ATOMIC STRUCTURE

ATOMIC STRUCTURE
INTRODUCTION (d) The mass of electron in motion is
expressed as
(a) The word atom was first introduced by
Ostwald (1803 - 1807) in scientific world. m
(b) According to him matter is ultimately made m´ =
F1 – v I
2
½

up of extremely small indivisible particles GH c JK

2
called atoms.
(c) It takes part in chemical reactions. where m’ = mass of the electron in motion
(d) Atom is neither created nor destroyed m = rest mass, v = velocity of the electron
DALTON’S ATOMIC THEORY c = velocity of light
Dalton proposed the atomic theory on the (e) In 1897, J.J. Thomson determined the e/
basis of the law of conservation of mass and m value (charge/mass) of the electron by
studying the deflections of cathode rays
law of definite proportions. He also proposed
in electric and magnetic fields. The value
the law of multiple proportion as a logical
of e/m has been found to be –1.7588 ×
consequence of this theory. The salient 108 coulomb
features of this theory are (f) The first precise measurement of the
(a) Each element is composed by extremely charge on the electron was made by
small particles called atoms. Robert A. Millikan. in 1909 by oil drop
(b) Atoms of a particular element are all alike experiment. Its value was found to be -
but differ with the atoms of other elements. 1.6022 x 10–19 coulomb.

(c) Atom of each element is an ultimate (g) The mass of electron can be calculated from
particle, and has a characteristic mass but is the value of e/m and the value of e which is
structureless. 9.1096 × 10–31 Kg.
(d) Atom is indestructible i.e. it can neither Cathode rays:
be destroyed nor created by simple chemical
To vacuum
reactions. Cathode rays Gas at low
pressure pump
(e) Atom of an element takes part in chemical
Cathode Anode
reaction to form molecule. - +
(f) Atoms of different elements combine in
fixed ratio of small whole numbers to form
c o m p o u n d
(now called molecules).
- +

FUNDAMENTAL PARTICLES High voltage

(a) The electron was discovered as a result of
1. Electron: the studies of the passage of electricity
(a) Electron was discovered by Sir J.J. through gases at extremely low pressures
Thomson known as discharge tube experiments.
(b) When a high voltage of the order of 10,000
(b) The charge on the electron is 1.6 × 10–19
volts or more was impressed across the
coulomb/gm (Millikan)
electrodes, some sort of invisible rays
(c) The molar mass of electron is 5.48 × 10– moved from the negative electrode to the
4 gm/mole positive electrodes these rays are called
as cathode rays
 ATOMIC STRUCTURE

Properties of Cathode rays: (c) It was observed that when a high potential
difference was applied between the
(i) Path of travelling is straight from the electrodes, not only cathode rays were
cathode with a very high velocity produced but also a new type of rays were
As it produces shadow of an object placed produced simultaneously from anode
in its path moving towards cathode and passed
through the holes or canal of the cathode.
(ii) Cathode rays produce mechanical effects.
These termed as canal ray or anode ray.
If a small pedal wheel is placed between
the electrodes, it rotates. This indicates Perforated
that the cathode rays consist of material cathode
part
(iii) When electric and magnetic fields are
Cathode Positive
applied to the cathode rays in the
+ ray rays
discharge tube, the rays are deflected
thus establishing that they consist of
charged particles. Cathode
(iv) Cathode rays produce X-rays when they Properties of Anode Rays
strike against hard metals like tungsten,
copper etc. (i) These rays travel in straight lines and
cast shadow of the object placed in
(v) When the cathode rays are allowed to
strike a thin metal foil, it gets heated up. their path.
Thus the cathode rays possess heating (ii) The anode rays are deflected by the
effect. magnetic and electric fields like
(vi) They produce a green glow when strike cathode rays but direction i s
the glass wall beyond the anode. Light is different that mean these rays are
emitted when they strike the zinc sulphide
positively charged.
screen.
(iii) These rays have kinetic energy and
(vii)Cathode rays penetrate through thin sheets
of aluminium and other metals. produces heating effect also.

(viii)They affect the photographic plates (iv) The e/m ratio of for these rays is
smaller than that of electrons
(ix) The ratio of charge to mass i.e. charge/
mass is same for all the cathode rays (v) Unlike cathode rays, their e/m value
irrespective of the gas used in the tube. is dependent upon the nature of the
2. Proton: gas taken in the tube.

(a) Proton was discovered by Goldestein (vi) These rays produce flashes of light on
Zn-S screen
(b) Proton carries a charge of +1.602 x 10–19
coulomb, i.e., one unit positive charge. (vii)These rays can pass through thin metal
foils
(c) Mass of proton is 1.672 x 10–27 kg or
1.0072 amu (viii)They are capable to produce ionisation
(d) A proton is defined as a sub-atomic particle in gases
which has a mass nearly 1 amu and a (ix) They can produce physical and
charge of chemical changes.
+1 unit
3. Neutron:
Positive Rays-Discovery of Proton
(a) It has been found that for all atoms except
(a) The existence of positively charged
particles in an atom was shown by E. hydrogen atomic mass is more than the
Goldstein in 1886 atomic number. Thus Rutherford (1920)
(b) He repeated the same discharge tube suggested that in an atom, there must be
experiments by using a perforated cathode. present at least a third type of fundamental
particle.
 ATOMIC STRUCTURE

