3.0 eV. What will be the wavelength a proton, if the proton is moving at half the velocity of the
of the line and in what region does the emission take. place? electron (mass of the proton = 167 x 10-27 kg; mass of the
[Ans. A 4140A; visible region] electron = 9.11 x 10-28 g).
[Hint: leV 1.6 X 10- 12 erg] [Ans. 9.2455 x 10-2 m]
h
51 •. The first ionisation energy of a certain atom took place with an [Hint: Applyde. Broglie equation, A
absorption of radiation of frequency 1.5 x 1018 cycle per
second. Calculate its ionisation energy in calorie per gram atom. Wavelength of electron
[Ans. 1.43 x 108 cal l
[Hint: 1 calorie 4.18 x 107 erg Wavelength of proton
Apply E = h x v x Avogadro's number]
52. Find the wavelength associated' with an electron which has 59. A moving electron has 2.8 x 10-25 J of kinetic energy.
mass 9.1 x 10-28 g and 'is moving with a velocity of lOS cm Calculate its wavelength.
sec-I. (Given h 6.625 x 10- 27
erg-sec) (Mass of electron = 9.1 x 10-3,1 kg)

[ Ans. A 7.28 x 10- ern] 5

[Ans. 9.2455 'X IO-7 m ]

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J20 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

66. An electron has mass 9.1 x 10-28 g and is moving with a

[Hint: v = ~2XKE
_.- - ':= 784.46 ms -;IA.= -h]
m . - mv velocity of 105 cm/sec. Calculate its kinetic energy and
60. Helium has ·mass number 4 and atomic number 2. Calculate wavelength when h = 6.626 X 10-27 erg-sec.
the nuclear binding energy per nucleon (mass of neutron [Ans. 4.55 x 10-8 erg; A =7.28 X 10-5 cm]
= 1.00893 amu and proton = 1.00814 amu, He = 4.0039 amu
67. Calculate the de Broglie wavelengths of an electron and a
and mass of electron is negligible).
proton having same kinetic energy of 100 eV
[Ans. 7.038 MeV]
[Ans. Ae = 123 pm; Ap = 2.86 pm]
61. Calculate the mass defect and binding energy per nucleon of
1~9v'hichhas a mass 15.99491 amu. 68. Work function of sodium is 2.5 eV Predict whether the
wavelength 6500 A is suitable for a photoelectron or not?
Mass of neutron = 1.008655 amu
[Ails. No ejection]
Mass of proton = 1.007277 amu
Mass of electron = 0.0005486 amu
69. Calculate the de Broglie wavelength associated with a helium
atom iri a helium gas sample at 27°C and 1 atm pressure.
1 amu = 931.5 MeV
[Ans. 7.3 x 10- 11 metre]
[Ans. 7.976 MeV/nucleon]
14
62. The circumference of the second Bohr orbit of electron in the 70. The threshold frequency for a certain metal is 3.3 x 10
hydrogen atom is 600 nm. Calculate the potential difference to cycle/sec. If incident light on the metal has a cut-off frequency
which the electron has to be subjected so that the electron 8.2 x 1014 cycle/sec, calculate the cut-off potential for the
stops. The electron had the de Broglie wavelength
photoeleetron;
corresponding to the circumferen~e. -
[Ans; 2 volt]
' N b· f . " Circumference
[Hint. urn er 0 waves n = - - - - - 71. .Can you locate the electron within 0.005 nm?
Wavelength
[Ans. No.]
nA= 2nr [Hint: Use uncertainty principle to determine uncertainty in
2A = 600 velocity.
A= 300nm h
Let stopping potential is Vo. ~v2:---
4nm&
eVo =-1 mv 2 . ... (i) On substitution, you will get,
1
_~v 2: 1.16x \07 ms- I
A=·~
mv Velocity of electron is therefore expeCted to be as high as velocity
h of light. We may say that the velocity of electron is uncertain
V=- ... (ii)
Am within 0.005 nm.]
From equations (i) and (ii), 72. The photoelectric cut~offvoltage in a certain experiment is 1.5
volt. - What is the maximum kinetic energy of the
.
eVo =.!.2 m (~)2
Am
photoelectrons emitted?
[Ans. 2.4 x 10-19 joule]
2 o
h 73. A proton is accelerated to one-tenth the velocity of light. If its
V. -
0- 2mA2e velocity can be measured with a precision of ± 1%, what must
(6.626 x 1O-34 i be its uncertainty in position?
2 x (9.1 x 10-31 ) x (300 x 10-9)2 x 1.6 X 10- 19 (h = 6.6 x 10-34 J-s; mass of proton = 1.66 x 10-27 kg)
[Ans. L05 x 10- 14 m]
= 1.675 X 10-5 V]
74. In a photoelectric effect experiment, irradiation ofa metal with
63. The velocity of an electron of mas~ 9.1 x 10-31 kg moving
light of frequency 5.2 x 1014 sec- I yields electrons with
round the nucleus in the Bohr orbit (diameter of the orbit is
maximum kinetic energy 1.3 x 10- 19 '1. Calculate the Vo of the
1.058 A) is 2.2 x 10-6 m sec-I. If momentum can be measured
metal.
within the accuracy of 1%, then calculate uncertainty in
position (&) of the electron. [Ans. 3.2 x 10 14 sec-I]
[Ans. 2.64 x 103 metre] 7S. Calculate the wavelength of a CO2 molecule moving with a
64. An electron wave has wavelengtIi 1 A. Calculate the potential velocity of 440 m sec-I.
with which the electron is accelerated. [Ans. 2.06 x 10- 11 metre]
[..ins. 0.0826 volt] 76. The predominant yellow line in the speCtrum of a sodium
65. Calc1)late the de Broglie wavelength .associated with an vapour lamp has a wavelength of 590 nm. What minimum
a~particle having an energy of 7.7 x 10- 13 J and a mass of accelerating potential is needed to excite this line in an
6.6-x" 10':'24 g. (h';" 6.6 x 10-34 J-s) - electron tube having sodium vapours?
13 [Ans. 2.11 volt]
[Alis. 6.56 x 10- cm]

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•
ATOMIC STRUCTURE. 121

77. Find out the wavelength of a track star running a 100 metre I mv 2 :::K Zex e
dash in 10.1 sec, if its weight is 75 kg. 2 ro
[Ans. 8.92 x 10-37 m] For Rutherford experiment,
78. At what velocity ratio are the wavelengths of an electron and a
proton equal? . .! mv 2 5.5 MeV::: 5.5 x )(1' x 1.6 x 10-19 J::: 8.8 X 10-13 J
2
24
(me == 9.1 x 10-28 g and mp = 1.6725 X 10- g) 9
8.8 X 10- 13 = 9x10 x2x79x
[Ans. Ve = 1.8 X 103] 10
vp 15
10 == 4.136 X 10- m
79. Through what potential difference must an electron pass to
have a wavelength of500 A? 10 = 4.IHm]
[Ans. 6.03 x 10-:4 eV ] 84. Photoelectrons are liberated by ultraviolet light of wavelength
h 3000 A from a metallic surface for which the photoelectric
[Hint: Use A== ~] threshold is 4000 A. Calculate de Broglie wavelength of
,,2eV m
electrons emitted with maximum kinetic energy.
80. Calculate the velocity of an ~-particle which begins to reverse
[Ans. A == 12 X 10-9 m]
its direction at a distance on X 10- 14 m from a scattering gold
nucleus (Z == 79). [Hint:
[Ans. 2.346 x 107 m! sec] KE = Quantum energy - Threshold energy. ..
81. Two hydrog~n atoms collide head.on and end up with zero 6.626 X 10-34 X3 X 10& 6.626 X 10-34 x 3 X 108
kinetic energy. E~ch then emits a ph:oton with a wavelength 3000x 4000 x
121.6 nm. Which transition leads to this wavelength? How
fast were the hydrogei1 atoms travelling before the collision? == 6.626 X 10- 19 4.9695 X 10-19
(Given, RH == 1.097 X 107 m- I and"'H == 1.67 X 10-27 kg) == 16565 X 10-19 joule
IAns. nl I; 112 = 2; 4.43 X 10 4
m sec-I]
[Hint: Wavelength is in UV region; thus nl will be 1.
.! mv 2 1.6565 x 10- 19
2
7
-----;;,::: 1.097 X 10 x 12 X'(112
12L6 x
- :~)
"1.
m4v 2 = 2 x 16565 X 10-19 x 9.1 X 10-31
mv = 5.49 x 10-25
112 2 34
I 2 he A ::: ...!:. = 6.626 X 10- = 1.2 X 10-9 m]
.- mv :::- mv 5.49 x
2 A
34 8 8S. Show that de Broglie wavelength of electrons accelerated V
-I x.
167 X 10-27 XV
2
:::
6.626 X 10- x 3 X 10
------;c-~-
volt is very nearly given by:
2 1216 x
v 4.43 X 104 m sec-I] AOn A) Iv50)112
(
82. Show that the wavelength of electrons moving at a velocity
very small compared to that of light and with a kinetic energy
of V electron volt can be written as,
[Hint: A= b
v'l,eVm
~
.A=~X
12.268 10-8 cm 2 ]1/2
",A= _h_ x 1020 A
.JV [ 2eVm
Use the relation, A =
[Hint: h
.J2Em =[ (6.626 x 10- )2 x 10
34
JI/2
20
[150J1I2]
Here, h = Planck's constant 2 x 16 X 10- 19 x V x 9.1 X 10-31 V
m ::: 9.1 X 10-28 g (mass of e- )
86. A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate
E Kinetic energy of electron
== V 12
, eV== V x 1.6 X 10- erg]
the distance of closest approach for head-on collision.
[Ans. 1.137 x 10- 13 m]
83. What is the distance of closest approach to the nucleus of an
a-particie whiCh undergoes scattering by 1800 in z,} .
Geiger-Marsden experiment? [Hint: d == . 2 ' Do hke Q.No. 83]
. 41tco(M mv )
[Ans. 10 = 4.13 fm]
87. What is the energy, momentum and wavelength of the photon
[Hint: For closest approach, emitted by a hydrogen atom when an electron makes a
transition from n == 2 to n I? Given that ionization potential
is 13.6 eV.

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•
122 .1 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

[Ans. 16.32 x lO-19 J, 5.44 X 10-27 kg m I sec, 1218 A]

[Hint: IE == Z2 x 2l.69 X lO-19 J
[Hint: E I ';' -13.6 eV
3
-13.6 V
11180 X 10 = Z2 x 21.69 X lO-19
E2 --e 6.023 X 12
4
3 Z'" 3] .
till= x13.6eV 90. Suppose 10- 17 J of light energy is needed by the interior of the
4
human eye to see an object. How many photons of green light
= 0.75 x 13.6 x 1.6 x lO-19 J = l.632 X lO-18 J
(A 550 nm) are needed to generate ihis minimum amount of
he l.632 x 10- 18
energy?
A [Ans. 28]
34 8 91. How many hydrogen atoms in the ground state are excited by
A=6.626XIO- X3XI0 =1218xlO- lo m=1218A means of monochromatic radiation of wavelength 970.6 A.
l.632 x How many different lines are possible in the resulting
Ao=!!. emission spectrum? Find the longest wavelength among these.
p [Ans. Six different lines, A 1215.6 A] 0=
34
-h -_ 6.626 X 10- . x 10- k.g-m I sec ]
_ 544 27
P - E _ he
A 1218 x . [Hint: En .1 - A
88. Calculate the orbital angular momentum of the following -2169 X 10- 19 21.69 x 10'::1-9 6.6.26x 10-34 x 3 x- j()s---
orbitals: - - - ;2 ; - - - + - - - -
n 1 970.6 X 10- 10
(a) 3p (b) 3d (c) 3s
[Ans. (a) Jili (b)..J6n (c) 0] n",,4 4
[Hint: (a) fil ~l(l + 1) h for 3p, I = I =Ji
2rc
n !
A
=RZ 2 (~ni ni1 J' '3 -+--.---f-r-..3

(b) III =..J6 n for 3d, I 2

(c) III = 0 for 3s, 1=0] 1 = 109677.77
A
x f II ~
1
_!J 4
2
1
89. A single electron system has ionization energy 11180 kJ mor 1• six different lines
Find the number of protons in the nucleus of the system. A ~ 1215.68A]
[Ans. Z == 3]

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ATOMIC STRUCTURE I 123

--!?OBlECTlVEQUESTIONS
Set-1: Questions with single correct answer (a) electrons (b) protons (c) neutrons (d) nucleus
13. Rutherford's scattering experiment is related to the size ofthe:
1. The ratio of elm for a cathode ray: (a) nucleus (b) atom (c) electron (d) neutron
(a) varies with a gas in a discharge tube 14. When alpha particles are sent through a thin metal foil, most of
(b) is fixed them go straight through the foil because:
(c) varies with different electrodes (a) alpha particles are much heavier than electrons
(d) is maximum ifhydrogen is taken (b) alpha particles are positively charged
2. Which of the following statements is wrong about cathode . (c) most part of the atom is empty space
rays? (d) alpha particles move with very high velocity
(a) They trav~1 in straight lines towards cathode 15. The radius of an atomic nucleus is of th~ order of:
(b) They produce heating effect [PMT (MP) 1991]
(c) They carry negative charge (a) 10- 10 cm (b) 10- 13 cm
(d) They produce X-rays when strike with material having
(c) 10-15 em (d) 10-8 cm
high atomic masses
3. Cathode rays are: 16. Atomic size is of the order of:
(a) electromagnetic waves (b) stream of a-particles (a) 10-8 cm(b) lO-IO cm (c) 10- 13 cm(dj l(r'6 cm
(c) stream of electrons (d) radiations 17. Atoms may be regarded ·as comprising of protons, neutrons
4. Cathode rays have: and electrons. If the mass attributed by electrons was doubled
(a) mass only (b) charge only and that attributed by neutrons was halved, the atomic mass of
(c) no mass and no charge (d) mass and charge both 12C would be:
S. Which is the correct statement about.proton? (a) approximately the same (b) doubled
(a) It is a nucleus of deuterium (c) reduced approx. 25% (d) approx. halved
(b) It is an ionised hydrogen molecule is. Positive ions are fonned from neutral atoms by the loss of:
(c) It is an ionised hydrogen atom (a) neutrons (b) protons
(d) It is an a-particle (c) nuclear charge (d) electrons
6. Neutron was discovered by: 19. The nitrogen atom has 7 protons and 7 electrons. The nitride
(a) 1.1. Thomson (b) Chadwick ion will have:
(c) Rutherford (d) Priestley (a) 10 protons and 7 electrons
7. The discovery of neutron came very late because: (b) 7 protons and 10 electrons
(a) it is present in nucleus (c) 4 protons and 7 electrons
(b) it is a fundamental particle (d) 4 protons and 10. electrons
(c)· it does npt move 20. A light whose frequency is equal to 6 x 1014 Hz is incident on a
(d) it docs not carry any charge metal whose work function is 2eV (h = 6.63 x 10- 34 Js,
leV'" 1.6 x. 1O- 19 J). The maximum energy of electrons
S. The fundamental particles present in equal numbers in neutral
emitted will be: (VlTEEE 2008)
atoms (atomic number 71) are:
(a) 2.49 eV (b) 4.49 eV (c) 0.49 eV (d) 5.49 eV
(a) protons and electrons (b) neutrons and electrons
[Hint: Absorbed energy = Threshold energy
(c) protons and neutrons (d) protons and positrons
+ Kinetic energy of photoelectrons
9. The nucleus of the atom consists of:
Absorbed energy hv .
(a) protons and neutrons
(b) , protons and electrons
= 6.626 x·lO- 34 x 6 X 10 14
(c) neutrons and electrons = 3.9756 x 10- 19 J
(d) protons, neutrons and electrons 19
3.9756 X 10- =2.49 eV
10. The absolute value of charge on the electron was determined by: 1.6 x 10- 19
(a) l1. Thomson (b) RA. Millikan
2.49 2 eV + Kinetic energy of photoelectron
(c) Rutherford (d) Chadwick Kinetic energy of photoelectron = 0.49 eV]
11. Atomic number of an element represents: (CBSE 1990) • 21. The size ofthe nucleus is measured in:
(a) number of neutrons in the nucleus (a) amu (b) angstrom
(b) atomic mass of an element (c) em (d) fermi
(c) valency of an element
22. The highest value of el m of anode rays has been observed
. (d) number of protons in the nucleus when the discharge tube is filled with:
12. Rutherford's experiment on .scattering of a-particles showed (a) nitrogen (b) oxygen (c) hydrogen (d) helium
for the first time that the atom has: [CMC (Vell~re) 1991]

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124 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

23. The particle with 13 protons and 10 electrons is: . (c) Dempster's mass.spectrograph .
(a) Al atom (b) A1 3 + ion (d) all of the above
(c) nitrogen isotope (d) none ofthese 36. Mass spectrograph helps in the detection of isotopes because
24. Which of the following atoms contains the least number of they:
neutrons? (a) have different atomic masses
(a) 2~iu (b) Z:U (b) have same number of electrons
(c) have same atomic number
(c) 2~jNp
(d) have same atomic masses
25. The number of neutrons in dip~sitive zinc ion (Zn 2+ , with 37. Which of the following statements is incorrect?
mass number 70) is: (a) The charge on an electron and proton are equal and
(a) 34 (b) 36 . (c) 38 (d) 40 opposite
26. Which of the properties of the elements is a whole number? (b) Neutrons have no charge
(a) Atomic mass (b) Atomic number (c) Electrons and protons have the same mass
(c) Atomic radius (d) Atomic volume (d) The mass of a proton and a neutron are nearly the same
27. Increasing order (lowest first) for the values of e / m 38. The charge on positron is. equal to the charge on:
(charge/mass) for electron (e 1 proton (p), neutron (n) and (a) proton (b) electron
a-particle (a lis: (c) a-particle (d) neutron
(a) e, p, n,a (b) n, p, e,a 39. Discovery ofthe nucleus of an atom was due to the experiment
(c) n,p,a,e (d) n,u,p,e carried out by: .
28. The mass of neutron is ofthe order of: (a) Bohr (b) Rutherford
(a) 10-27 kg (b) 10-26 kg (c) Moseley (d) Thomson
(c) 10-25 kg (d) 10-24 kg 40. Isobars are the atoms of: (CBSE 1991)
29. The. atomS of various isotopes of a particular element differ (a) same elements having same atomic number
from each other in the number of: (b) same elements having same atomic mass
(a) electrons in the outer shell only (c) different elements having same atomic mass
.(b) protons in the nucleus (d) none ofthe above
(c) electrons in the inner shell only 41. Which of the following pairs represents isobars?
(d) neutrons in the nucleus (a) ~Reand iRe fi
(b) Mg and ~iMg
30. Isotopes of the same element have: (c) t~K and ~gCa (d) tgK and ?gK
(a) same number of neutrons 42. Na + ion is isoelectronic with: (CPMT 1990)
(b) same number of protons (a) U+ (b) Mg2+ (c) Ca 2+ (d) Ba 2+
(c) same atomic mass
43. The triad .of nuclei that is isotonic is:
(d) different chemical properties
31. Which of the following conditions is incorrect for a well (a) I:C, IjN, I~F (b) l~C, IjN, IgF
behaved wave function (\if)? [EAMCE;T (Engg.) 2010]
(a) \if must be finite (b) \if must be single valued
. (c) \if must be infinite (d) \if must be continuous 44. Sodium atoms and sodium ions:
32. Atomic mass of an element is not a whole number because: (a) are chemically similar
(a) it contains el~trons, protons and neutrons (b) both react vigorously with water
(b) . it contains isotopes (e) have same number of electrons
(c) it contains allotropes (d) have same number of protons .
(d) aU of the above fi g
45. In Cl and Cl, which of the following is false?
33. Nucleons are: (a) Both have 17 protons
(a) protons and neutrons (b) Both have 17 electrons
(b) neutrons and electrons (e) Both have 18 neutrons
(c) protons and electrons (d) Both show same chemical properties
(d) protons, neutrons and electrons 46. Which of the following is isoelectronie with neon?
34. Isotopes of an element have: . (a) 0 2 - (b) F + (c) Mg (d) Na
(a) different chemical and physical properties 47. Neutrino has:
(b) similar chemical and physical properties (a) charge + I, mass 1 (b) charge 0, mass 0
(c) similar chemical but different physical properties (c) charge - L mass 1 (d) charge 0, mass 1
(d) .similar physical and different chemical properties 48. Positronium is the name given to an atom-like combination
35. Isotopes are identified by: . formed between: (JIPMER 1991)
(a) a positron and a proton
(a) positive ray analysis
(b) a positron and a neutron
(b) Astons' mass spectrograph

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I
ATOMIC STRUCTURE 125

(c) a positron and an a-particle 61. A 600 W mercury lamp emits monochromatic radiation of
(d) a positron and an electron wavelength 331.3 nm. How many photons are emitted from
49. An isotone of ~~Ge is: the lamp per second? [PET (Kerala) 2010]
(a) ~iGe. (b) j~As (c) I!Se (d) j!Se (h == 6.626 x 10-34 Is, velocity oflight 3x 108 ms-I)
50. Which of the following does not characterise X-I1lYs? (a) Ix 1019 (b) Ix 102°
1
(lIT 1992) (c)lxl02 (d) Ix 1023
(a) The radiations can ionise gases 2
(e) Ix 102
(b) It causes ZnS to fluorescence
[Hint: Power Energy
(c) Deflected by electric and magnetic fields Time
(d) Have wavelengths shorter than ultraviolet rays
600= nhc
51. X-rays are produced when a stream of electrons in an X-ray Axlsec
tube: 34 8
600 == n X 6.626 10- x 3 X 10
(a) hits the glass wall ofthe tube
33L3x
(b) strikes the metal target 21
n==lx10 ]
(c) passes through a strong magnetic field v
62. Out of X-rays, visible, ultraviolet, radiowaves, the largest
(d) none of the above
frequency is of:
52. Radius of a nucleus is proportional to:
(a) X-rays (b) visible
(a) A (b) A I/3 (c) A2 (d)A 2/3 (c) ultraviolet' (d) radiowaves
53. The nature of positive rays produced in a vacuum discharge 63. The wave nnmber 'which,corresponds to electromagnetic
tube depends upon: radiations of 600 nm is equal to:
(a) the nature of the gas filled (a) 1.6 x 104 cm- I (b) O.l6x 104 cm- 1
(b) nature ofthe material of cathode
(c) nature of the material of anode (c) 16 x 104 cm- I (d) 160 x 104 cm- i
(d) the potential applied across the electrodes 64. Line spectrum is characteristic of:
. 54. Electromagnetic radiation with maximum wavelength is: (a) molecules (b) atoms
(MLNR 1991) (c) radicals (d) none of these
(a) ultraviolet (b) radiowaves 65. Which one of the following is not the characteristic of
(c) X-rays (d) infrared Planck's quantum theory of radiation? (AIIMS 1991)
55. The ratio of energy of radiations of wavelengths 2000 A and (a) The energy is not absorbed or emitted in whole number
4000 A is: . (eDSE 1994) multiple of quantum' .
(a) 2 (b) 4 (c) 1/2 . (d) 1/ 4 (b) Radiation is associated with energy
56. The ratio of the diameter of the atom and the diameter of the (c) Radiation energy is not emitted or absorbed continuously
nucleus is: . but in the form of small packets called quanta
(a) 10 5
(b) ItY (cPO (d) 10- 1 (d) This magnitude of energy associated with a quantum is
proportional to the frequency
57. The ratio of the volume of the atom and the volume of the
nucleus is: 66. Which of the following among the visible colours has the
minimum wavelength? .
(a) 1010 (b) 1012 (c) 1015 (d) 102°
(a) Red (b) Blue
58. Which of the following statements is incorrect?
(a) The frequency of radiation is inversely proportional to its (c) Green (d) Violet
67. The spectrum of helium is expected to be similar to that of:
wavelength
(b) Energy of radiation increases with increase in frequency (a) H (b) Na (c)He+ , (d) Li+
(c) Energy of radiation decreases with increase in wavelength 68. According to classical theory. if an electron' is moving in a:
(d) The frequency of radiation is directly proportional to its.• circular orbit around the nucleus:
wavelength . (a) it will continue to do so for sometime
59. Visible light consists of rays with wavelengths in the (b) its orbit will continuously shrink
approximate range of: (c) its orbit will continuously enlarge
(a) 4000A to 7500 A (d) it will continue to do so for all the time
(b) 4 x 10-3 cm to 7.5 x 10-4 cm 69. Bohr advanced tlJe idea of:
(a) stationary electrons (b) stationary nucleus
(c) 4000 nm to 7500 nm
(c) stationary orbits (d) elliptical orbits
(d) 4 x 10-5 m to 7.5 x 10-6 m
70. On Bohr stationary orbits:
60. Which of the following statements concerning light js false? (a) electrons do not move
(a) It is a part of the electromagnetic spectrum (b) electrons move emitting radiations
(b) It travels with same velocity, i.e., 3 x 1(Yo cm/ s (c) energy ofthe electron remains constant
(c) It cannot be deflected by a magnet
(d) angular momentum ofthe electron is..!!...
(d) It consi,ts A photons of same energy . 2n

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126 G.R.B. PHYSICAL CHEMISTRY FOR CoMPETITIONS

7.1. Energy ofEohr orbit: (DPMT 1991) 2nme2

(a) increases as we move away from the nucleus (d) E" =:
n 2h2
(b) decreases as we move away from the nucleus .
81. According to Bohr theOlY, the angular momentum for an
(c) remains the same as we move away from the nucleus
electron of 5th orbit is:
(d) none of the above
(a) 5h/n (b) 2.5h/n (c) 5n/ h (d) 25h/n
72. Which of the following statements does not form part of
82. The value of Bohr radius of hydrogen atom is: (CBSE 1.991)
Bohr's model ofthe hydrogen atom?
(a) 0.:529 x 10-7 cm (b) 0.529 x 10-8 cm
(a) Energy of the electron in the orbit is quantized
(b) The electron in the orbit nearest to the nucleus has the (c) 0.529 x 10-9 cm (d) 0.529 x 10- 10 cm
lowest energy 83. The energy of an electron in the nth Bohr orbit of hydrogen
(c) Electrons revolve in different orbit nucleus atom is: . (CBSE 1992)
(d) The position and velocity of the electron in the orbit (a) - 13.6 eV (b) 13.6 eV (c) 13.6 eV (d) _ 13.6 eV
cannot be determined simultaneously n4 n3 n2 . n
73. Which of the following statements does not form a pa11 of 84. Which ofthe following electron ·transitions in hydrogen atom
Bohr's model of hydrogen atom? (DCE 2005) will require largest amount ofenergy? . (MLNR 1992)
(a) Energy of the electrons in the orbit is quantised (a) from n = Ito n 2 (b) from n = 2 to n 3
(b) The electron in the orbit nearest to theJlUcleus has the- (c) from n 00 to n =: 1 (d) from n = 3 to n 5
lowest energy <". .
85. For a hydrogen atom, the energies-that an electron canhave~
(c) Electrons revolve in different orbits around the nucleus are given by the expression, E 13.58/n 2 eV, where n is an
(d) The position and velocity of the electrons in the orbit integer. The smallest amount of energy that a hydrogen atom
cannot be determined simultaneously in the ground state can absorb is:
74. The radius of the first orbit of H-atom is r. Then the radius of (a) 1.00 eV (b) 3.39 eV (c) 6.79 eV (d) 10.19 eV
the first orbit ofLi2+ will be: [AMU-PMT 2009] 86. The energy of hydrogen atom in its ground state is -13.6 e V.
r (b) r The energy of the level Corresponding to n == 5 is:
(a) - (c) 3r (d) 9r
9 3 (CBSE 1990)
2
[Hint: r= n xO.S29A] (a) - 0.54 eV (b) 5.40 eV (c) -0.85 eV (d) - 2.72eV
z 87. Ell = - 313.6/ n 2 kcallmol. If the value of E 34.84
75. The energy liberated when an excited electron returns to its .. kcallmol, to which value does 'n' correspond?
ground state can have: (a) 4 (b) 3 (c) 2 (d) I
(a) any value from zero to infinity 88. The ratio of the difference between 1st and 2nd Bch orbits
(b) only negative values energy to that between 2nd and 3rd orbits energy is:
(c) only specified positive values (a) 112 (b) 1/3 (c) 27/5 (d) 5127
(d) none of the abovc 89. Bohr's model can explain:
76. On the basis of Bohr's model, the radius of the 3rd~orbit is: (a) spectnim of hydrogen atom only
(a) equal to the radius of first orbit (b) spectrum of any atom or ion having one .electron only
(b) three times the radius of first orbit (c) spectrum of hydrogen molecule
( c) five times the radius of first orbit (d) solar spectrum
(d) nine times the radius of first orbit . 90. The energy difference between two electronic states is 43.56
77. The ratio of 2nd, 4th and 6th orbits of hydrogen atom is: kcal/mol. The frequency of light emitted when the electron
(a) 2 : 4 : 6 (b) I : 4 : 9 drops from higher orbit to lower orbit, is:
(Planck's constant = 9.52 x 10- 14, kcallmol)
(c) I : 4 . 6 (d) I : 2 : 3
78. Which point does not pertain to Bohr's model of atom? (a) 9.14 x I(j4 cycle/sec (b) 45.1 x 1014 cycle/sec
(a) Angular momentum is an integral multiple of h/ (2it) (e) 91.4 x 1014 cycle/sec (d) 4.57 x 1014 cyc1e/se~
(b) The path of the electron is circular 91. Which of the following transitions of an electron in hydrogen
(c) Force of attraction towards nucleus = centrifugal force of
atom emits radiation the lowest wavelength?
(d) The energy changes are taking place continuously [EAMCET (Engg.) 2010]
79. The distance between 3rd and 2nd orbits in the hydrogen atom (a) n2 == to nl 2
00 (b) n2 4ton1 = 3
is: (c) n2 = 2 to nl "" 1 (d) n2 5 to n l == 3
(a) 2.646 x 10-8 cm (b) 2.116 x 10-8 cm 92. The wavelength of a spectral line for an electronic transition is
(c) 1.058 x 10-8 cm . (d) 0.529 x 10-8 cm inversely related to:
(a) number of electrons undergoing transition
80. The correct expression derived for the energy of an electron in
(b) the nuclear charge of the atom
the nth energy level in hydrogen atom is:
( c) the velocity of an electron undergoing transition
2n 2 me4 2n 2 me4
(a) En (b) En = (d) the difference in the energy levels involved 111 the
2h2
---n 2
, nh transition

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I
ATOMIC STRUCTURE 127

93. The ionisation energy of the electron in the ls-orbital of the (a) 15,200cm- I (b) 60,800cm- 1
hydrogen atom is 13.6 eY. The energy of the electron after
(c) 76,000 cm- 1 (d) 1,36,800cm-:- 1
promotion to 2s-orbital is: IISC (Bihar) 1993)
(a) -3.4eV (b) -13.6eV 103. "The position and the velocity of a small particle like electron
(c) -27.2eV (d) O.OeV cannot be simultaneously determined." This statement is:
94 .. Which electronic level would allow the hydrogen atom to (a) Heisenberg uncertainty principle
absorb a photon but not to emit it? (b) Pauli's exclusion principle
(a) Is (b) 2s (c) 3s (d) 4s (c) aufbau's principle
95. The spectral lines corresponding to the radiation emitted by an (d) de Broglie's wave nature ofthe electron
electron jumping from 6th, 5th' and 4th orbits to second orbit 104. de Broglie equation describes therelationship of wavelength
belong to: associated with the motion of an electron and its:
(a) Lyman series (b) Balmer series (a) mass (b) energy (c) momentum (d) charge
(c) Paschen series (d) Pfund series 105. If the magnetic quantum number of a given atom is
96. The spectral lines corresponding to the radiation emitted by an represented by then what will be its principal quantum
electron jumping from higher orbits to first-Orbit belong to: number? [BHU (Pre.) 2005)
(a) Paschen series (b) Balmerseries (a) 2 (b) 3 (c) 4 (d) 5
(c) Lyman series (d) None of these J06. Which of the following relates to photons both as wave motion
and as a stream of particles? (IlT 1992)
97. In a hydrogen atom, the transition takes plac.efrOll1n 3 to
n = 2. If Rydberg constant is 1.097 x 107 m-I, the wavelength (a) Interference (b) Diffraction
of the emitted radiation is: (c) E == hv (d) E = me2
(a) 6564A (b) 6064 A 107. If uncertainty in the positiOil of an electron is zero, the
(c) 6664A (d) 5664 A uncertainty in its momentum would be:

[Hint: Apply.!. = R
A
[~
x j} Ca) zero,
(c) > h/(4n)
(b) < hI(4'n)
(d) infinite
108. Which one of the following explains light both as a stream of
98. The speed ofthe electron in the 1st orbit of the hydrogen atom
particles and as wave motion?
in the ground state is (e is the velocity oflight):
(a) Diffraction (b) A = h/ p
e e e e
(a) 1.37 (b) l370 (c) 13.7 (d) 137 (c) Interference (d) Photoelectric effect
109. A body of mass x kg is moving with velocity oflOO msec- I . Its
[Hint: Velocity of electron in the 1st orbit, v = h/(2nmr)
de Broglie wavelength is 6. 62 X 10-35 m. Hence x is:
, = 2.189 X 108 em/sec., velocity of light, c = 3 X 10 10 ern/sec. (h == 6.62x 10-34 J sec) [CET (Karnataka) 2009)
Ratio c/v l37] (a) 0.25 kg (b) 0.15 kg
99. , Find the value of wave number v in terms of Rydberg's (c) 0.2 kg (d) 0.1 kg
constant, when transition of electron takes place between two 110. A 200 g cricket ball is thrown with a'speed of 3.0 x 103 cm
levels ofHe+ ion whose sum is 4 and difference is 2. sec-I. What will be its de Broglie's wavelength?
(a) 8R (b) 32R (II =6.6 X 10-27 g cm2 sec-I ) [CET (Gujftrat) 2008)
9 9 (a) 1.1 x 10-32 cm (b) 2.2 x 10- 32 cm
(c) 3R (d) None of these (c) 0.55 x 10-32 cm (d) Il.Ox 10-32 cm
4
Ill. The electronic configuration of a dipositive ion M 2+ is 2, 8, 14
[Hint: nl +~ 4; n2 -nj 2 and its atomic mass is 56. The number of neutrons in 'the
V RZ2[~_~J nucleus would be:
nl ni ( (a) 30 (b) 32 (c) 34 (d) 42
112. An element with atomic number 20 will be placed in which
=R X 22 [~- ~J
2
32R] period of the periodic table?
1 3 9
(a) 5th (b) 4th (c) 3rd (d) 2nd
100. With the increasing principal quantum number, the energy
113. The frequency of radiation emitted when the ekc~on falls
difference between adjacent energy levels in hydrogen atom:
from n 4 to n 1 in a hydrogen atom will be (Given
(a) increases (b) decreases
ionisation energy of H=2.18x 10- 18 Jatom- 1 and
(c) is the same (d) none of these h = 6.626 X 10-34 Js): [Manipa, (Me d.) 2007)
101. An electron in an atom: rCEET (Bihar) 1992] (a) 1.54 x 1015 S-I (b) 1.03 x 1015 8- 1
(a) moves randomly around the nucleus
(b) has fixed space around the nucleus
(c) 3.08 X 101 5 Cd) 2 x 1015 s-i
(c) is stationary in various energy levels 114. In a multi-electron atom, which of the following, orbitals
(d) moves around its nucleus in definite energy levels described by the three quantum numbers will have the same
102. The wave number of first line of Balmer series of hydrogen is energy in the absence of magnetic and electric fields?
15200 cm- I • The wave number ofthe first Balmer line ofLi 2+ (AIEEE2005)
ion is: [liT !}~f:r:'i'''~';,,~~ 1992] (i) 17 = I, I=: 0, m 0 (ii) n == 2, I 0, m = 0

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I
128 I G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

(iii) n =: 2, I I, m =: I (iv) n =: 3, I=: 2, m =: I 127. Any p-orbital can accommodate up to: (MLNR 1990)
(v) n =: 3, I ::= 2, m =: 0 (a) 4 electrons
(a) (i) and (ii) (b) (ii) and (iii) (b) 2 electrons with parallel spins
(c) (iii) and (iv) (d) (iv) and (v) (c) 6 electrons
115. Which of the ions is not having the configuration ofNe? (d) 2 electrons with opposite spins
(a) Cl- (b) F - (c) Na + (d) Mg2+ 128. How'many electrons can fit into the orbitals that comprise the
116. Which ofthe following has the maximum number of unpaired 3rd quantum shell n =: 3?
d-electrons? (KeET 2008) (a) 2 (b) 8 . (c) 18 (d) 32
(a) Ni 3+ (b)Cu+ (c)Zn2+ (d) Fe2 + 129. The total number of orbitals in a principal shell is:
117." Which of the following expressions gives the de Broglie (a) n (b) n2 (c) 2n 2 (d) 3n 2
relationship? IJEE (WB) 2008) 130. Two electrons in K-shell will differ in:
h h (a) principal quantum number
(a) p =: - (b) A
mv mv (b) spin quantum number
(c) A =: h
h (c) azimuthal quantum number .
(d) Am=
mp p (d) magnetic quantum number .
118. The principal quantum number of an atom is related to the: 131. Which one of the following orbitals has the shape of a
(MLNR 1990) baby-boother ?
(a) size of the orbital (a) dxy (b) d x 2 - y 2 (c) d z 2 (d) Py
(b) orbitafangular momentum
132. Which one of the following represents an impossible
(c) spin angular momentum
arrangement? (AIEEE 2009)
(d) orien.tation ofthe orbital in space n I m s n I m s
119. The magnetic quantum is a number related to:
(a) 3 2 - 2 112 (b) 4 0 o 1/2
(a) size (b) shape
(c) 3 2 . -3 112 (d) 5 3 o 112
(c) orientation (d) spin
120. The principal quantuIn number represents: (CPMT 1991)
(a) shape of an orbital
(b) nU!l).ber of e~ectrons in an orbit
(c) distance of electron from nucleus
(d) numbei: of orbitals in an orbit
121. The quantum number not obtained from the Schrodinger's
wave equation is: (lIT 1990)
(a) n (b) 1 (c) m (d) s
122. In a given atom, no two electrons can have the same values for
all the four quantum nU!l).bers. This is called: (CPMT 1990)
(a) Hund's rule (b) Pauli's exclusion principle
(c) Uncertainty principle (d) aufbau principle
123. The atomic orbital is:
(a) the circular path of the electron
(b) elliptical shaped orbit .
(c) three-dimensional field around nucleus
(c) 21 + I (d) 41 + 2
. (d) the region in which there is maximum probability· of
finding an electron [Hint: Number of electrons with same spin
·124. If the ionization energy for hydrogen atom is 13.6 eV, then the 1 x Total no. of electrons
ionization energy for He+ ion should be: 2
1 .
I PMT (Haryana) 2004] =: - x 2 (21 + 1) =: (21 + 1)]
(a) 13.6 eV (b) 6.8 eV 2
(c) 54.4 eV (d) 72.2 eV 136.. Which of the following represents the correct set of four
125. Principal,. azimuthal and magnetic quantum numbers are quantum numbers of a 4d-e1ectron? (MLNR 1992)
respectively related to: (a) 4,3,2, + 112 (b) 4,2, 1,0
(a) si?e,shape and orientation (c) 4,3, - 2, + 112 (d) 4,2, i, 1/2
.(b) shape, size and orientation 137. Values of magnetic orbital quantum number for an electron of
(c) size, orientation and shape M-shell can be: (PET (Raj.) 2008]
(d) none of the above (a) 0, 1,2 (b)-2,-1,0,+1,+2
126. Energy of e1ectron·in the H-atom is determined by : (c) 0, 1,2,3 (d) 1,0,+1
(a) only n (b) bothn and I 138. Correct set of four quantum numbers for the outermost
(c) n, I and m (d) all the four quantum numbers . electron of rubidium (Z = 37) is:

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ATOMIC STRUCTURE 129

(a) 5,0,0,1/2 (b) 5, 1,0, 112 149• The energy ofaa~lectron of 2p y' orbital is:
.(c) 5,1, 1.112 (d) 6,0,0,112 (a) greater than 2px orbital
139. Which one of the following subshells is spherical in shape? (b) less than 2pz orbital
(a) 4s (b) 4p (c) 4d (d) 4f (c) equalto 2sorbitil
140. In hydrogen atom, the electron is at a distance of 4.768 A from (d) same as that of2px and 2pz orbitals
the nucleus. The angular momentum of the electron is : 150• The two electrons occupying the same orbital are
. [EAMCET (Med.) 2010] distinguished by: .
(a)· 3h . (b) ~ (a) principal quantum.number
211: 211: (b) azimuthal quantum number
h (d) 3h (c) magnetic quantum number
(c) -
11: 11: (d) spin quantum number
n2 151. The maximum number of electrons in a subshell is given by
[Hint: r = - x 0.529 A
z the expression: (AIEEE 2009)
n2 (a) 41 + 2 (b) 4/- 2
4.768 = x 0.529
1 (c) ()./ + I (d) 2n2
n=3 152. The electronic configuration of an atom/ion can be defined by
nh __ 3h] which of the following1
:. Angular m()mentum (mvr)
211: 211: (a) A1.dbau PriI1ciple
141. Total number of m values for n 4 is: (b) Pauli's exclusion principle
(a) 8 (b) 16 (c) 12 (d) 20 (c) Hund's rule of maximum mUltiplicity
142. What is the total number of orbitals in the shell to which the (d) All of the above
g-subshell first arise?
153. An electron has a spin quantum number + 112 and a magnetic
(a) 9 (b) 16 (c) 25 . (d)36 quantum number 1. It cannot be present in:
[Hint: For g-subshell, / = 4 (a) d-orbital (b).forbital (c) s-orbital (d) p-orbital
:. It will arise in 5th shell. 154. The value of azimuthal quantum number for electrons present
Total number of orbitals in 5th shell == n2 = 25] in 4 p-orbitals is:
143. In Bohr's model, if the atomic radius of the first orbit 1j, then (a) 1
radius offourth orbit will be : [BHU (Screening) 2010] (b) 2

(a) 4'1 (b) 6'1 (c) 161j(d) !L

°
(c) any value between and 3 except I
16 (d) zero
144. Which of the following statements is not correct for an 155. For the energy levels in an atom which one of the following
electron that has quantum numbers n = 4 and m = 2? statements is correct?
(MLNR 1993) (a) The 4s sub-energy level is at a higher energy than the 3d
, (a) The electron may have the q. no. I = + 1/2 sub-energy level
(b) The electron may have the q. no. I = 2 (b) The M -energy level can have maximum of 32 electrons
(c) The electron may have the q. no. 1= 3 (c) The second principal energy level can have four orbitals
(d) The electron may have the q. no. I = 0, I, 2, 3 and contain a maximum of 8 electrons
145. The angular momentum of an electron depends on: (d) The 5th main energy level can have maximum of 50
(a) principal quantum number electrons
(b) azimuthal quantum nwnber 156. A new electron enters the orbital when:
( c) magnetic quantum number (a) (n + /)is minimwn (b) {n + J)is maximum
(d) all of the above (c) (n+m)isminimum (d) (n+m)ismaximum
146. The correct set of quantum numbers for the unpaired electron 157. For a given value of n{principal quantum number), the energy
of a chlorine atom is: (DPMT 2009) of different subshells can be arranged in the order of: .
1 I (a) !>d> p>s (b) s> p>d>!
(a) 2,0,0, + (b) 2, I, -1, + (c)!>p>d>s (d)s>!>p>d
2 2
I 1 158. After filling the 4d-orbitals, an electron will enter in:
(c) 3,1, I,±- (d) 3,0,O,±- (a) 4p (b) 4s (c) 5p (d) 4f
2 2
159. According to t\ufbau principle, the correct order of energy of
147. The magnetic quantum number for valency electron of sodium
atom is:
3d,4s and 4 p-orbitals is: [eEl' (J&K) 2006)
(a) 3 (b) 2 (c) I (d) zero (a) 4p<3d<4s (b) 4s<4p<3d
148. The shape of the orbital is given by: (c) 4s < 3d < 4 p (d) 3d < 4s < 4 p
[PET (Raj.) 2008J
(a) spin quantum number 160. Number of p-electrons in bromine atom is:
(b) magnetic quantum number [PMT (Haryana) 20041
.(c) azimuthal quantum number (a) 12 (b) 15
(d) principal quantum number (c) 7 (d) 17

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130 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

161. [Ar] 3dlO 4sl electronic configuration belongs to: . 169. The radial probability distribuuon curve obtained for an
[PET (MP) 20081 orbital wave function ('If) has 3 reaks and 2 radial nodes. The
(a) Ti (b) Tl (c) Cu (d) V valence electron of which one of the following metals does
162. How many unpaired electrons are ther~ in Ni2+ ? (Z 28) this wave function ('If ) correspond to ?
(a) Zero (b) 8 (c) 2 (d) 4 [EAMCET (Med.) 20101
163. The electronic configuration of chromium (Z 24) is: (a) Co (b) Li (c) K (d) Na
[PMT (MP) 1993; BHU (Pre.) 20051 [Hint: Na ll ls2, 2s22l, 3s 1
'----v----'
(a) [Ne] 3i3 p 6 3d 44i (b) [Ne] 3s2 3p 6 3d 54s1 Valence electron
Number of radial node =: n-l-I (n == 3)
(c) [Ne] 3i3p 6 3d l 4i (d) [Ne] 3i 3p64i4p4
3-0-1 = 2]
164. The number of d-electrons in Fe 2+ (At. No. 26) is not equal to 170. ~pton (At. No. 36) has the electronic configuration [A]
that of the: (MLNR 1993) 4s 3d 104 p 6. The 37th electron will go into which one of the
(a) p-electrons in Ne (At. No. 10) . following sub-levels?
(b) s-electrons in Mg (At. No. 12) (a) 4/ (b) 4d
(c) d-electrons in Fe atom (c) 3p (d) 5s
(d) p -electrons in Cl- ion (At. No. 17) 171. An ion which has 18 electrons in the outermost shell is:
165. If the electronic structure of oxygen atom is written as (CBSE 1990)
f--2p---+ (a) K + Q » Cu.+ (c) Cs+ (d) Th 4 +
172: Which ofthe following has non-spherical shell of electronJa
ls2, 2s2 !i.j, 1i.j, 1 . 1; it would violate: fISC (Bihar) 1993]
(lIT 1993)
(a) Hund's rule (a) He (b) B (c) Be (d) Li
173. Which one of the following sets of quantum numbers is not
(b) Pauli's exclusion principle
possible for an electron in the ground state of an atom with
(c) both Hund's and Pauli'sprincipJes
atomic number 19? [PET (Kerala) 2006;
(d) none of the above
CET (Karnataka) 2009]
166. The orbital diagram in which 'aufbau principle' is violated,is:
(a) n = 2, I 0, m == 0 (b) n = 2, I = 1, m 0
2s f - - 2 p---+
(c) n = 3, I =: I, m -1 (d) n 3,1=2,m=±2
(a) [ill I i.j,1 i I (e) n 4,1=0,m=0
174. Helium nucleus is composed of two protons and two neutrons.
(b) If the atomic mass is 4.00388, how much energy is released
when the nucleus is constituted?
(c) :1 i.j, i (Mass of proton == 1.00757, Mass of neutron == 1.00893)
(a) 283 MeV (b) 28.3 MeV
(d) :1 i.j, :1 (c) 2830 MeV (d) 2.83 MeV
175. Binding energy per nucleon of three nuclei A, B and Care
167. The manganese (Z = 25)has the outer configuration: 5.5, 8.5 and 7.5 respectively. Which one of the following
4s ( 3p ) nuclei is most stable?
(a) [illi i.j, II i il i l i II i !I (a) A (b) C
(c) B (d) Cannot be predicted
(b) [ill 176. The mass of jLi is 0.042 less than the mass of 3 protons and
4 neutrons. The binding energy per nucleon in j Li is:
(c) (ill :1 i .j, il j.j, :1 i .(BHU1992)
(a) 5:6 MeV (b) 56 MeV (c) 0.56 MeV (d) 560 MeV
(d) 0 II j.j, II i.j, II i i i 177. Meson was discovered by:
(a) Powell (b) Seaborg
168. Which of the following elements is represented by the (c) Anderson (d) Yukawa
electronic configuration? 178.. In most stable elements, the number of protons and neutrons
f--2p-c-? are:
(a) odd-odd (b) even-even
(c) odd-even (d) even-odd
179. Nuclear particles responsible for holding all nucleons together
are:
(a) electrons (b) neutrons
(c) positrons (d) mesons
(a) Nitrogen (b) Fluorine 180~ The introduction of a neutron into the nuclear composition of
(c) Oxygen (d) Neon an atom would lead to a change in: (MLNR 1995)