(b) It should be electrically neutral and posses RUTHERFORD’S ATOMIC MODEL

mass nearly equal to that of proton. He
Rutheford gold foil experiment :
proposed its name as neutron.
(c) Chadwick (1932), bombarded beryllium with Rutherford carried out experiment on the
a stream of -particles and observed bombardment of atoms by high speed
electrically and magnetically neutral positively charged  - particles emitted from
radiations. radium and gave the following observations,
which was based on his experiment.
(d) There were neutral particles which was
Very few Thin gold Few
called neutron. Nuclear reaction is as follows Radioactive
substance foil
4Be
9
+ 2He4  6C12 + 0n1
(e) A neutron is a subatomic particle which
Most
has a mass 1.675 x 10–24g, approximately
Lead block
1 amu, or nearly equal to the mass of Lead plate
proton on hydrogen atom and carrying no with hole Few
electrical charge. Observation:
Fundamental Particles Table: (a) Most of the  - particles (nearly 99%)
Atoms are made up of three fundamental continued with their straight path.
particles. The charge and mass of these (b) Some of the  - particles passed very close
fundamental particles are as follows: to the centre of the atom and deflected by
Electron Proton Neutron small angles.
o – 1 1
Symbol e or -1e or e P or 1p n or 0n
-31 -27 -27
(c) Very few particles thrown back (180º) .
Kg 9.109534 × 10 1.6726485 × 10 1.6749543 × 10
Atom of metal foil
Mass

-4
amu 5.4858026 × 10 1.007276471 1.008665012
1
Few
Relative 1 1
1837
-19 -19
Beam of –particles Majority
of –rays
Actual (in C) 1.6021892 × 10 1.6021892 × 10 0

Charge
relative –1 +1 0

One unit charge = 4.80298 × 10–10 esu Very Few

= 1.60210 × coulombs 10–19
Majority
1 of –rays
one amu = × mass of 6C12 atom
12 Few
Atom of metal foil
THOMSON’S ATOMIC MODEL Atomic model :
It states the arrangement of electrons and (a) Most of the  - particles were continued
protons in an atom. The main principles are their straight path that means most of the
(a) After discovery of electron and proton space of the atom is empty.
attempts were made to find out their
arrangement in an atom. The first simple (b) The centre of an atom has a positively
model was proposed by J.J. Thomson charged body called nucleus which repel
known as Thomson’s atomic model. positively charged  - particles and thus
(b) He proposed that the positive charge is explained the scattering phenomenon.
spread over a sphere of the size of the (c) Whole mass of an atom is concentrated
atom (i.e. in its nucleus and very few throw back
10–8 cm radius) in which electrons are means the size of the nucleus is very
embedded to make the atom as whole small 10–13 cm. It showed that the nucleus
neutral. is 10–5 times small in size as compared
(c) This model could not explain the to the total size of atom.
experimental results of Rutherfords -
particle scattering, therefore it was
rejected.
 ATOMIC STRUCTURE

APPLICATIONS OF RUTHERFORD Drawbacks of rutherford model:-

MODEL (1) This theory could not explain stability of atom.
According to Maxwell electron loses it energy
On the basis of scattering experiments, Rutherford
proposed model of the atom, which is known as continuously in the form of electromagnetic
nuclear atomic model. According to this model - radiations. As a result of this, the e- should loss
(i) An atom consists of a heavy positively charged energy at every turn and move closer and closer to
nucleus where all the protons and neutrons are the nucleus following a spiral path. The ultimate
present. Protons & neutrons are collectively reffered result will be that it will fall into the nucleus, thereby
to as nucleons. Almost whole of the mass of the making the atom unstable.
atom is contributed by these nucleon. The
magnitude of the +ve charge on the nucleus is Nucleus
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 diameter of the order of 10–12 to 10–
–
13
cm and the atom has a diameter of the order of e
10–8 cm (2) If the electrons loss energy continuously, the
DA 8 observed spectrum should be continuous but the
Diameteroftheatom 10
  13 = 10 5 , actual observed spectrum consists of well defined
DN Diameterofthenucleus 10 lines of definite frequencies. Hence, the loss of
DA = 105 DN energy by electron is not continuous in an atom.
Thus diameter (size) of the atom is 105 times the Some Important Definitions : -
diameter of the nucleus. Mass Number : It is represented by capital A. The sum
 The radius of a nucleus is proportional to the cube of number of Neutrons and protons is called the
root of the number of nucleons within it. mass Number of the element. i.e. A = number of
protons + number of Neutrons
R A1/3  R = R0A1/3 cm
Where R0 = 1.33  10-13(a constant) and A = Atomic Number : It is represented by Z. The number of
protons present in the Nucleus is called atomic
mass number (p + n) and R = radius of the nucleus.
number of an element. It is also known as nuclear
charge.
R = 1.33 × 10–13 A1/3 cm
For neutral atom : Number of proton = Number of
(iii) There is an empty space around the nucleus called electron
extra nuclear part. In this part electrons are For charged atom : Number of e– = Z – (charge on
present. The number of electrons in an atom is atom), Z= number of protons only
always equal to number of protons present in the Ex. 17Cl35  n = 18, p = 17, e = 17
nucleus. As the nuclear part of atom is responsible Two different elements can not have the same
for the mass of the atom, the extra nuclear part is Atomic Number
responsible for its volume.
Number of Neutrons
The volume of the atom is about 1015 times the
= Mass number – Atomic number
volume of the nucleus.
= A – Z = (p + n) – p = n
volumeoftheatom (10–8 )3 Representation of element  ZXA (where X
  1015
volumeofthenucleus (10–13 )3 symbol of element)
(iv) Electrons revolve around the nucleus in closed Isotopes : Given by Soddy, are the atoms of a given
orbits with high speeds. The centrifugal force acting element which have the same atomic number but
on the revolving e- is being counter balanced by different mass number i.e. They have same Nuclear
the force of attraction between the electrons and charge but different number of Neutrons.
the nucleus. Ex.1 Cl37 Cl37
17 17
 This model was similar to the solar system, the
n = 18 n = 20
nucleus representing the sun and revolving
electrons as planets. e = 17 e = 17
p = 17 p = 17
 ATOMIC STRUCTURE

Ex.2 6
C12 6
C13 C14]
6
Isotones/ Isonuetronic species / Isotonic :
e=6 e=6 e=6 They are the atoms of different element which have
p=6 p=6 p=6 the same number of neutrons.
n=6 n=7 n=8 Ex1 1H3 2
He4
Ex.3 (Proteium Dueterium Tritium) p=1 p=2
n=2 n=2
1
H1 1
H2 1
H3
e=1 e=2
e=1 e=1 e=1
Ex. 2 19 K 39 20
Ca40
p=1 p=1 p=1
p = 19 p = 20
n=0 n=1 n=2 e = 19 e = 20
1
1
H is the only normal hydrogen which have n = 0 n = 20 n = 20
i.e. no nuetrons Isosters : They are the molecules which have the same
Duetrium is also called as heavy hydrogen. It number of atoms & electrons.
represent by D Ex.1 CO2 N2O
 Isotopes have same chemical property but different Ex.2 CaO KF
physical property. Atoms = 1 + 2 Atoms = 2 + 1
 Isotopes do not have the same value of e/m. Atoms 2 2
Isobars : Given byAston, isobars are the atoms of different = 3 =3
element which have the same mass number but Electrons 20 + 8 19 + 9
different Atomic number i.e They have different Electrons = 6 + 8 × 2 Electrons = 7 × 2 + 8
number of Electron, Protons & Neutrons But sum 28 e– 28 e–
of number of neutrons & Protons remains same. = 22 e– = 22e–
Ex.1 1H3 He3 Isoelectronic Species :They are the atoms, molecules
2

p=1 p=2 or ions which have the same number of electrons.

Ex.1 Cl– Ar
e=1 e=2
Electron 18 e– 18 e–
n=2 n=1
Ex.2 H2O NH3
p+n=3 p+n=3 e=2+8 e=7+3
Ex.2 19 K40 20
Ca40 10 e– 10 e–
p = 19 p = 20 Ex.3 BF3 SO2
n = 21 n = 20 e = 5 +9 × 3 16 + 8 × 2
e = 19 e = 20 5 + 27 16 + 16
n+p = 40 n + p = 40 32 e– 32 e–
 Isobars do not have the same chemical & physical CONCEPT APPLICATION
property. EXERCISE-1
Isodiaphers : They are the atoms of different element 1. If S1 be the specific charge (e/m) of cathode
which have the same difference of the number of rays and S2 be that of positive rays, then which
is true ?
Neutrons & protons.
(1) S1 = S2 (2) S1 < S2
Ex1 5 B 11 6
C13 (3) S1 > S2 (4) Either of these
p=5 p=6
2. An increasing order (lowest first) for the values
n=6 n=7 of e/m for electron (e), proton (p), neutron (n)
e=5 e=6 and alpha () particle is :
n – p =1 n – p =1 (1) e, p, n,  (2) n, , p, e
(3) n, p, e,  (4) n, p, , e
Ex.2 7 N15 9
F 19
3. Select iso electronic set :-
p=7 p=9 (a) Na+, H3O+, NH4+
n=8 n = 10 (b) CO3–2, NO3–, HCO3–
e=7 e=9 (c) P–3, HCl, C2H6, PH3
n – p =1 n – p =1 (d) N–3, O–2, F
(1) a, b, d (2) b, c, d
(3) a, b, c, d (4) a, b, c
 ATOMIC STRUCTURE

4. The ratio of specific charge of a proton and Wavelength () (Lambda) : It is defined as the distance
an –particle is :– between two nearest crest or trough.
(1) 2 : 1 (2) 1 : 2 It is measured in terms of a Å (Angstrom), pm
(3) 1 : 4 (4) 1 : 1 (Picometre), nm (nanometer), cm(centimetre), m
5. An isotone of 76 (metre)
32Ge is :-
(i) 32Ge77 (ii) 33As
77 1Å = 10–10 m, 1 Pm = 10–12 m,
(iii) 34Se77 (iv) 34Se78 1nm = 10–9 m, 1cm = 10–2m
(1) (ii) & (iii) (2) (i) & (ii) Frequency () (nu) : Frequency of a wave is defined as
(3) (ii) & (iv) (4) (ii) & (iii) & (iv) the number of waves which pass through a point in
13 12
1 second.
6. 6 C and 6 C differ from each in respect of
It is measured in terms of Hertz (Hz ), second–1 , or
number of cycle per second (cps) (1 Hertz = 1 second–1 )
(1) electrons (2) protons Time period (T) : Time taken by a wave to pass through
(3) neutrons (4) none of these
1
7. Atomic weight of Ne is 20.2. Ne is mixture of Ne20 one point. T  second

and Ne22, Relative abundance of heavier isotope is :-
Velocity (c) : Velocity of a wave is defined as distance
(1) 90 (2) 20
covered by a wave in 1 second
(3) 40 (4) 10
c =  or = c/ orc = (s–1) × (m)
ELECTROMAGNETIC WAVES or c = (m s )
–1