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ATOMIC STRUCTURE 1.131

(a) its atomic inass (a) 3 (b) 4 (c) 2 (d) 1
(b) its atomic number 191. A certain negative ion has in its nucleus 18 neutrons,and
(c) the chemical nature of the atom 18 electrons in its extranuclear structure. What is the mass
(d) number of the electron also number of the most abundant isotope of X ?
181. Which of the following has highest orbital angular (a) 36 (b) 35.46 (c) 32 (d) 39
momentum? 193. Which of the following statements is not correct?
(a) 4s (b) 4p (c) 4d (d) 4f
(a) The shape of an atomic orbital depends on the azimuthal
181. Which of the following has maximum number of unpaired
quantum number .
electrons? (PMT (Raj.) 2004; BHU (Pre.) 2005)
(b) The orientation of an atomic orbital depends on the
(a) (b) Fe 2+ (c) C0 2+ (d) C0 3+ magnetic quantum number
183. An electron isnot deflected on passing through a certain (c) The en~rgy of an electron in an atomic orbital of
region,because: multielectron atom depends on the principal : quantum
(a) there is no magnetic field in that region number
(b) there is a magnetic field but velocity of the electron is (d) The number of degenerate atomic orbitals of one type
parallel to the direction of magnetic field depends on the values of azimuthal and magnetic qUa)1tum
(c) the electron is a chargeless particle numbers
(d) none ofthe aboye 194. Gases begin to conduct electricity at low pressure because:
184. In MiHik1m's oil drop experiment, we make~use of:
(CSSE 1994)
(a) Ohm's law (b) Ampere's law (a) at low pressures gases tum to plasma
(c) Stoke's law (d) Faraday's law .(b) colliding electrons can acquire higher kinetic energy due to
185. A strong argument for the particle nature of cathode rays is: increased mean free path leading to ionisation of atoms
(a) they can propagate in vacuum (c) atoms break up into electrons and protons
(b) they produce fluorescence (d) the electrons in atoms can move freely at low pressure
(c) they cast shadows
195. An electron of mass m and charge e, is accelerated from rest
(d) they are deflected by electric and magnetic fields through a potential difference V in vacuum. Its final speed will
186. .As the speed of the electrons increases, the measured value of be: (eBSE W94)
charge to mass ratio (in the relativistic units):
(a) increases (a) ~ (b) 2eVIm
(b) remains unchanged (c) J(eVI2m) (d) J(2eVlm)
(c) decreases .
(d) first increases and then decreases 196. The difference in angular momentum associated with .the
187. Which of the following are true for cathode rays? electron in the two successive orbits of hydrogen atom is:
(a) It travels along a straight line (a) hln (b) hl2n (c) hl2 (d) (n -1)hI2lt
(b) It emits X -rays when strikes a metal 197. Photoelectric effect can be explained by assuming that light:
(c) It is an electromagnetic wave (a) is a form of transverse waves
(d) It is not deflected by magnetic field (b) is a form of longitudinal waves
ISS. Three isotopes of an element have masl' numbers, M, (M + 1)
(c) can be polarised
and (M + 2). If the mean mass number is (M + 0.5) then
which of the following ratios may be accepted for (d) consists of quanta .
M, (M + 1), (M + 2)in that order? 198. The photoelectric effect supports quantum nature of Jight
(a) I : 1 : I (b) 4: 1 : 1
because:
(a) there is a minimum frequency of light below which no
(c) 3 : 2 : 1 (d) 2 : 1 : 1
photoelectrons are emitted .
189. The radii of two of the first four Bohr orbits of the hydrogen
(b) the maximum kinetic energy of p\1otoelectrons depends
atom are in the ratio I : 4. The energy difference between them
only on the frequency oflight and not on its intensity
maybe: . .. .
(a) either 12.09 eV or 3.4 eV (6) .either 2.55 eV or 10.2 eV (c) even whet! metal surface is faintly illuminated the
photoelectrons leave the surface immediately
(c) either 13.6 eV or 3.4 eV (d) either 3.4 eV or 0.85 eV
(d) electric charge of photoelectrons is quanti sed
190. Photoelectric emission is .observed from a surface. for
frequencies VI and v 2 ofthe incidentradiafion (VI> V2 ).Ifthe 199. The mass ofa proton at rest is: .(CB.SE 1991)
maximum kinetic energies of the photoelectrons in the two (a) zero (b) 1.67 x 10-:-35 kg
cases are in the ratio 1 : k then the threshold frequency V 0 is (c) one amu (d) 9 x 10-31 kg
given by:
200. Momentum of a photon of wavelength A is: (eBSE 1993)
(a) v 2 - VI (b) k VI ~V2 (c) _k--=.._..!.. (d) v 2 - VI 2
k-l k-l k k (a) hlA (b) zero (c) hAle (d) hAle
191. The number of waves made by a l30hr electron in an orbit of 201. When X-rays pass through air they:
maximum magnetic quantum number +2 is: . ( a) produce light track in the air
(b) ionise the gas

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.....
132 G.R.B. PHYSlpAL C~EMISTRY .FOR COMPETITIONS

(c) produce fumes in the air h

(a) L= nh (b) L=1(1+1)-
(d) accelerate gas atoms. 2n 2n
202. X.rays:· (CPMT 1991) mg h
(a) are deflected in a magnetic field (c) L= (d) L=-
2n . 4n
(b) are deflected in an electric field
214. Ground state electronic configuration of nitrogen atom can be
(c) remain undeflected by both the fields
represented by: (lIT 1999)
(d) are deflected in both the fields
203. Find the frequency of light that corresponds to photons of
energy 5.0x 10-5 erg: (AIIMS 2010)
1. [ill DIJ 11 11 11 I 2. ill] [ill 11 1 ~ 11 1
(a) 7.5xlO-21 sec-1
(c) 7.5xl02 1 sec~l
(b) 7.5 x 10-21 sec
(d) 7.5 x 1&1 sec
3. DO l!J 11 I ~ I ~ 1 4. [!] [!] 1J 1J 1J 1
E 5xlO-s erg (a) 1 only (b) 1,2 only (c) 1,4 only (d) 2,3 only
[Hlnt:v=-= .
. h 6.63 X 10-27 erg sec 215. Which of the following statement{s) are correct?
1. Electronic configuration ofCr is [Ar] 3d 5481 (At. No. of
= 7.54 xl02 1 sec-I]
Cr=24)
204. The energy of an electron in the flISt Bohr orbit of H·atom is 2. The magnetic quantum number may have negative value
-:-13.6 eV. The possible energy value(s) of the excited state(s)
3. In silver atom, 23 electrons have a spin of one type and 24
for electr~ns in Bohr orbits of hydrogen is/are: (lIT 1998)
of the opposite type (At. No~()fA,g== 47L _.........__. ___ _
(a) ..:.3.4 tV (b) - 4.2 eV 4. The oxidation state. of nitrogen in HN3 is -3 (lIT 1998)
(c) -6.8 eV (d) +6.8 eV (a) 1,2,3 (b) 2,3:4 (c) 3,4 (d) 1,2,4
205. The electrons identified by quantum numbers n and I, (i) n == 4, 216. The electronic configuration of an element is
1= 1 Oi) n = 4, 1= 0 (iii) n = 3, I = 2 (iv) n = 3,1 = 1 can be ls22i 2p6, 3i 3p6 3d 5, 4}. This represents: (lIT 2000)
placed in order of increasing energy, from the lowest to highest . (a) excited state (b) ground state
as: (lIT 1999) (c) cationic state (d) anionic state
(a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii) 217. The quantum numbers + 1 and ~ ~ for the electron spin
(c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii) represents: 2 2 (liT 2000)
206. The wavelength of the radiation emitted when an electron falls (a) rotation of the electron in clockwise and anticlockwise
from Bohr orbit 4 to 2 in hydrogen atom is: (lIT 1999) directions respectively
(a) 243 nm (b) 972 nm (b) rotation of the electron in anticlockwise and clockwise
(c) 486 nm (d) 182 nm directions respectively
207. The energy of the electron in the flISt orbit of He+ is (c) magnetic moment of electron pointing up and down
- 871.6 x 10-20 J..The energy of the election in the first orbit respectively
of hydrogen would be: (lIT 1998) (d) two quantum mechanical spin states which have no
20 20 classical analogues
(a) -871.6x 10- J (b) - 435 X 10- J
(c) -217.9x 10-20 J (d) -108.9x 10-20 J 218. Rutherford's experiment, which established the nuclear model
of the atom, Osed a beam of: (lIT 2002)
208. The wavelength associated with a golf ball weighing 200 g and (a) j3.particles, which impinged on a metal foil and got
moving with a speed of 5 mIh is of the o~ of: (lIT 2000) absorbed
(a) 10-10 m (b) 10-20 m (c) 10-30 m (d) 10-40 m (b) y·rays, which impinged on a metal foil and ejected
209. Who modified Bohr theory by introducing elliptical orbits for electrons
electron path? (CBSE 1999) (c) helium atoms, which impinged on a metal foil and got
(a) Hund (b) Thomson scattered . . >

(c) Rutherford (d) Sommerfeld (d) helium nuclei, which impinged on a metal foil and got
210. The uncertainty in momentum of an electron is 1 x 10-5 kg scattered
ms -1. The uncertainty in its position will be: 219. How many moles of electrons weiflt one kilogram?
(h = 6.62 x 10-34 kg.m2 .s) (CBSE 1999; BHU 2010) (Mass of electron = 9.108 x 10- J kg, Avogadro's number
(a) 1.05 x 10-28 m (b) 1.05 x 10-26 m = 6.023 x 1023 ) (lIT 2002)
(c) 5.27x 10-30 m (d) 5.25 x 10-28 m (a) 6.023 x 1023 (b) _1_ X lerl
9.108
211. The Bohr orbit radius for the hydrogen atom (n = 1) is
approximately 0.530 A. The radius for the first excited state (c) 6.023 x 10S"4 (d) I x 108
9.108 9.108 x 6.023
(n = 2) orbits is: (CBSE 1998) 7
(a) 0.13 A (b) 1.06 A (c) 4.77 A (d) 2.12 A 220. If the electronic configuration of nitrogen had ls , it would
have energy lower than that of the normal ground state
212. The number of nodal planes in px.orbital is: (lIT 2000)
configuration ls22i2p 3 because the electrons would be
(a) one . (b) two (c) three (d) zero
closer to the nucleus. Yet ls7 is not observed because it
213. The angular momentum (L) of an electron in a Bohr orbit is violates: (liT 2002)
given as: (lIT 1997)

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I 133
(a) Heisenberg uncertainty principle 230. In ground state, the radius of hydrogen atom is 0.53 A. The
(b) Hund's rule radius ofLi 2 + ion (Z =3) in the same state is: .
(c) Pauli's exclusion principle (PET (Raj.) 2007]
(d) Bohr postulates of stationary orbits (a) 0.17 A (b) 1.06 A (c) 0.53 A (d) 0.265 A .
221. The orbital angular momentum of an electron in 2s-orbital is: 231. How many d-electrons in Cu + (At. No. = 29) can have the spin
(lIT 1996; AIEEE 2003; PMT (MP) 2004) quantum number (- ~ f! (SCRA 2007)
I h (a) 3 (b) 7 (c) 5 (d) 9
(a)+-- (b) zero 232. Which of the following electronic configurations, an atom has
22'1t
the lowest ionisation enthalpy? [CBSE (Med.) 2007)
h
(c)- (d)Ji !!.... (a) li2s 2 2p 3 (b) li2s'22p6 3s1
2'1t 2'1t
(c) ti2s 2 2p 6 (d) li2s 2 2p s
222. Calculate the wavelength (in nanometre) associated with a
proton moving at I x I (fl m sec -I. 233. The measurement of the electron position is associated with an
uncertainty in momentum, which is equal to 1 x 10- 18 g cm s - I.
(mass of proton = 1. 67x 10-27 kg, h::::; 6.63 x 10-34 J sec)
The uncertainty in electron velocity is: (mass ofan electron is
(AIEEE 2009)
9 x 10- 28 g) ICBSE-PMT (Pre.) 2008]
(a) 0.032 nm (b) 0.40 nm
(a) Ix lOS ems- I (b) Ix lotI cms- I
(c) 2.5 nm (d) 14 nm
(c)lxI09 cms-' . (d)lxI0 6 cms- 1
-[1IiDt ~A=~,/~ 16~'~~~"~~:nj3~ 234. The ionization enthalpy of hydrogen atom is L312 x 1 0 ! J - - -
mol-I. The energy required to excite the electron in the atom
::::; 0.397x 10-9 m 0.4 nm] from n = 1 to n = 2 is : (AIEEE 2008)
223. The value of Planck's constant is 6.63 x 10-34 J-s. The (a) 9.84 x lOS J mol- I (b) 8.51 x lOS J mor l
velocity of light is 3 x I rf mI sec. Which value is closest to the (c) 6.56 x lOS J mol-I (d) 7.56 x lOS J mol- I
wavelength in nanometer of a quantum of light with frequency [Hint: El ::::; - I.312 X 106 J mol-'
of8x lotS sec-I? . (CBSE(PMT) 2003] 6
E _EI _ 1.312xlO J 1- 1 ...
18 2- 2 - - 4 mo
(a) 5 x 10- (b) 4 x 10' 2
(c) 3 X 107 (d) 2 x 10-25 t.E= (E2 - EI)::: 1.312 X 10 (l-~) 6

224. Which of the following statements in relation to the hydrogen

atom is correct? (AIEEE 2005) =~ x 1.312 X 106 =9.84 x 105 J mol-']
4
(a) 3s-orbital is lower in energy than 3p-orbital
235. The wavelengths of electron waves in two orbits is 3.: 5. The
(b) 3p-orbital is lower in energy than 3d~orbital
ratio of kinetic energy of electrons will be: (EAMCET 2009)
(c) 3s--and 3p -orbitals are oflower energy than 3d-orbital
(a) 25: 9 (b) 5 : 3
(d) 3s, 3p -and 3d-orbitals all have the same energy
(c) 9:25 (d) 3:5
225. The number of d-electrons in Ni (At. No. ::: 28) is equal to that
of the: (CPMT (UP) 2004] [Hint: We know, A = ~
'I2Em
(a) sand p-electrons in F-
~:::~
(b) p-electrons in Ar (At. No. = 18)
(c) d-electrons in Ni 2 + 1.2 vE;
(d) total number of electrons in N (At. No. = 1) ~ ~
226. The number of radial nodes of 3s- and 2p-orbitals are 5 VE,~
respectively: [lIT (Screening) 2005) .. E,:E2 =25:9]
(a) 2,0 (b) 0, 2 (c) 1,2 (d) 2, I 236. Electrons with a kinetic energy of 6. 023 x 104 l/mol are
227. Which of the following is not permissible? (DCE 2005) evolved from the surface of a metal, when it is exposed to
(a) n = 4, 1= 3, m = 0 (b) n = 4, I = 2, m = 1 radiation of wavelength of 600 nm. The minimum amount of
=
(c) n 4, I = 4, m 1 = (d) n = 4, I = 0, m 0 = energy required to remove an electron from the metal atom is :
228. According to Bohr theory, the angular momentum of electron d ,. . •• (E~.l~fCE1:20f)1!)
in 5th orbit is: (AIEEE 2006) (a)· 2.3125x 10:Ci9
l ~ (b) 3~.10-19}'i , - ' . ~ ~ , ...
h h h h
(a) 25- (b) 1- (c) 10- (d) 2.5- (c) 6.02x 10- 19 J (d) 6.62x 10-34 J
'It 'It 'It 'It
229. Which of the following sets of quantum numbers represents [Hint: Absorbed energy = Threshold energy + kinetic energy
the highest energy of an atom? (AIEEE 2007) of photoelectron
(a) n = 3, I = 0, m ::::; 0, S + ~ = he
-r=Eo+KE
(b) n = 3, I = I, m = I, s = + ~
8 4
(c) n = 3, I = 2, m = I, s = + ~ 6.62x 10-34 X 3 x 10 E. 6.023 x 10 JI
(d) n = 4, I::: 0, m = 0, s = + ~ 600 x 10-9 = 0 + 6.023 X atom 1023

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134 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

3.31x.1O- 19 ";'Eo+lxI0- 19 2. Which IS the correct relationship?