(EM WAVES) THEORY OR RADIANT Since c is constants  i.e. frequency is inversely

ENERGY propotional to
According to this theory the energy transmitted Wave number (  ) ( nu bar) : It is the reciprocal of the
from one body to another in the form of waves and wave length that is number of waves present in
these waves travel in the space with the same 1
1cm 
speed as light ( 3 × 108 m/s) and these waves are 
known as Electro magnetic waves or radiant It is measured in terms of cm–1, m–1 etc,
energy. The radiant Energy do not need any Amplitude (a) : The amplitude of a wave is defined as
medium for propogation. the height of crust or depth of through.

c
Ex : Radio waves, micro waves, Infra red rays, visible
Important note :   c
rays, ultraviolet rays, x–rays, gama rays. 

 1
 The radiant Energy have electric and magnetic fields    
and travel at right angle to these fields. The upper  
most point of the wave is called crest and the lower
most portion is called trough. Some of the terms ELECTROMAGNETIC SPECTRUM
employed in dealing with the waves are described
Electromagnetic wave or radiation is not a
below. single wavelength radiation, but a mixture of
various wavelength or frequencies. All the
frequencies have same speed.
Vibrating source

Crest Crest If all the components of Electromagnetic


a Radiation (EMR) are arranged in order of
a Direction decreasing or increasing wavelengths or
of propogation frequencies, the pattern obtained is known as

Trough Trough
Electromagnetic Spectrum. The following table
shows all the components of light.
 ATOMIC STRUCTURE

S.No. Name Wavelength Frequency(Hz) Source Important postulates of Bohr’s Model

1. Radio wave 3 × 107 – 3 × 1014 1 105 –1109 Alternating
current of (1) Electrons revolve around the nucleus along
high
frequency certain circular paths known as “ORBIT” or
2. Microwave 6 × 106 – 3 × 107 1 109 – 5 1011 Klystron tube
“SHELLS”
3. Infrared (IR) 7600 – 6 × 10 6
5 1011 – 3.95 1016 Incandescent
objects
4. Visible 3800 – 7600 3.95  1016 – 7.9 1014 Electric
K, L, M, ................. are shells
bulbs, sun
rays K 1st orbit
5. Ultraviolet(UV) 150 – 3800 7.9 1014 – 2 1016 Sun rays, arc
lamps with
mercury L 2nd orbit
vapours
6. X-Rays 0.1– 150 2 1016 – 3 1019 Cathode rays M 3rd orbit
striking metal
plate
(2) Electron is associated with a fixed energy in a
7.  -Rays 0.01 – 0.1 3  1019 – 3 10 20 Secondary
effect of particular orbit. The change in electronic energy
radioactive
decay is possible only when; electron changes its orbit
number.
PLANCK'S QUANTUM THEORY
(a) When an electron goes to higher orbit from
According to planck's quantum theory: lower orbit, then energy is absorbed.
1. The radiant energy emitted or absorbed by a body E1 – E2 = E
not continuously but discontinuously in the form of
small discrete packets of energy and these
packets are called quantum.
2. In case of light, the smallest packet of energy is (Absorbed energy)
called as 'photon' but in general case the smallest
packet of energy called as quantum.

3. The energy of each quantum is directly proportional (b) If an electron jumps to lower orbit from higher
to frequency of the radiation i.e. orbit, then energy is radiated.

hc  c E2 – E1 = E
E   E = h or E =    
   

h is proportionality constant or Planck's constant

h = 6.626 × 10–31 kJ s or 6.626 × 10–34 J s (Emission of energy)

(3) The electrostatic force of attraction acting
or 6.626 × 10–27 erg s between the electron and the nucleus is
counterbalance by the centrifugal force
2
4. Total amount of energy transmited from one body mV 2 Ze
to another will be some integral multiple of energy =K 2
of a quantum. E = nh Where n is an integer and n
r r
= number of quantum

nhc
E = nhν = = nhcν
λ

Bohr’s Atomic Model:

Ze2
or mv2 = K ......................(i)
Bohr developed a model for hydrogen and r
hydrogen like atoms one-electron species
(hydrogenic species). He applied quantum
1 Ze 2
= . .............(ii)
theory in considering the energy of an electron 4 0 r
bond to the nucleus. [ 0 = 8.8 × 10–12 C2N–1 m–2]
 ATOMIC STRUCTURE

(4) Electrons can revolve only in those orbits in which

nh nh
h As we know that mvr = or v =
angular momentum is integral multiple of 2 2mr
2
....(2)
Now putting the value of v from eqn.(2) to eqn.(1)
2
 nh  KZe2 n2h2 KZe2
 2mr   mr or 
  42m2r2 mr

h n2h2
mvr = n. r
2 42mKZe2 ....(3)
where, h = Planck’s constant = 6.62 × 10–34 J
s–1 = 6.62 × 10–27 erg. sec–1 Putting the value of , h, m, K, & e (Constants) in
m = mass of electron the above eqn. (3)
v = Electronic velocity
n =orbit number n2
r = 0.529 × 10–8 × cm { 1Å = 10–10m = 10–8 cm}
Z
Note: If the energy supplied to hydrogen atom is less
than 13.6 eV, it will accept or absorb only
n2
those quanta which can take it to a certain rn  0.529  Å
higher energy level i.e., all those photons having Z
energy less than or more than a particular This formula is only applicable for hydrogen and
energy level will not be absorbed by hydrogen hydrogen like species i.e. species contains single
atom. But if energy supplied to hydrogen atom electron.
is more than 13.6 eV then all photons are
absorbed and excess energy appear as kinetic 2. Velocity of an electron :
energy of emitted photo electron. Since Coulombic force = Centrifugal force

APPLICATION OF BOHR'S MODEL or

KZe2
=
mv2
or KZe2 = (mvr)(v)
2
1. Radius of various orbits (shell): r r

n2h2 now putting the value of Angular momentum m.v.r.

r
42mKZe2 nh nh
Derivation : =  KZe2 = (v)
2 2
Kq1q2 K.Ze.e KZe2
Columbic force = = = 2 2KZe2
r2 r2 r v =
nh
(Tangential now putting the value of k, e & h
velocity)
Z
v = 2.188 × 108 × cms-1
n
r 3. Energy of an electron
+ Ze e–
Coulabic or Let the total energy of an electron be E. It is the
Nucleus Electrostatic sum of kinetic and potential Energy.
force Kq1q2
( ) i.e. E = K.E.+ P.E.
r2
 1 2   Kq1q2   Kze2 
E =  mv      P.E.   
2   r   r 
(Where K is Constant) K = 9 × 109 Nm2 /coulumb2 1 2 K.Ze.(e) 1 2 KZe2
E= mv  = mv 
As we know that Coulombic force = Centrifugal 2 r 2 r
force now putting the value of mv2 from eq. (1)
KZe2 mv2 KZe2 KZe2 KZe2 KZe2
 or v2
 ....(1) E=  = –
r2 r mr 2r r 2r
now putting the value of r from eq. (3)
 ATOMIC STRUCTURE

KZe2  42mKZe2 n = 6 or P E6 = – 0.38 eV

En = 
2n2h2 n = 5 or O E5 = –0.54 eV
2 2 m × K 2 Z 2 e 4 n = 4 or N E4 = – 0.85 eV E5 – E4 = 0.36 eV
or En = –
n2 h2
n = 3 or M E3 = – 1.51 eV E4 – E3 = 0.66 eV
now putting the value of , K, e, m, h, we get :
n = 2 or L E2 = – 3.4 eV E3 – E2 = 1.89 eV
21.69 × 10 -19 × Z 2
En = – J / atom n = 1 or K E1 = – 13.6 eV E2 – E1 = 10.2 eV
n2
i.e. (E2 – E1) > (E3 – E2) > (E4 – E3) > (E5 – E4).....
Z2
or En = –13.6 × 2 eV/ atom Energy Definations:
n
(a) Ground state (G. S.)
Z2 The lowest energy state of an atom or ion or
En = –1312 × kJmol–1
n2 molecule
G. S. for H-atom n = 1
Z2 He+ ion n=1
or En = –313.6× kcalmol–1
n2
This formula is applicable for hydrogen atom & hy- (b) Excited state (E. S.)
drogen like species i.e. single electron species. The energy states above the ground state are
Since n can have only integral values, it follows referred to as an excited
that total energy of the e– is quantised. n=2 1st E. S.
The –ve sign indicats that the e–is under attraction n=3 2nd E. S.
towards nucleus.
n=4 3rd E. S.
Some extra points :
Total E. S. = (n – 1)
KZe2 1 (c) Excitation Potential (E. P.) :
(i) K.E= i.e. K.E. 
2r r The amount of energy needed to promote an
On increasing radius, K.E. decreases. electron from ground state to ‘n’ level
1 1st I. P. = E2 – E1 = –3.4 – (13.6) = 10.2 eV
KZe2
(ii) P. E. = – i.e. P.E. – 2nd E. P. E3 – E1 = –1.51 – (–13.6) = 12.09
r r
On increasing radius, P.E. increases. (d) Ionization potential (I. P..)/Energy/Enthalpy
1 Corresponding energy required to remove the
KZe2
(iii) E = – i.e. E. – electron from G.S. to the infinite excited state.
2r r
Results : On incresing radius, total energry increases. I. P. = E– E1
= 0 – (–13.6) = 13.6 eV
P.E = 2KE KE = E P.E = 2E
Energy diffrance between two energy levels : (e) Separation Energy (S. E.)
The amount of energy needed to remove to infinity
1 1
En2  En1 = – 13.6 × Z2  2  2  the electron from the excited state.
 n2 n1 
e.g., S. E. = E – E2 = 0 – (–3.4) = 3.4 eV
Energy level for H atom can be represented as follows: also S. E. = E – E3 = 0 – (–1/51) = 1.51 eV
Shell O E5
Shell N
Shell M
E4 CONCEPT APPLICATION
Shell L E3
EXERCISE-2
Shell K E2
1. If the radii of first orbits of H, He+, Li+2 and Be+3
E1
Nucleus + are r1, r2, r3 and r4 respectively, then their correct
Shell 1 decreasing order will be:
Shell 2 (1) r1 > r2 > r3 > r4 (2) r3 < r2 > r4 < r1
Shell 3
Shell 4 (3) r1 < r2 < r3 > r4 (4) Radius of all are equal
Shell 5
 ATOMIC STRUCTURE