Eo = 231xlO- J] 19 (a) EI of H 112 E2 of He+= 1/3E30f U 2 +== 1I4E4 of Be 3+

237. : :::::['nl~~ili]:~alue n n:::;~:;::~:~ of

l
and
(b) El (H) == E2 (He + ) E3 (Li 2 +) ,E4(Be 3+ )
(c) EI(H) = 2E2 (He+) == 3E3(Li2+) == 4E4(Be 3+)
(d) No relation
nl n2

(a) nl I, n2 = 2, 3, 4 .. .
3. Which is correct for any kind of species?
(b) n 1 == 2, n2 =3,4,5 .. . (a) (E2 - EI ) > (E3 - E 2 ) > (E4 - E 3 )
(b) (E2 -E,)«E3 E 2 )«E4 E 3 )
(c) nl == 3,n2 4, S, 6 .. .
(d) nl == 4, n2 5,6,7... (c) (E2 -E,)=(E3 E 2 )=(E4 E 3 )
138. Ionization energy of He+is 19.6 x 10- 18 J atom-I. The energy (d) (E2 - El ) = 1/4(E3 - E2 ) 1/9(E4 - )
oftne first stationary state (n == I) of-Li 2+ is: (AIEEE 2010) 4. No. of visible lines when an electron returns from 5th orbit to
:.../ (a) -22x 10-15 J atom- l (b) 8:82 x 10-17 J atom-I ground state in H spectrum is:
(a) 5 (b) 4 (c) 3 (d) 10
(c) 4.4lxlO-16 J at9m-' (d) -4.41 x 10- 17 J atom-'
S. Quantum numbers I 2 and m 0 represent which orbital?
Z2
[Hint: =-1 (a) dxy (b) dxz _ y2 (c) d z2 (d) d zx
lLi&!- zi 6. If n and I are principal and azimuthal quantum numbers
respectively, then the expression fo,{ calculating the total-
lLi&!- . numbers of electrons in allY energy level is:
I 11 I n-1
9 2(21 + I)
f ,&!-= - x19.6x 10- 18 (a) L 2(21 + 1) (b) L
LI 4 1 0 1= 1
I=n+' I=n-I
= 44.1 X 10- 18
(c) L 2(21 + 1) (d) L 2(2/ + 1)
=4.41 X 10-'7 J atom- 1
1=0 1=0
E ,&!- = -4.41 x 10- 17 J atom-I] 7. Order of no. ofrevolutionlsec YI' Y2, Y3 and Y4 forI, II, III and
@LI

139. The energy required to break one mole ofCI-C1 bonds in Cl 2

IV orbits is:
is 242 kJ mor'. The longest wavelength of light capable of (a)Y'>Y2>Y3>Y4 (b)Y4>Y3>Y2>Y,
breaking single CI-Cl bond is: (AIEEE 2010) (c) Y1 > Y2 > Y4 > Y3 (d) Y2 > Y3 > Y4 > Y1
(c == 3 X 108 m sec-I, N A 6.023 x Hr3 mol-') 8. Consider the following statements:
(a) 700nrn (b)494nrn (c)594nm (d)640nm (A) Electron density in the xy-plane in 3d 2 Z orbital is zero
x - y
rHO t 242 (B) Electron density in the xy-plane in 3d 2 orbital is zero
l-_Ift : Bond energy of single bond = 23 z
6.023 x 10 (C) 2s-orbital has one nodal surface
=4.017xI0-22 kJ
(D) For 2pz-orbital yz is the nodal plane,
=4.017xlO- 19 J
Which are the correct statements?
E= he (a) (A) :-nd (C) (b) (B) and (t)
A
(c) Only (B) (d) (A), (B), (e) and (D)
6.626 x 10-34 x 3 X 108
4.017 X 10- 19 9. The first emission line in the H-atom spectrum in the Balmer
A series appears at:
A = 4.94 X 10-7 m = 494 run1
240. In Sommerfeld's modification of Bohr's theory, the trajectory (a) 5R cm- 1 (b) 3R cm- 1 (c) 7R cm- I (d) 9R m1
of an electron in a hydrogen atom is: (JEE (WB) 20101 36 4 144 400
(a) perfect ellipse
10. I BM is equal to:
(b) a closed ellipse like curve, narrower at the perihelion (a) ~4 (b) ~ (c) ihe ., (d) eke
position and flatter at the aphelion position • m1te 41tm ,4m 1tm
(c) a closed loop on spherical surface 11. Radial probability distribution curve
(d) a rosette is shown for s-orbital. The curve is:
(a) Is
Set-2: The questions given below may have more
(b) 2s
than one corTeCI answers
(c) 3s
I. Correct order of radius of the 1st orbit of H, He+, Li 2 + and (d) 4s
Be 3+ i s : ' Time-
12. dz 2 orbital has:
(a) H > He+ > Li 2+ > Be3 + (a) a lobe along z-axis and a ring along xy-plane
(b) Be3+ > Li 2+ > He+ > H (b) a lobe along z-axis and a lobe along xy-plane
(c) He+ > Be 3+ > Li 2+ > H (c) a lobe along z-axis and a ring along yz -plane
(d) a lobe and ring along z-axis
(d) He+ > H > Li 2+ > Be3 +

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ATOMIC STRUCTURE 135

13. When a light of frequency vI is incident ona metal surface the 20. CQnsider the followip.g sets, of quantum numbers:
photoelectrons emitted have twice the kinetic energy as did the ',; "n ' '. ','/;T; :. '. , . • m s
photoelectron emitted when the same metal has irradiated with (A) 3 0 0 + Yz
light of frequency V2' What will be the value of threshold (B) 2 2 + Yz
frequency?
(C) 4 3":'2 - Yz
(a) Vo = VI v2 (b) Vo VI 2V2
(d) Vo = VI + v 2
(D) I 0 -1 - Yz
(c) Vo = 2v, - v 2
14.. Heisenberg's uncertainty principle is not valid for:
(E) 3 2 3 +Yz
Which of the following sets of quantum numbers is not
(a) moving electrons (b) motor car possible? [eBSE (Med.) 2007]
(c) stationary particles (d) all of these (a) (A), (B), (C) and (D) (b)(B), (D) and (E)
15. Consider these electronic configurations for neutral atoms; (c) (A) and (C) (d) (B), (C) and (D)
(i) li2s2 2p 6 3s1 (1i) li2s2 2p 6 4sl 21. For three different metals A, B, C photo-emission is observed
Which of the following statements is/are false? one by one. The graph of maximum kinetic energy versus
(a) Energy is required to change (i) to (il) frequency of incident radiation are sketched as :
(b) (i) represents 'Na' atom (BHU (Screening) 2010)

~~ KL~-."-.
(c) (i) and (ii) represent different elements
(d) More energy is required to remove one electron from (i) than
(ii).
16. For the energy levels in an atom which one of the following
statements is/are correct?
(a) There are seven principal electron energy levels v- v-
(b) The second principal energy level can have 4 subenergy (a) (b)
levels and contain a maximum of 8 electrons
(c) The M energy level can have a maximum of 32 electrons
(d) The 4s subenergy level is at a lower energy than the 3d
subenergy level
17. Which of the following statements are correct for an electron
that has n = 4 and m :::: -:- 2 ?
(a) The electron may be in ad-orbital v- v-
. (c) (d)
(b) The electron is in the fourth principal electronic shell
(c) The electron may be in a p-orbital 22. For which of the following species, the expression for the
(d) The electron must have the spin quantum number = + 1/2
18. The angular momentum of electron can have the value(s): energy of electron in the nth [En:::: 13.6 Z2 J
atom-I has

(a) h (b) h
the validity? [SHU (Mains) 2010)
2n n
(a) Tritium (b) Li 2 +
(c) 2h (d)~~
n 22n (c) Deuterium (d) He 2+
19. Which ofthe following statements is/are wrong?
(a) If the value of 1= 0, the electron disttibution is spherical
(b) The shape of the orbital is given by magnetic quantum no.
(c) Angular moment of Is, 2s, 3s electrons are equal
(d) In an atom, all electrons travel with the same velocity

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136 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Assertion-Reason TYPE QUESTIONS

Set-1 S.t-2
The questions given below consist of an 'Assertioa' (A) and
the 'ReasGII' (R). Use the following keys for the appropriate The questions given below consist of two statements as 'Assertion'
answer: (A) and 'Reasoa" (R); while answering these choose anyone of
them:
(a) If both (A) and (R) are correct and (R) is the correct reason for
(a) If (A) and (R) are both correct and (R) is the correct reason for
(A).
(A).
(b) If both (A) and (R) are correct but (R) is not the correct
(b) If (A) and (R) are both correct but (R) is not the correct reason
explanation for (A).
for (A).
(c) If (A) is correct but (R) is incorrect.
(c) If (A) is true but (R) is false.
{d} If(A) is incorrect but (R) is correct.
(d) Ifboth (A) and (R) are false.
I. (A) F-atom has less electron affinity than cr atom. 11. (A) A special line will be seen for 2px - 2py transition.
(R) Additional electrons are repelled more effectively by 3 p (R) Energy is released in the form of wave of light when the
electrons in CI atom than by 2p electrons in F-atom. electron dropsfrom2px to -2py orbital. (AHMS 1996}--
2. (A) Nuclide rgAI is less stable than ~Ca. U. (A) Ionization potential of Be (At. No. = 4) is less than B (At.
No. =5).
(R) Nuclide having odd number of protons and neutrons are
generally unstable. . (lIT 1998) (R) The first electron released from Be is of p-orbital but that
from B is of s-orbital. (AIlMS 1997)
3. (A) The first IE of Be is greater than that of B.
13. (A) In Rutherford's gold foil experiment, very few a-particles
(R) 2p-Olbital is lower in energy than 2s.
are deflected back.
4. (A) The electronic configuration of nitrogen atom is
(R) Nucleus present inside the atom is heavy.
represented as:

If rt Iii
14. (A) Limiting line in the Balmer series has a wavelength of
DO
and not as:'
[J[] 364.4mm.
(R) Limiting line is obtained for a jump of electron from
n = 00.
[U 15. (A) Each electron in an atom has two spin quantum numbers.
(R) Spin quantum numbers are obtained by solving
(R) The electronic configuration of the ground state of an Schrodinger wave equation.
atom is the one which has the greatest mUltiplicity. 16. (A) There are two spherical nodes in 3s-orbital.
5. (A) The atomic radii of the elements of oxygen family are (R) There is no planar node in 3s-orbital.
smaller than the atomic radii of corresponding elements of 17. (A) In an atom, the velocity of electron in the higher orbits
the nitrogen family. keeps on decreasing.
(R) The members of oxygen family are all more (It) Velocity of electrons is inversely proportional to radius of
electronegative and thus have lower value of nuclear the orbit.
charge than those of the nitrogen family. II. (A) If the potential difference applied to an electron is made 4
6. (A) For n = 3, I may be 0, I and 2 and may be 0, ±l and 0, ±l times, the de Broglie wavelength associated is halved.
and±2. (R) On. making potential difference 4 times, velocity is
(R) For each value of n, there are 0 to (n - 1) possible values doubled and hence d is halved.
of I; for each value of I, there are 0 to ±l values of m. 19. (A) Angular momentum of Is, 2s, 3s, etc., all have spherical
7. (A) An orbital cannot have more than two electrons. shape.
(R) The Iwo electrons in an orbital create opposite magnetic (R) Is, 2s, 3s,etc., all have spherical shape.
field. 10. (A) The radial probability of Is electron first increases, till it is
I. (A) The configuration ofB-atom cannot be 1s2 2s2 . maximum at 53 A and then decreases to zero.
(R) Hund's rule demands that the configuration should disph~! (R) Bohr radius for the first orbit is 53 A.
maximum multiplicity. 11. (A) On increasing the intensity of incident radiation, the
9. (A) The ionization energy ofN is more than that ofO. number of photoelectrons ejected and their KE increases.
(R) Electronic configuration of N is more stable due to half- (R) Greater the intensity means greater the energy which in
filled 2p-orbitals. tum means greater the frequency of the radiation.
18. (A) p-orbital is dumb-bell shaped. ll. (A) A spectral line will be seen for a 2px 2py transition.
(R) Electron present in p.:orbital can have anyone of the three (R) Energy is released in the form of wave of light when the
values of magnetic quantum number, i.e., 0, +1 or-I. electron drops from 2px to 2py orbital. (VMMC 2007)

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ATOMIC STRUCTURE I 137

• Set-1
1. (b) 2. (a) 3. (c) 4. (d) 5. (c) 6. (b) 7. (d) 8. (a)
9. (a) 10. (b) 11. (d) 12. (d) 13. (a) 14. (c) 15. (b) 16. (a)
17. (c) 18. (d) 19. (b) 20. (c) 21. (d) 22. (c) 23. (b) 24. (a)
25. (d) 26. (b) 27. (d) 28. (a) 29. (d) 30. (b) 31. (c) 32. (b)
·33. (a) 34. (c) 35. (d) 36. (a) 37. (c) 38. (a) 39. (b) 40. (c)
41. (c) 42. (b) 43. (d) 44. (d) 45. (c) 46. (a) 47. (b) 48. (d)
49. (d) SO. (c) 51. (b)· 52. (b) 53. (a) 54. (b) 55. (a) 56. (a)
57. (c) 58. (d) 59. (a) 60. (d) 61. (c) 62. (a) 63. (a) 64. (b)
65. (a) 66. (d) 67. (d) 68. (b) 69. (c) 70. (c) 71. (a) 72. (d)
73. (d) 74. (b) 75. (c) 76. (d) 77. (b) . 78. (d) 79. (a) 80. (d)
81. (b) 82. (b) 83. (c) 84. (a) 85. (b) 86. (a) 87. (b) 88. (c)
89. (b) ...94)~J4!L. !1.. _(at .. !2.Jd), 93. (a) 94. (a) 95. (b) 96. (c)
97. (a) 98. (d) 99. (b) 100. (b) 101. (d) 102. (d) 103. (a) 104. (c)
105. (c) 106. (c) 107. (d) 108. (b) 109. (d) 110. (a) 111. (a) ~ 112. (b)
113. (c) 114. (d) 115. (a) 116. (d) 117. (b) 118. (a) 119. (c) 120. (c)
121. (d) 122. (b) 123. (d) 124. (c) 125. (a) 126. (a) 127. (e) 128. (e)
129. (b) 130. (b) 131. (c) 132. (c) 133. (c) 134. (b) 135. (c) 136. (d)
137. (b) 138. (a) 139. (a) 140. (a) 141. (b) 142. (c) 143. (c) 144. (a)
145. (b) 146. (c) 147. (d) 148. (c) 149. (d) 150. (d) 151. (a) 152. (d)
153. (c) 154. (a) 155. (c) 156. (a) 157. (a) 158. (c) 159. (c) 160. (d)
161. (c) 162. (c) 163. (b) 164. (b) 165. (a) 166. (b) 167. (b) 168. (b)
169. (d) 170. (d) 171. (b) 172. (b) 173. (d) 174. (b) 175. (c) 176. (a)
177. (d) 178. (b) 179. (d) 180. (a) 181. (d) 182. (a) 183. (a, b) 184. (c)
185. (a) 186. (a) 187. (b) 188. (b) 189. (b) 190. (b) 191. (a) 192. (c)
193. (c) 194. (b) 195. (a) 196. (a) 197. (d) 198. (a) 199. (c) 200. (a)
201. (a) 202. (c) 203. (c) 204. (a) 2OS. (a) 206. (b) 207. (c) 208. (c)
209. (d) 210. (c) 211. (d) 212. (a) 213. (a) 214. (e) 215. (a) . 216. (b)
217. (d) 218. (d) 219. (d) 220. (c) 221. (b) 222. (b) 223. (b) 224. (d)
225. (c) 226. (a) 227. (c) 228. (d) 229. (c) 230. (a) 231. (c) 232. (b)
233. (c) 234. (a) 235. (a) 236. (a) 237. (c) 238. (d) 239. (b) 240. (c)
• Set-2
1. (a) 2. (b) 3. (a) 4. (c) 5. (c) 6. (d) 7. (a) 8. (a)
9. (a) 10. (a) 11. (a) 12. (a) 13. (e) 14. (b, c) 15. (c, d) 16. (a, d)
17. (b, c) 18. (a, b, c) 19. (c) 20. (b) 21. (d) 22. (a, b, c)

1. (c) 2. (a) 3. (c) 4. (a) 5. (c) 6. (a) 7. (b) 8. (c)

9. (c) 10. (a) n. (d) 12. (d) 13. (b) 14. (a) 15. (d) 16. (b)
17. (c) 18. (a) 19. (b) 20. (b) 21. (d) 22. (d)

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138 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

r"~~;~7"'.i;l:lmii~ __it·]O$'i~GJu;t·]:j·4$~ftI~~~~~
IOBJECTIVE QUESTIONS] for ( lIT ASPIRANTS' .~ .I
The following questions contain a single correct option:
4. Maximum value (n + I + m) for unpaired electrons in second
1. The configuration of Cr atom is 3d 54s1 hut not 3d 44s 2 due to excited state of chlorine l7Ci is:
reason RI and the configuration of Cu atom is 3d 10 4S1 hut not (a) 28 (b) 25
3d 94s 2 due to reason R 2 , RI and R2 are: (c) 20 (d) none of these
(a) R 1: The exchange energy of 3d 54s1 is greater than that of [Hint: Configuration in second excited state may be given as:
3d 44s 2 •
R 2 : The exchange energy of 3d 10 4s1 is greater than that
of3d 94s 2 •
(h) R1: 3d 54s1 and 3d 44s 2 have same exchange energy but
3d 54s1 is spherically symmetrical.
n 3 3 3 3
2 3]
2 . Total n + 1+ m = 25]
R 2 : 3d 104s1 is also spherically symmetricaL m -I o +1 +2 +1
( c) R 1 : 3d 5 4s 1 has greater exchange energy than 3d 4 4i.
5. Which of the following is correctly matched?
R 2: 3d 104s1 has spherical symmetry.
(a) Momentum ofH-atom when electrons
(d) R 1 : 3d 54s1 has greater energy than 3d 4 4i.
R 2: 3d 104s1 has greater energy than 3d 94i·. return from n := 2 to n I 3Rh
4
[Hint: 3d 54s I is correct because it has greater exchange (b) Momentum of photon Independent of wavelength
possibilities of unpaired electrons. of light
1-2, 1-3, 1-4, 1-5
11 I 111 11 11 I :~~~~~lf~es:
(10)
2-3,2-4,2-5
3-4,3-5
(c) elm ratio of anode rays Independent of gas in the
discharge tube
2 3 4 5 (d) Radius of nucleus (Mass no.)1/2
4-5
[Hint: 1 = R [~ ~J := 3R
3d I04s 1 is correct because 3d to-orbitals are spherically A 12 22 4
symmetrical. ] h
A=-
2. Which of the following graphs is incorrect? p
p h = h x 3R 3Rh]
A 4 4
t
v
6. In hydrogen spectrum, the third line from the red end
corresponds to which one of the following inter-orbit jumps of
I the electrons from Bohr orbit of hydrogen?
(a) 4 ~ I (b) 2 ~ 5 (c) 3 ~ 2 (d) 5 ~ 2
7. In which of the following pairs is the probability of finding the
electron in xy-plane zero for both orbitals?
(a) (b)
(a) 3d yz , 4dx 2 - y 2 (b) 2Pz, d z 2
(c) 4dzx',3pz (d) None of these
tv tv 8. In which of the following orbital diagrams are both Pauli's
exclusion principle and Hund's rule violated?
I I (a)@] I I II J, I I (b)
r----,---.-:----.
@] I t It J, II
z ----+ n--+
(c) (d) (C)@] IH Itt I J, I (d) @] IHI II I
3. Which among the following is correct of 5B in normal state?
9. The distance between 3rd and 2nd Bohr orbits of hydrogen
2s 2p atom is:
(a) 0.529 x 10-8 cm (b) 2.645 x 10-8 cm
(a) [ill I I I Against Hund's rule
(c) 2.1l6xI0-8 cm (d) 1.058xlO-8 cm
2
[Hint: 1j r2 = (3 - 22) x 0.529 X 10-8 cm]
. (b)[i] L......!--'--'--'---' Against aufbau principle
as well as Hund's rule 10. Which diagram represents the best appearance of the line
(c) [ill '---'--'---'-----" Violation ofPau!i's exclusion spectrum of atomic hydrogen in the visible region?
principle and not Hund's rule (PET (Kerala) 2007)
(d) [ill I'----!...--"---,
L-I
Against aufbau principle

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I
ATOMIC STRUCTURE 139
Increasing wavelength 17. A photosensitive material would emit electrons if excited by
photons beyond a threshold. To overcome the threshold, one
(a) would increase: (VITEEE 2007)
(a) the voltage applied to the light source
(b) the intensity of light
(b)
(c) the wavelength of light
(d) the frequency oflight .
(c) 18. Which of the following electronic configurations have the
highest exchange energy?
3d 4s
(d)
1 I I (a) I iii i I IITJ
(b) ITJ
(e)
11111 II II III (c) i
11. The 'm' value for an electron in an atom is equal to the number
of m values for 1= I. The electron may be present in: (d) IT]
~a) 3dx2 _ y2 . (b) 5~(X2_y2) 19. Which of the following graphs correspond to one node?
(c) 4/3
x Iz
(d) noneofthese
[Hint: Total values of m (21 + I) = 3 for 1 = 1 t
'If
m 3 is for f-subshellorbitals.]
12. If m = magnetic quantum number, I = azimuthal quantum
number, then:
(a) m=I+2 (b) m 2P + 1 -lao -lao
m-I (a) (b)
(c) 1= - (d) 1= 2m+ 1
2
[Hint: Magnetic quantum number 'm' lies between (-I, 0, + I);
thus total possible values of 'm' wiH be (21 + I). t t
m -I 'If 'If
m 21+1, i.e.,. 1= 2 ]
-lao
13. What are the values of the orbital angular momentum of an
electron in the orbitals Is, 38, 3d and 2p?
(0) (d)
(a) 0,0, .J61i, J2 Ii· (b) I, I, J4 Ii, J2 n
(c) 0, 1,.J6 Ii, J3 Ii (d) 0,0,50 1i,.J61i 20. Angular distribution functions of all orbitals have:
(a) I nodal surfaces (b) (l-l)nodal surfaces
[Hint: Orbital angular momentum = ~l(l + 1) h = ~1(1 + 1)il] (c) (n + I) nodal surfaces (d) (n 1- I)nodal surfaces
21£
21. If uncertainty in position and momentum are equal then
14. After np-orbitals are filled, the next orbital filled will be:
uncertainty in velocity is: [CBSE.PMT (Pre.) 2008]
(a) (n + l)s (b) (n + 2)p (c) (n + l)d (d) (n + 2)s
IS. The ratio of (E2 - Ej ) to (E4 E3 ) for the hydrogen atom is (a) fI
V;:
(b) fh
V"2n
(c) _1
2m
fI
V;:·
(d).!
m
fI
V;:
approximately equal to:
(a) ·10 (b) 15 (c) 17 (d) 12 h
[Hint: tJ,x·Ap;;:-
[Hint: 41£

(-~)-(-n 1 1
"9 - 16 _ 7 4
----x
7 1
Ap~ H.: when tJ,x = IIp

108 '" 15 ]
(-±) 3 144 3 mAv;;: rT
-(-I) 4 f41t
16. Which ofthe following electronic configurations hal1 zero spin Av;;:_I_
2m
fE]
V;:
multiplicity?
22. The number of waves made by a Bohr electron in an orbit of
(a) It Ii maximum magnetic quantum number 3 is:
(a) 3 (b) 4 (c) 2 (d) 1
(c) I i I J, [Hint: m == 3, I = 3, n = 4
[Hint: Spin multiplicity =(2u + 1)] For, n = 4, number of waves will be 4.]

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140 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

23. The number of elliptical orbits excluding circular orbits in the h -.

mv9x=3-
N·shell ofan atom is: 21t
003 004 002 WI h = 61tx
[Hint: For, N·shelI, n = 4. This shell will have one Circular and mv
three elliptical orbits.] A.=6ru:]
24. From the electronic configuration of the given elements K, L, 30. How many times does light travel faster in vacuum than an
M and N, which one has the highest ionization potential? electron in Bohr first orbit of hydrogen atom?
(a) M = [Ne] 3i3p2 (b) L=[Ne] 3s 23p 3 (a) 13.7 times (b) 67 times (c) 137 times (d) 97 times
(c) K=[Ne]3i3/ (d) N=[Ar]3d 1O
,4i4p 3 [Hint: v = ~ x 2.188 X 108 cmlsec
n
[Hint: L has half·filled p·subshell and it is smaller than N,
hence, L wiII have the highest ionization potential.] VI = ~ x 2.188 X 108 cmlsec
I
25. Which of the following pairs of electrons is excluded from an
Velocityoflight 3xHYo
atom? ---~---'=--- = 2.188 = 137'
tImes
]
I 1 Velocity of electron x 108
(a) n = 2, 1= 0, m = 0, s + - and n = 2, 1= 0, m = 0, s = + .;...
2 2 31. A compound of vanadium has a magnetic moment of 1.73 BM.
I . The electronic configuration of vanadium ion in the
(b) n = 2, I = I, m = + I, s = +- compound is:
-------- - --- . 2 -r'- (a) [Ar]3d 2 (b) [Ar]3d 14s0 (c)[Ar]3d 3{d) [Ar]3d °4sl-
and n ::: 2, I = 1, m =- 1, s = + -
2 [Hint: Magnetic moment = In(n + 2) BM
1 I
(c) n = 1, I::: 0, m:::.0' s::: + 2 and n = 1, I = 0, m = 0's:::: - 173 = In(n + 2)
2
I
Jj::: In(n + 2)
(d) n =3, I =2, m - 2, s = + - n:::: I (number of unpaired electrons)
2
V 22 ~3d 24i
I
andn 31=Om=Os=+
, , , 2 V 3+ -t 3d '4i (has one unpaired electron)]
[Hint: Both 2s electrons have same spin, hence excluded from 32. The orbital angular momentum of an electron in p·orbital is:
the atom.] [PET (Kerala) '2006]
(d) .!.~
26. Given set of quantum numbers for a multielectron atom is: (b) _h_ h
(a) zero (c) -
n I m s .fi1t 21t 221t
2 0 0 + 112
2 0 0 -1/2 (e) ~
2-v 21t
What is the next higher allowed set of 'n' and'/' quantum 33. When a hydrogen atom emits a photon of energy 12.1 e V, the
numbers for this atom in the ground state? orbital angular momentum changes by:
(a) n = 2, I = 0 (b) n = 2, I = I (a) L05 x 10-34 J sec (b) 211 x 10-34 J sec
(c) n=3,1=0' (d) n=3,I=I
27. In how many elements does the last electron have the quantum (c) 3.16 x 10-34 J sec (d) 4.22 x 10-34 J sec
numbers of n 4 and I = I ? [Hint: Emission of photon of 12.1 eV corresponds to the
(a) 4 (b) 6 (c) 8 (d) 10 transition from n =- 3 to n = L
[Hint: n::: 4, I::: I represent 4p·subshell containing six :. Change in angular momentum
electrons. Thus, there will be six elements having 4pl to 4l h
=-(~ -nl)-
electronic configuration.]
21t
28. If there are three possible values (-112, 0, + 112) for the spin =(3 -l)~=!:
quantum, then electronic configuration ofK (19)will be: 21t 1t
6.626 X 10-34
(a) Is3 , 2s3 2p 9, 3s3 3pl (b) Is2 , U2 p 6, 3i3p 6, 4S1 =
3.14
(c) Ii, U2 p 9, 3i 3p4 (d) none of these
= 2.11 x 10-34 J sec]
29. lethe radius offirst Bohr orbit of hydrogen atom is 'x' then de
34. The total energy of the electron of H·atom in the second
Broglie wavelength of electron in 3rd orbit is nearly:
quantum state is - E2 • The total energy of the He + atom in the
(d) ~ third quantum state is:
(a) 2m; (b) 61tx (c) 9x
[Hint: r" =n 2rJ 3
(a) -(~)E2 (b) G)E2 (c) (~)E2 (d) -C:)E2
13::: 91j = 9x
h [Hint: Energy of electrons in n th state
mvr=n Z2
21t =- -2 X 13.6eV
n

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ATOMIC STRUCTURE 141

E2(H) = - 13.6 eV (a) 1..= 1.23 (b) 1..= 1.23m (c) 1.23 nm
(d) 1..= 1.23
1 .r;;, Jh .JV V
E 3 (He+ ) -
- 13.6 x 4 eV 40. An excited electron ofH-atoms emits of photon of wavelength
-- -9-
A and returns in the ground state, the principal quantum
number of excited state is given by :

For negative value of E2 , E3 will also be negative.] (a) ~AR (AR - I) (b) ~ (A~~ I) .
35. What is the ratio of the Rydberg constant for helium to·
hydrogen atom? (c) I (d) ~(AR - I)
(a) 1/2 (b) lI4 (c) lI8 (d) 1/l6 ~AR (AR -I) AR

[Hint:
-2n 2mZ 2e4
R = --,,-----
ch 3 [Hint: ±=R[~~ - ~] =RU- ~J
RHe = 2 X 22 = 8
RH Ixe ~=~ A.RA.R-I ]
RH =.!.] 41. A dye a!>sorbs a photon of wavelength A and re-emits the same
RHe 8 energy into two photons of wavelength AI and 1..2 respectively.
36. If the kinetic energy ofa particle is doubled, de Broglie The wavelength A is related to AI and A2 as : _____ _
wavelength becomes: (a) 1..= 1..11..2 (b) 1..= AI + 1..2
(a) 2 times (b) 4 times (c) .J2 times (d) _1_ times (AI + A2 i 1..11..2
.J2 (c) 1..= 1..11..2 (d) A~ A~
[Hint: 1..= _h_ , where, E = Kinetic energy of the particle AI + A2 AI + 1..2
.J2Em 42. The radii of maximum probability for 3s, 3p and 3d-electrons
h h
AI = - - ; 1..2 = ---;::::==- are in the order :
.J2Em J2x2Em (a) (rmax) 3$ > (rmax) 3p > (rmax )3d
Al =..fi i.e. A =~] (b) (rmax) 3$ = (rmax) 3p =(rmax )3d
1..2 ' , 2 ..fi
(c) (rmax) 3d > (rmax) 3p > (rmax )3$
37. Imagine an atom made up of a proton and a hypothetical
particle of double the mass of the electron but having the same (d)(rmax) 3d> (rmax) 3$ > (rmax )3p
charge as the electron. Apply the Bohr:s atomic model and Following questions may have more than one correct options:
consider all possible transitions of this hypothetical particle to 1. Select the correct relations on the basis of Bohr theory:
the fIrst excited level. The largest wavelength photon that will
be emitted has wavelength A (given in terms of the Rydberg (a) velocity of electron oc.!. (b) frequency of revolution oc ~
n n
constant R for the hydrogen atom) equal to: 1
9 . 36 18 (c) radius of orbit oc n 2Z 4
(d) force on electron
(a) 5R (b) 5R (c) 5R (d) i n
oc

R
[Hint: Energy is related to mass: 2. To which of the following species, the Bohr theory is not
applicable?
En oem
The longest wavelength Amax photon will correspond to the (a) He (b) Li 2+ (c) He 2+ (d) H-atom
transition of particle from n = 3 to n = 2 3. The magnitude of spin angular momentum of an electron is
given by: .
A~x =2R(;2~ ;2) (a)S = ~s(s + 1) in (b) S =:s 2:
A
max
=!!]
5R J3 h "1 h
(c) S =-x- (d) S =±-x-
38. What is ratio of time periods (11. I T2 ) in second orbit of 2 2n . 2 2n
hydrogen atom to third orbit ofHe+ ion? [Hint: Spin angular momentum == Js(s + I) ~
(a) -.! (b) 32 (c) 27 (d) 27 2n
27
n3
27 32 8
S=~.!.22(.!. + I) ~
2n
=.J3
22n
x~ ]
[Hint: T 00-
Z2 4. Select the correct confIgurations among the following:
T n
.-1.=_1_
3 x Z2 23 X_
_2 = _ 22 =_]
32 (a) Cr (Z= 24):[Ar] 3d 5,4sl
T2 zf ni
x 12 X 33
27 (b) Cu (Z = 29): [Ar] 3d 10, 4s1
39. The de Broglie wavelength of an electron accelerated by an (c) Pd (Z= 46):[Kr] 4d 1O,5so
electric fIeld of V volt is given by :
(d) Pt (Z = 78): [Xe] 4d 10 4i