2. An electron in an atom jumps in such a way (1) Emissions spectrum : When the radiation
x emitted from Incandescence source (eg. from the
that its kinetic energy changes from x to .
4 candle, sun, tubelight, burner, bulb, or by passing
The change in potential energy will be : electric discharge through a gas at low pressure,
3 3 by heating some substance at high temprature)
(1) + x (2) − x
2 8 is passed directly through the prism and the
3 3 pattern obtained on the screen is known as
(3) + x (4) − x
emission spectrum
4 4
(a) Emission continuous spectrum or continuous
3. The distance between 4th and 3rd Bohr orbits
spectrum:
of He+ is :
When a narrow beam of white light is passed
(1) 2.645 × 10–10m (2) 1.322 × 10–10m through a prism, it is dispersed into 7 colours
–10
(3) 1.851 × 10 m (4) None from violet to Red.
Narrow beam Screen U V Region
4. What atomic number of an element “X” would
of white light V
have to become so that the 4th orbit around X I
would fit inside the 1st Bohr orbit of H atom? B
(1) 3 (2) 4 G Visible
region
(3) 16 (4) 25 y

5. The energy levels for zA(+z–1) can be given by : O

(+z–1) 2 R
(1) En for A = Z × En for H
Infra red region
(2) En for A(+z–1) = Z × En for H
(b) Emission line spectrum : When an atomic gas
1
(3) En for A(+z–1) = × En for H is raised to incandescence source or subjected to
z2
1 electrical excitation, it first absorbs energy & then
(4) En for A(+z–1) = × En for H
z gives it out as radiation. On examining these
radiation through a spectro scope a spectrum is
6. The angualr momentum of electron of H-atom
obtained which have well defined lines,each
is proportional to :
corresponding to a definite wave length & these
1
(1) r2 (2) lines are separated from each other by dark space.
r
1 This type of Emission spectrum is called as
(3) r (4) Emission line spectrum.
r
7. The ratio of magnetic moments of Fe (III) and
Co (II) is :
(1) 5 : 7 (2) 35 : 15
(3) 7 : 3 (4) 24 : 15
Special Note :
SPECTRUM 1. No two Elements will have identical line spectrum
Spectrum: When a radiation is passed through a since no two elements have identical energy level
spectroscope (prism) for the dispersion of the therefore the line spectrum of the elements are
radiation, the pattern (photograph) obtained on described as finger prints differing from each other
the screen (photographic plate) is called as like the finger prints of the human beings.
spectrum of the given radiation 2. Since line spectrum is obtained by the Emission
Classification of Spectrum of Energy through the atoms of the element there-
fore line spectrum is also called as atomic
(1) Emission (2) Absroption spectrum.

(a) Continuous (b) line (c) band (a) line (b) band
 ATOMIC STRUCTURE

(c) Emission band spectrum : If molecular form of 7 Q

the gas is used, it first absorbs energy for not only 6
Far I.R. region P
Humphery series
electron transition but for rotational, vibrational and 5 O
I.R. region
electron translation then emits radiations. P fund series
4 N
On examining these radiations through a I.R. region
Bracket series
spectroscope a spectrum is obtained on the 3 M
screen, which consists group of closely packed Infra Red region
or
lines called Bands, therefore this type of Emis- Paschen series
sion spectrum is called as emission band spectrum. 2 L
Visible region
Bands are separated from each other by dark or
Balmer series
space. 1
Ultra violet region
K
or
Lyman series

Note: Since band spectrum are caused by

molecules therfore band spectrum are also called
as molecular spectrum.
(2) Absorption spectrum : When white light is Rydberg formula : In 1890, Rydberg gave a very
first passed through a solution or vapours of a
simplest theoretical Equation for the calculation of the
chemical substance or gas and then analyzed by
wavelength of various lines of hydrogen like spectrum and
spectroscope, it is observed that some dark lines
are obtained in otherwise continuous spectrum. the equation is
This type of spectrum is called as Absorption
1 2 1 1
spectrum.   RZ  2  2 
  n1 n2 
Screen
Coloured
Coloured where R = Rydberg constant
Incandescence
Gas
source
Coloured = 109678 cm–1 ~ 109700 cm–1
Coloured
1
 If white light is passed through atomic gas then –6
R = 9.12 × 10 cm = 912 Å
the obtained spectrum is called as Absorption line
spectrum. n1 and n2 are orbits and for a particular series n1 is
constant and n2 varies.
 If white light is passed through molecular gas then
for Lyman n1 = 1, n2 = 2, 3, 4,.....
the obtained spectrum is called as Absorption band
for Balmer n1 = 2, n2 = 3, 4, 5,....
spectrum.
for Paschen n1 = 3, n2 = 4, 5, 6,....
HYDROGEN LINE SPECTRUM OR for Bracket n1 = 4, n2 = 5, 6, 7,...
HYDROGEN SPECTRUM for Pfund n1 = 5, n2 = 6, 7, 8,...
When an electric excitation is applied on hydro- for Humphery n1 = 6, n2 = 7, 8, 9,...
gen atomic gas at Low pressure,a bluish light is
Derivations of Rydberg formula :
emitted. when a ray of this light is passed through
a prism, a spectrum of several isolated sharp line E = En2 – En1
is obtained.The wavelength of various lines show
E =
that spectrum lines lie in visible, Ultraviolet and
Infra red region. These lines are grouped into 22mK2Z2e4  22mK2Z2e4 
 