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I

142 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

5. Which among the following statements is/are correct? 11. The maximum kinetic energy of photoelectrons is directly
(a) lJI2 represents the atomic orbitals proportional to ... of the incident radiation. The missing word
can be:
(b) The number of peaks in radial distribution is (n - I)
(a) intensity (b) wavelength
(c) Radial probability density Pnl (r) 41t? R;/(r}
(c) wave number (d) frequency
(d) A node is a point in space where the wave function (lJI ) has 12. Rutherford's experiment established that:
zero amplitude
(a) inside the atom there is a heavy positive centre
6. Select the correct statement(s} among the following: (b) nucleus contains protons and neutrons
(i) Total number of orbitals in a shell with principal quantum
(c) most of the space in an atom is empty
number 'n' is n 2
(d) size of nucleus is very small
(ii) Total number of subs hells in the n th energy level is n
13. Which of the following orbital(s) lie in the xy-plane?
(iii) The maximum number of electrons in a subshell is given
by the expression (41 + 2)
(a) d x2 _ y2 (b) dX), (c) dxz (d) Fa
(iv) m = 1+ 2, where 1 and m are azimuthal and magnetic 14. In which ofthe following sets of orbitals, electrons have equal
. quantum numbers .orbital angular momentum?
(a) (i), (iii) and (iv) are correct (a)13and2s (b) 2sand2p (c) 2p and 3p(d) 3pand3d
(b) (0, (ii) and (iii) are correct 15. Which ofthe following orbitals have no spherical nodes?
(c) (ii), (iii) and (iv) are correct (a) 13 (b) 2s (c) 2p (d) 3p
(d) (i), (iiyand (iv) are correct Hi~ For a'snellofpriiiCipalquaniiirri Ilurnber'n~~4, tnere are:~
7. Which among the following are correct about angular (a) 16 orbitals (b) 4 subshells
momentum of electron? (c) 32 electrons (maximum) (d) 4 electrons with 1= 3
17. The isotopes contain the same number of:
(a) 2 n (b) 1.5!: (c) 2.5ft (d) 0.5!:
1t 1t (a) neutrons (b) protons
8. Which of the following is/are incorrect for Humphrey lines of (c) protons + neutrons (d) electrons
hydrogen spectrum? 18. Which of the following species has less number of protons
(a) n2 7 -7 n! = 2 (b) n 2 10 -7 n! = 6 than the number of neutrons?
(c) n2 5 -7 n! = 1 (d) n2 11-7 n! = 3 (a) !~C (b) 19F (c) nNa (d) fiMg
9. In the Bohr's model ofthe atom: 19. The angular part of the wave function depends on the quantum
(a) the radius of nth orbit is proportional to n 2 numbers are:
(b) the total energy ofthe electron in the n th orbit is inversely (a) n (b) I . (c) m (d) s #

proportional to 'n' 20. Which of the following species are expected to have spectrum
(c) the angular momentum of the electron is integral multiple ,similar" to hydrogen? '
of h/21t (a)He+ (b) He 2+ (c)'U 2 + (d)U+
(d) the magnitude of potential energy of an electron in an orbit 21. Which of the following statements is/are correct regarding a
is greater than kinetic energy hydrogen atom?
10. Which among the following series is obtained in both (a) Kinetic energy ofthe electron is maximum in the first orbit
absorption and emission spectrums?
(b) Potential energy of the electron is maximum in the first orbit
(a) Lyman series (b) Balmer series
(c) Radius of the second orbit is four times the radius of the
(c) Paschen series (d) Brackett series first orbit
(d) Various energy. levels are equally spaced

[;/I~ ..-:- - - - - - - - - - - - - - - - - - - - - 1
• Single ,correct option
1. (e) 2. (d) 3. (c) 4. (b) :5~ (a) 6. (d) 7. (d) 8. (d)
9. (b) 10. (c) 11. (b) 12. (e) 13. (a) 14. (a) 15. (b) 16. (c)
17. (d) 18. (b) 19. (b) 20. (a) 21. (c) 22. (b) 23. (a) 24. (b)
25. (a) 26. (b) 27.. (b) 28. (a) 29. (b) 30. (c) 31. (b) 32. (b)
33. (b) 34.. (e) 35. (e) 36. (d) 37. (e) 38. (b) 39. (e) 40. (b)
41. (e) 42. (a)

• One or more than one correct ,options '4.

1. (a, b, d) 2. (a, e) 3. (a, c) (a,b,e) 5. (a, b, c, d) 7. (a, b, d) 8. (a, c, d)
.6. (b)
9. (a, c, d) 10. (a) 11. (e, d) 12. (a, e, d) 13. (a, b) 14. (a, c) 15. (a, e) 16. (a, b, e)
17. (b, d) 18. (b, c) 19. (b, e) . .20. (a, c) 21. (a, c}
.. "

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ATOMIC STRUCTURE 143

Integer Answer TYPE QUESTIONS

This section contains 10 questions. The answer to each 48.24 nm. From the same orbit, wavelength corresponding to a
of the questions is a single digit integer, ranging from transition to third orbit is 142.46 nrn. The value of n2 is :
oto 9. If the correct answers to question numbers X, Y, 9. The energy corresponding to one of the lines in the Paschen
Z and W (say) are 6, 0, 9 and 2 respectively, then the series for H-atom is 18.16 x 10-20 J. Find the quantum numbers
correct darkening of bubbles will look like the figure: for the transition which produce this line.

[Hint: AE= 2. 18XlO-18[~-J..J

nf ~
1. For Li 2+, when an electron falls from a ~gher orbit to nth
orbit, all the three types of lines, i.e., Lyman, Balmer and
2o
18.l6xlO- = 2. 18XlO-
18
[i- :2]
Paschen was found in the spectrum. Here, the value of' n' will On solving, n = 6 ]
be: 10. The angular momentum of electron in the shell in which the
2. The emission lines of hydrogen contains ten lines. The highest . g-subshell first appears is x x ~. The value of x will be :
orbit in which the electron is expected to be found is : 21t
[Hint: Number of lines == n(n -1) = 10 [Hint: I = 4 for g-subshell
.. ..... . .2--- Thus, the subshell will first appear in (n = 1+ 1<= 5) 5th shell.._ -
.• n=5] h
Angular momentum (mur) = n -
3. Total number of nodes present in 4d orbitals will be: 21t
4. Spin mUltiplicity of nitrogen in ground state will be : h
=5-
5. Orbital frequency of electron in nth orbit of hydrogen is twice .2n:
that of 2nd orbit. The value of n is : n= 5]
6. If kinetic energy of an electron is reduce by (1/9) then how
many times its de Broglie wavelength will increase.
7. If electrons in hydrogen sample return from 7th shell to 4th
shell then how many maximum number of lines can be
observed in the spectrum of hydrogen.
8. An electron in Li 2+ ion is in excited state (n2) . The
wavelength corresponding to a transition to second orbit is

[~. ~'
.... ~
....
I
1. (1) 2. (5) .3. (3) 4.. (4) 5. (1) 6. (3) 7. (6) 8. (5)
9~ (6) l(t (5)

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144 I G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

• Passage 1 5. Which of the following is not correctly matched?

The observed wavelengths in the line spectrum of hydrogen atom (a) 6569 A (Red) (b) ~ - 4861 A (Blue)
were first expressed in terms ofa series by Johann Jakob Balmer, a (c) liy- 4340 A (Orange) (d) He - 4101 A (Violet)
Swiss teacher.
s
Balmer empirical formula is: • Passage 2
Aformula analogous to the Rydbergformula applies to the series
i=RH [; - nI2]n=3,4,5, ...
of spectral lines which· arise from transitions from higher energy
level to the lower enetID' level ofhydrogen atom.
R H = 109678 cm-\ is the Rydberg constant.
A muonic hydrogen atom is like a hydrogen atom in which the
Niels Bohr derived this expression theoretically in 1913. The electron is replac~d by a heavier particle, the 'muon '. The mass of
formula is generalised to anyone electron atom/ion. the muon is about 207 times the mass ofan electron, while the charge
Answer the following questions: remains same as that of the electron. Rydberg formula for hydrogen
1. Calculate the longest wavelength in A (l A 10- 10 m) in the atom is:
Balmer series of singly ionized helium He +. Select the correct 1
artswer.-Ignore the nuClear monon iIi your calCulation. A
(a) 2651 A (b) 1641.1 A
(c) 6569 A (d) 3249 A Answer the following questions:

[Hint: _1-
AHe+
=RHZ2 [J.. -J..J
22 3 2
1. Radius of first Bohr orbit of muonic hydrogen atom is:
(a) 0.259 A (b) 0.529 A
207 207
= 109678 x 4 [:6J (c) 0.529 x 207 A (d) 0.259 x 207 A
2. Energy of first Bohr orbit of muonic hydrogen atom is:
AHe+ =1641.1 A]
(a) - 13.6 eV (b) -13.6 x 207eV
2. How many lines in the spectrum will be observed when 207
electrons return from 7th shell to 2nd shell?
(c) + 13.6 eV (d) +13.6 x 207 eV
(a) 13 (b) 14 (c) 15 (d) 16 207
[Hint: Number of lines in the spectrum 3. Ionization energy of muonic hydrogen atom is:
= (1~ - nl )(~ - n) + I) (a) +13.6 eV (b) +13.6 x 207eV
2 207
_ (7-2)(7-2+ 1) IS
(c) - 13.6 ~V (d) -13.6 x 207eV
2 207
7- 4. Angular momentum of 'muon' in muonic hydrogen atom may

ll1~~~
be given as:
6-
S- 15 lines in the spectrum.] (a) h (b) ~
(c) h (d) h
'IE 2n 4'IE 61t
4-
5. Distance between first and third Bohr orbits of muonic
3- hydrogen atom will be:
2- 2
(a) 0.529 x 2A (b) 0.529 x 7A
3. The wavelength of first line of Balmer spectrum of hydrogen 207 . 207
will be: (c) 0.529 x 8A (d) 0.529 A
(a) 4340 A (b) 4101 A (c) 6569 A (d) 4861 A 207 207
[Hint: .!.A = RH [J.. -~J
n2
22 • Passage 3
for fIrst line n = 3,
I [ I I]
-=109678 - - -2
A. 22 3
Nuclei that have 2,8,20,28,50,82 and 126 neutrons or protons
are more abundant and more stable than other nuclei ofsimilar mass.
It is suggested that in the nuclear structure ofthe numbers 2, 8, 20,28,
A= 6569A] 50, 82 and 126, which have become known as magic, nu11ll}(!rs, the
4. In which region of electromagnetic spectrum does the Balmer nuclei possessing magic numbers are spherical and have zero
series lie? quadruple moment and hence they are highly stable. Nuclear shells
(a) UV (b) Visible are filled when there are 2, 8, 20, 28, 50, 82 and 126 neutrons or .
protons in a nucleus. In even-even nuclei all the neutrons and protons
(c) Infrared (d) Far infrared
are paired and cancel out spin and orbital angular momenta.

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ATOMIC STRUCTURE 145

Answer the followhig questions regarding the stability of The only way to explain the results, Rutherford found,. was to picture
nucleus: an atom as being composed of a tiny nucleus in which its positive
1. Which of the following element(s) is/are stable though having charge and nearly all its mass are concentrated. Scattering of
odd number of neutrons and protons? a-particles is proportional to target thickness and is inversely
(a) ~Li (b) I~B (c) iRe (d) IjN . proportional to the fourth power of sin %' where, (-) is :'cattering
2. Stable nuclei having number of neutrons less than number of
angle. Distance ofclosest approach may be calculated as:
protons are:
2
(a) iH (b) ~He Z;Z2e
1:. =-.--
mill 4nEoK
3. Doubly magic nucleus is ...
(a) 2~iPb (b) 2~Pb (c) 2~~Pb (d) 2~;Bi where, K kinetic energy ofa-particles.
Answer the follOwing questions:
4. Which among the following has unstable nucleus?
1. Rutherford's a'particle scattering experiment led to the
(a) IjN (b) l~N (c) I~N (d) 1~0 . con~lusion that: '
5. Which of the following has zero spin and angular momentum? (a) mass and energy are related
(a) ~~Ca (b) iH (c) I~C (d) {jCI (b) mass ,and positive charge of an atom are concentrated in
the nucleus .
(c) neutrons are present in the nucleus
• Passage4
(d) atoms are electrically rieutrill·
The substances which contain species wi{h unpaired electrons in 2. From the a-particle scattering experiment, Rutherford
their orbitals behave as paramagnetic substances. Such. substances concluded that:
are weakly attracted by the magnetic field. The paramagnetism is (a) a-particles can approach within a distance of the order of
expressed in terms of magnetic moment. The magnetic moment is 10'-14 m of the nucleus
related to the number of unpaired electrons according to the
(b) the radius of the nucleus is less than 10- 14 ~
following relation:
(c) scattering follows Coulomb's law
Magnetic moment,!l ~n(n + 2) EM (d) the positively charged parts of the atom move with
extremely high velocities
where, n number of unpaired electrons. 3. Rutherford's scattering formula fails for very small scattering
BM stands for Bohr magneton, a unit of magnetic moment. angles because:
(a) the gold foil is very thin
IBM = eh = 9.27 x 10-24 Am2 orJT- 1
4nmc (b) the kinetic energy of a-particles is very high
(c) the full nuclear charge of the target atom is partially
Answer the following questions: screened by its electron
1. Which of the. following ions has the highest magnetic (d) there is strong repulsive force between thea-particles and
moment? nucleus of the target
(a) Fe 2+ '(b) Mn 2+ (c) Cr 3+ (d) V 3+ 4. Alpha particles that come closer to the nuclei:
2. Which of the following ions has magnetic moment equal to (a) are deflected more (b) are deflected less
that ofTi 3+: . .
(c) make more collision (d) are 'slowed down more
(a) Cu 2+ (b) Ni2+ (c) C0 2+ Cd) Fe 2+ 5. Which of the following quantities will be zero for alpha
3. An ion of a d-block element has magnetic moment 5.92 BM particles at the point of closest approach to the gold atom, in
Select the ion among the following: Rutherford's scattering of alpha particles?
(a) Zn 2+ (b) Sc 3+ (c) Mn 2+ (d) Cr 3+ " (a) Acceleration (b) Kinetic energy
4. In which ofthese options do both constituents of the pair have (c) Potential energy (d) Electrical energy
the same magnetic moment?
r
(a) Zn 2+ and Cu + (b) C0 2+ and Ni 2+ • Passage 6.
(c) Mn4+ andCo + 1 (d) Mg2+ and Sc+
The splitting ofspectral lines by a magnetic field is called the
5. Which of the following ions are diamagnetic? Zeeman effect 'after the Dutch physicist Pieter Zeeman. The Zeeman
(a) H~2+ (b) Sc 3+ (c) Mg2+ . (d) 0 2- effect is a vivid confirmation of space quantization. Magnetic
quantum number 'm' was introduced during the study of Zeeman
• Passage5 effect. 'm' can have the (21 + 1) values (-1,0, + I). Magnetic
quantum number represents the orientation of atomic orbitals in
At the suggestion of Ernest Rutherford, Hans Geiger and Ernest three-dimensional space. The normal Zeeman effect consists of the
Marsden bombarded a thin gold foil by a-particles from a polonium splitting ofa spectra/line offrequency Vo into three components, i.e.,
source. It was expected that a-particles would go right through the
e . e
roil with hardly any deflection. Although, most of the alpha particles --B; v 2 Vo +--B
indeed were not deviated by much, a few were scattered through very 4nm 4nm
large angles, Some were even scattered in the backward direction. Here, B is magnetic field.

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1.46 .1 G.RB. PHYSICAL CHEMISTRY FOR COMPETITIONS

Answer the following questions: .' 3. Which of the following quantum numbers is not derived from
1. Which of the following statements is incorrect with reference SchrOdinger wave equation?
to the Zeeman effect? ., (a) Principal. (b) Azimuthal
(a) In a magnetic field, the energy of a particular atomic state (c) Magnetic (d) Spin
. depends on the values of'm' and 'n' . 4. In any subshell, the maximum number of electrons having
. (b) Zeeman effect is us~d to calculate the elm ratio for an same value of spin quantum number is:
electron
(c) Individual Spectral lines split into separate lines. The (a)~l(l+l) (b) 1+2 (c) 21 + I (d) 41 + 2
distance between them is independent ofthe magnitude of 5. The orbital angular momentum for a 2p-electron is:
.the magnetic field .
(d) Tile Zeeman effect involves splitting of a spectral line of
(a) J3 Ii (b) 16 Ii (c) zero (d) J2'!!'-'
2n
frequency Vo into tbree components
2. A d-subshelt· in an atom in the presence and absence of • Passage 8
magnetic field is: .
Dual nature ofmatter was proposed by de Broglie in 1923, it was
lea) five~fold degenerate, non-degenerate
experimentally verified by Davisson and Germer by diffraction
(b) seven-fuld degenerate, non-degenerate
experiment. Wave character of matter has significance only for
(c) five-fold degenerate, five.:fold degenerate microscopic particles. de Broglie wavelength or wavelength of
·'(d) non-degenerate, five-fold degenerate matter wave can be calculated using the following relation:
3~ Which among the·. following. is/are . correct about the
orientation of atomic- orbitals in space? A= h
(a) s-orbitals has single orientation mv
(b) d-subs4ell orbitals have three orientations along
where, 'm' and v' are the mtl$S and velocity of the particle.
x, yand z directions .
de Broglie hypoth~sis suggested that electron. waves were being
(c) f-subshell have seven orientations in their orbitals
diffracted by the target, much as X-rays are. diffracted by planes of
(d) None of the above·
atoms in the crystals.
t Zeeman effect explains splitting of spectral'lines in: Answer the following questions:
(a) magnetic field .(b) electric field
1. Planck's constant has same dimension as that of:
(c) both (a)' and (b) (d) none of these
(a) vvork (b) energy
i. Inpresence of magnetic field, d-suborbit is:
(c) power (d) angular momentum
(a) five-fold degenerate {b) three-fold degenerate
2. Wave nature of electrons is shown by:
(c) seven-fold degenerate (d) non-degenerate
(a) photoelectric effect (b) Compton effect
(c) diffraction experiment Cd) Stark effect
Passage 7 3. de 'Broglie equation is obtained by combination of which of
Spin angular . momentum of an electron has no analogue in the following theories?
classical mechanics. However, it turns out that the treatment ofspin (a) Planck's quantum theory
angular momentum is closely analogous to the treatment of orbital (b) Einstein's theory of mass-energy equivalence
angular momentum. (c) Theory of interference
Spin angular momentum ~ s(s + 1) Ii (d) Theory of diffraction'
4. Which among the following is not used to calculate the de
Orbital anE,ular momentum = ~'l(l + I) It
Broglie wavelength?
Total spin of an atom or ion is a multiple ofl.. .Spin mUltiplicity is
. . . 2
(a) A';'~ (b) A h
"v mV
a factor to confirm the electronic configuration of an atom of' ion. h . h
(c) A"" r;:;::;- (d) A"" - _ .
. Spin multiplicity =(n:s + 1) v2Em ~2qVm
-. . . $
5. The wavelength .of matter waves associated with a body of
Answer the following qurstiolls: mass 1000g moving with a velocity of 100 mlsec is:'
1. Total spinofMn 2+ (Z 25)ion will be: . (a) 6.62 x 1O-39 .cm (b) 6.62 x 10-36 ern .
. (c) 6..626 X 10- 36 m (d) 3.31 x 10-32 m
(aY ~ (b) 1 (c) 5
2 ·2 2 6. An electron microscope is used to probe the atomic
2. which of the following electrq'~ic configurations have fo~r arrangements to a resolution of 5 A. What should be the electric
spin multiplicity? potential to which the electrons need to .be accelerated ?
(VITEEE 2008)
(a) 2.5 V (b) 6 V
(c) 2.5 kV (d) 5 kV

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I

. ATOMIC STRUCTURE 147

• Passage 9 . . • Passage10
Orbital is the region in an atom where th~ probability ()Jfinding:
the electron is maximum. It represents three-dimensional motion oj The hydrogen-like species Li2+ is in a spherically symmetric state. .
an electron around the nucleus. Orbitals do ,not specify a definite S I with one radial node. Upon absorbing light the ion undergoes·
path according to the uncertainty principle. An orbital is described transition to a state S 2' The state S 2 has one radial node an? its
withthe,help o/waveJunction 'If. Whfmever an electron is described energy is equal to the ground state energy oj the hydrogen atom.
by a wave jUnction., we say that an electron occupies that erbital. (liT 2010)
Since, many wave Junctions are possible Jor an electron, there are Answer the following question;:
many atomic orbitals in an atom. Orbitals have different shapes; 1. The state S I is :
except s-orbitals. all other orbitals have directional character. OOb ~~ ~~ ~~.
Number oj spherical nodes in an orbital is equal to (n 1- 1). [Hint: 2.s: iss~etrical having one radial node.]
Orqital angular momentum oj an electro'n is ~ I (l + I) 'Ii. 2. Energy of the. state S I in uilits of the hydrogen atom ground
Answer the following questions: state energy is: . '
1. Which of the following orbitals is not cylindrically (a) 0.75 (b) 1.50 (c) 2.25 Cd) 4.50 .
. 9 ~
symmetrical about z-axis? E .2+(2s) --x 13.6
(apd z2 (b)4pz (c) 6s (d) 3dyz ~~~ u =2~]
EH -13.6 "
2. The nodes present in 5p-orbital are:
(8,) one planar, five spherical (b) one planar, four spherical 3. The orbital angular momentUm quantum mimber of the state
(c) one planar, three spherical(d) four spherical S2 is:
3. When an atom is pla~ed in a magnetic field, the possible (a) 0 (b) I (c) 2 (d),3
number of orientations for ~n· orbital .of azimuthal quantum [Hint: Orbital angular momentuniquantum number .of 3p
number 3 is: subsheU, i.e., I == 1
(a) three (b) one (c) five (d) seven
4. Orbital angular momentum of J-electrons is:
(a).fi'li' (b)/31i (c)JJi'li (d)21i
5. Which of the following orbitals haslhave two nodal planes?
(a)dxy (b)d yz (c)dx2 _ y2 (d)Allofthese

[/I~
Passage 1. 1, (b) 2, (c) 3. (c) 4. (b) 5. (c)
Passage 2. 1. (b) 2. (b) 3. (b) 4. (b) 5. (c)
Passage 3. 1. (a, d) 2. (a, b) 3. (c) . 4. (c) 5. (a)
Passage 4. i. (b) .2. (a) "3. (c) 4. (a, c) S. (b, c, d)
Passage 5. 1. (b) 2. (a, b, c) 3: (c,.d) 4. (a) 5. (b)
Passage 6. 1. (b) 2. (d) 3. (a,c) 4. (a) S. (d)
Passage 7. 1. (c) 2. (a) 3. (d) 4. (c) 5. (d)
Passage 8. 1. (d) 2. (c) .3. (a, b)
• 4. (~) S. (c) 6. (b)
Passage 9. 1. (d) 2. (c) 3. (d) 4. (c) S. (d)
Passage 10. 1. (b) 2. (c) 3. (b)

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148 I G.R.B. PHYSICAl- CHEMISTRY FOR COMPETITIONS

<@>.SELF ~SESSMENT <:>

ASSIGNMENT NO.2
SECTION-I 11. Which phenomenon best supports the theory that matter has a
. Straight OJJjective Type Questions wave nature? (VITEEE 2008)
This section contains II multiple choice questions. Each (a) Electron momentum (b) Electron diffraction'
.questionhas 4 choices (a), (b), (c) and (d), out of which only (c) Photon momentum (d) Photon diffraction
. one is correct.
. 1. Which. one of the following leads to third line of Balmer SECTION-II
spectrum from red end (For hydrogen atOm)? Multiple Answers Type Objective Questions
(a) 2 5 (b)5~2 (c)3----72 (d)4~1 12. Which of the following is/are correct?
2. The orbital angular momentum and angular momentum (classical (a) An electron in excited state cannot absorb a photon
. analogue) for the electron of 4s-orbital are respectively, equal to:
(b) Energy of electroris depends only on the principal
i;;:. h h 2h
(a) 'V 12 - and (b) zero and- quantum numbers
21t 21t 1t (c) Energy of electrons depends only on the principal
(c).J6h and 2h (d).J2 ~ and 3h quantum number for hydrogen atom
.1t 21t 21t (d) Difference in poteritial energy of two shens is equal to the-
3. A sample of hydrogen atom is excited to n 4 state. In the difference in kinetic energy of these shells
spectrum of emitted radiation, the number of lines in the 13. Which of the following statements is/are correct?
ultraviolet and visible regions are respectively:
(a) Energy of 4s, 4p, 4d and 4J are same for hydrogen
(a)3:2 (b)2:3 (c)I:3 (d) 3: I
(b) Angular momentum of electron fro
.. 4. Number of de Broglie waves made by.a Bohr electron in an
(c) For all values of 'n', the p-orbitals have the same shape
orbit of maximum magnetic quantum number + 2 is:
(d10rbital angular momentum = nh/21t
(a) I (b) 2 (c) 3 (d) 4
14. Which of the following orbitals are associated with angular
. 5. First line of Lyman series of hydrogen atom occurs at A x A.
nodes?
.The corresponding line of He + will occur at:
(a) f (b)d (c) p (d)s
(a)4x (b)3x (c)x/3 (d)x/4
15. The correct statement(s) among the following is/are:
6. Electronic transition in He + ion takes from n2 to nl shell such
(a) All d-orbitals except d _2 have two angular nodes
that;
2n2 + 3n] 18 • . ...0) (b) d x 2 - y-., , d z ,:tie
...
on the ~xes
2n2 - 3n] = 6 .... (ii) (c) The degeneracy of p -orbitals remains unaffected in the.
then what will be the total number of photons emitted when· presence of external magnetic field
electrons transit to n] shell? .(d) d-orbitals have 3-fold degeneracy
(a) 2 I (b) IS (c) 20 (d) 10
7 .. Which of the following sets of quantum numbers is not SECTION-III.
. possible for an electron? [PET (Raj.) 2008] Assertion-Reason Type Questions .Fi
(a)n :;: I, I = 0, ml == 0, ms = 1/2 This section contains 5 questions. Each question contains
. . (b) n .= 2, I :;: I, mJ 0, ms = - 1/2 Statement-l (Assertion) and Statement-2 (Reason). Each
'i (c) n == 1, l I, mz 0, ms = + 1/2 question has following 4 choices (a), (b), (c) and (d), out of
(d)n =2,1=I,mJ :;:O,m, +112 which only one is correct
8.' Th~ average life of an excited state of hydrogen atom is of the ,ja) Statement-I is true; statement-2 is true; statement-2 is a
order of .10-8 sec. The number of revolutions made by an' correct explanation for statement-I.
electron when it returns from n :;: 2 to n :;: 1is: (b) Statement-l is true; statement-2 is true; statement-2 is not
,(a)2.28x 106 (b) 22.8x 106 (c) 8.23 x 106 (d)2.82x 106, a correct explanation for statement-I.
9. ~e wave number of a particular spectral line in the atomic (c) Statement-l is true; statement-2 is false.
. spectrum of a hydrogen like species increases 9/4 times when (d) Statement-! is false; statement-2 is true .
deuterium nucleus is introduced into its nucleus, then which of 16. Sbltement-l: Kinetic energy of photoelectrons increases with
the following will be the initial hydrogen like species? increase in the frequency of incident radiation,
Because
(a) Li 2+ (b) Li+ (c) He+ (d) Be3+
Statement-2: The number of photoelectrons ejected increases
10. Energy of electron in the first Bohr orbit of H-atom is - 313.6 with increase in intensity of incident radiation.
kcal mol-I; then the energy in seconq Bohr orbit will be: 17. Statement-I: Photoelectric effect is easily pronounced by
(a) + 313.6 kcal morl. (b) 78.4 kcal mol-I caesium metal.
(c) - 34.84 kcal mol-I (d) -!2.5 kcal mol- 1 Because
Statement-2: Photoelectric effect is easily pronounced by the
metals having high ionization energy.

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ATOMIC STRUCTURE :149

18. Statement~I:Electrons in K-shell revolve in circular orbit. nh

( c) Wavelength of matter wave (r)
. Because 2n
Statement-2: Principal quantum number 'n' is equal to I for (d) Quantised value(s) (~) hlp
the electrons in K-shelL 23. Match the Column-I with Column-II:
19. Statement-I: Orbit and orbital are synonymous. Column-I Column-II
Because
(a) Elctrons cannot exist (p) cde Broglie wave
Statement~2: Orbit is the path around the nucleus in which
in the nucleus
electron revolves. .
(b) ".Microscopic particles (q) Electromagnetic
20. Statement-I:· C 6 = 1s2 , 181 , 2p 3 .is the electronic
in motion are wave
configuration in first excited state.
associated with
Because
Statement-2: Maximum energy by an electron is possessed in (c) No medium is required (r) Uncertainty principle
its ground state. for propagation
(d) Concept of orbit was (s) Transverse wave
SECTION-IV replaced by orbital
Matrix-MatchingJype Questions
This section contains 3 questions. Eac~ question contains SECTION-V
statements given in two columns which have to be matched. Linked Comprehension Type Questions
Statements (a, b, c and d) in Column-I have to be matched with A chemist was performing an experiment to study the effect of~
statements (p, q, rand s) in Column~II. The answers to these varying voltage on the velocity and de Broglie wavelength of the
questions have to be appropriately bubbled as illustrated iil the electrons. In first experiment, the electron was accelerated
following examples: through a potential difference of I kV and in second experiment,
If the correct matches are (a-p,s); (b-q,r); (c-p,q) and (d-s); it was accelerated through a potential difference 0(2 kV.
then the correctly bubbled 4 x 4 matrix should be as follows:' The wavelength of de Broglie waves associated with electron
p q r s is given by:

aft®0 A=_h_
J2qVm

b®_00 where, V is the voltage through which an electron is

accelerated.

eft_CD0 Putting the values of h , m and q we get:

A = 12.3 A
.jV .
d®®00 .....
Answer the following questions:
21. Match the Column-I with Column-II: 24. The wavelength of electron will be:
Column-I Column-II (a) 1.4 times in first case than in second case
(a) 48 . (p) Circular orbit around nucleus (b) 1.4 times in second case than in first case
(q) Non-direction orbitals (c) doubJe in second case than in first case
(b) 4p
. . 2h (d) double in first case than in second case
(c) Is (r) Angular momentum = 25. In order to get half velocity of eJectrons in second case, the
n
applied potential will be:
(d) . 3d (s) Number of radial node = 0
(a) 0.25 kV (b) 2 kV (c) 0.5 kV (d) 0.75 kV
22. Match the properties of Column-I with the formulae in
26. .The velocity of electron will be:
Column-I!:
(a) same in both cases
Column-I Column-II
(b) 1.4 times in second experiment than in first ewetiment
(a) Angular momentum of electron . (p) Jl(l + 1) 2: (c) double in second experiment than in first experiment
'(d) four times in the second case than in first case
(b) Orbital angular momentum (q) /(0

1. (b) 2. (b) 3. (a) 4. (c) 5. (d) 6. (d) 7. (c) 8. (c)

9. (d) 10. (b) 11. (b) 12. (a, c,d) 13. (a, b, c) 14. (a, b, c) "15. (a, b, c) . 16. (b)
17. (c) 18. (b) 19. (d) 20. (c) 21: (a-p, q,r) (b-r) (c-l?,q) (d-s)
22.' (a-q,r) (b-p) (c-s) (d~q,r) 23. (a-r)
.,. (b-p) (c-q,s)
. (d-r) 24.. Ca) 25. (a) 26. (b)

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