different series. n22h2  n12h2 
 ATOMIC STRUCTURE

4. Bohr's theory does not explain the fine structure of

22mK2Z2e4 22mK2Z2e4
=  the spectral lines. Fine structure of the spectral
n12h2 n22h2
line is obtained when spectrum is viewed by spec-
troscope of more resolution power.
 hc 
 E  h    5. Bohr theory does not explain the spiliting of spectral
 
lines in the presence of magnetic field (Zemman's
effect) or electric field (Stark's effect)
hc 2 mK Z e
2 2 2 4 1 1
  2  2 WAVE MECHANICAL MODEL OF AN
 h2  n1 n2 
ATOM

1 2 mK e Z  1  1  This model consists three model.

2 2 4 2

or    2 2
 ch3  n1 n2  1. de-Brogle concept (Dual nature of Matter)
2. Heisenberg's Uncertainity principle.
22mK2e4
where is a constant which is equal to 3. Schrodinger Equation.
ch3
1. The Dual nature of matter (The wave
rydberg constant (R).
nature of electron)
1 1 1 (i) In 1924, a French physicist, Louis de-Broglie
= RZ2  2  2 
  n1 n2  suggested that if the nature of light is both that of a
particle and of a wave, then this dual behavior should
be true for the matter also.
Calculation of number of spectral lines
(ii) The wave nature of light rays and X-rays is proved
(a) Total number of spectral lines = 1 + 2 + ..... on the basis of their interference and diffraction and
(n2 – n1)(n2 − n1 + 1) many facts related to radiations can only be ex-
(n2 – n1) = plained when the beam of light rays is regarded as
2
composed of energy corpuscles or photons whose
if n1 = 1(ground state) velocity is 3 × 1010 cm/s.
Total number of spectral lines (iii) According to de-Broglie, the wavelength  of an
electron is inversely proportional to its momentum p.
=
(n2 − 1) n2 = n (n − 1)
2 2 h
1
 or  (Here h = Planck's
(b) Number of spectral lines which falls in a particular p p
series (n2 – n1) constant, p = momentum of electron)
where n2 = higher energy series, n1 = lower energy  Momentum (p) = Mass (m) × Velocity (v)
series
Limitation of the Bohr's model : h
 λ =
mv
 Bohr's theory does not explain the spectrum of multi
electron atom. (iv) The above relation can be confined as follows by
2. Why the Angular momentum of the revolving using Einstein's equation, Planck's quantum theory
and wave theory of light.
nh
electron is equal to , has not been explained Einstein's equation, E = mc2 where E is energy, m
2
is mass of a body and c is its velocity.
by Bohr's theory.
E = h(According to Planck's quantum theory)
3. Bohr interrelate quantum theory of radiation and
and c =  (According to wave theory of light)
classical law of physics with out any theoritical
explanation.
 ATOMIC STRUCTURE

number of waves made by electron in one complete

c c
   E=h×
  2π r
revolution = = n (orbit number)
But according to Einstein's equation E = mc 2 λ
c h 2. Heisenberg uncertainity principle:
E = mc2 = h × or mc =
  In 1927, Werner Heisenberg presented a principle
known as Heisenberg uncertainity principle which
h h
or p= or = states as: "It is impossible to measure simultane-
 p
ously the exact position and exact momentum of
(v) It is clear from the above equation that the value of
a body as small as an electron."
decreases on increasing either m or v or both.
The wavelength of many fast-moving objects like The uncertainity of measurement of position (x)
an aeroplane or a cricket ball, is very low because and the uncertainity of momentum (p) or mv),
of their high mass. are related by Heisenberg's relationship as :
Bohr's Theory and de-Broglie concept :
h
(i) According to de-Broglie, the nature of an electron x . p >
moving around the nucleus is like a wave that flows 4
in circular orbits around the nucleus.
(ii) If an electron is regarded as a wave, the quantum h
or x . (mv) >
condition as given by Bohr in his theory is readily 4
fulfilled.
(where h is Planck's constant.)
(iii) If the radius of a circular orbit is r, then its
circumference will be 2r. For an electron of mass m(9.10 × 10–28 g), the
nh product of uncertainity is quite large.
(iv) We know that according to Bohr theory, mvr =
2 h
nh x . v >
2r  4m
or (mv = p momentum) or
mv 6.624  1027
x . v >
4  3.14  9.10 1028
nh  h 
2r     de-Broglie equation = 0.57 erg second per gram approximately
p  p 
x v is called uncertainity product
 2r = n(where n = total number of waves 1, When x = 0, v= and vice-versa.
2, 3, 4, 5, ..... and = Wavelength
In the case of bigger particles (having considerable
nh nh mass), the value of uncertainity product is
(v)  2r = or mvr =  mvr =
mv 2 negligible. If the position is known quite accurately,
Angular momentum i.e., x is very small, v becomes large and vice-
Thus mvr = Angular momentum, which is a integral
h versa.
multiple of .
2
(v i) It is clear from the above description that according h
Et > (for energy and time)
to de-Broglie there is similarity between wave theory 4
and Bohr theory.
On the basis of this principle, therefore, Bohr's
picture of an electron in an atom, which gives a
fixed position in a fixed orbit best we can think of
in terms of probability of locating an electron with
a probable velocity in a given region of space at a
given time. The space or a three-dimensional region
figure : Similarity between de-Broglie waves and Bohr's orbit round the nucleus where there is maximum
probability of finding an electron of a specific energy
is called an atomic orbital.
 ATOMIC STRUCTURE

3. Schrodinger Equation 4. Line spectra is characteristic of :

(1) Molecules (2) atoms
The nature of electron in atom can be discussed in (3) radicals (4) none of these
terms of the solution of schrodinger wave equation
5. Splitting of spectral lines under the influence of
magnetic field is called :
2 2 2 82m
   (E  v)  0 (1) Zeeman effect (2) Stark effect
x2 y2 z2 h2 (3) Photoelectric effect (4) None of these

 - wave function 6. Which statement relating to the spectrum of H

atom is false?
(1) The lines can be defined by quantum
 - probability of finding electron number
(2) The lines of longest wavelength in the
E - total energy Balmer series corresponds to the transition
between n = 3 and n = 2 levels
(3) The spectral lines are closer together at
v - potential energy longer wavelength
The wave function of an electron ( in the field of (4) A quantum occurs at n = 
nucleus of atom is called atomic orbital. It is a 7. Which of the following series of lines in the
three dimensional amplitude. atomic emission spectrum of hydrogen is in
the visible region?
 is the probability of finding the electron in a (1) Lyman (2) Paschen
volume element  surrounding the nucleus of an (3) Balmer (4) Brackett
atom. 8. If ‘RH’ is the Rydberg constant, then the energy
Solution of schrodinger equation provide set of three of an electron in the ground state of hydrogen
atom is :
qauantum number (n, l, m)
RHc 1
(1) (2) R ch
CONCEPT APPLICATION h H

EXERCISE-3 hc
(3) R (4) –RHhc
H
1. What electronic transition in Li2+ produces the
radiation of the same wavelength as the first line QUANTUM NUMBER
in the Lyman series of hydrogen? As we know to search a particular person in this
(1) n = 4 to n = 2 (2) n = 9 to n = 6 world 4 things are needed:-
 Country to which the person belongs.
(3) n= 9 to n = 3 (4) n = 6 to n = 3  The city in that country where the person is
2. The first Lyman transition in the hydrogen residing
 The area in that city
spectrum has E = 10.2 eV. The same energy
 House number
change is observed in the second Balmer Similarly to locate the position of an electron in
transition of : the atom 4 identification number are required and
these identification number are called as quantum
(1) Li2+ (2) Li+
mumber.
(3) He+ (4) Be3+ 1. Principal quantum number (n)  Shell (Orbit)
3. A hydrogen atom in the ground state is excited 2. Azimuthal quantum number (l) Sub shell
3. Magnetic quantum number (m) Orbital
by monochromatic radiation of wavelength  Å.
4. Spin quantum number (S) Spin of e–
The resulting spectrum consists of maximum
15 different lines. What is the wavelength ? 1. Principal quantum number : Given By 
(RH = 109737 cm ) –1 Bohr's
It represents the name, size and energy of the
(1) 937.3 Å (2) 1025 Å shell to which e– belongs
(3) 1236 Å (4) None of these
 ATOMIC STRUCTURE

 The value of n lies between 1 to  h

i.e n = 1,2,3,4– – – – – corresponding name of  The orbital angular momentum = (  1) or
2
shells are K, L, M, N, O, – – – – –
 Greater the value of n, greater is the distance  h
from the nucleus.     1      {  is called as 'hash' }
 2 
n2 Orbital angular momentum : For s subshell = 0
r = 0.529 × Å
z h
r1 < r2 < r3 < r4 < r5 – – – – – – – –
For p Subshell = 2
2
or 2
 Greater the value of n, greater is the energy of  The number of electron in a particular subshell is
shell equal to 2(2+1)
z2 for s subshell number of electrons = 2 e—
E = – 13.6 × eV/atom for p subshell number of electrons = 6 e—
n2
for d subshell number of electrons = 10 e—
E1 < E2 < E3 < E4 – – – – – – – – for f subshell number of electrons = 14 e—
 The angular momentum of a revolving electron is  Shape of the subshell :
nh s spherical
mvr =
2 p dumb bell shape
Where n = Principal quantum number. d double dumb bell shape
f complex shape
 The number of electrons in a particular shell is 3. Magnetic quantum number /
equal to 2n2
Orientation quantum number (m) :
2. Azimuthal quantum number / Angular
given by lind
quantum number / Secondary quan-
tum number /Subsidiary quantum It represents the shape of different orbitals and
the orientation of electron cloud (orbital)
number:
 Under the influence of magnetic field each
Represented by '' (Given by – Sommerfeld)
subshell is further subdivided into orbitals ( The
 It represents the shape of the subshell and orbital
electron cloud is knwon as orbital)
and orbital angular momentum
Magnetic quantum number describe these
 Value of  between 0 to (n – 1)
different distribution of electron cloud.
i.e.  = 0,1,2 – – – – – – – – – (n–1)  Value of m = all integral value from –  to + 
 = 0(s Subshell) including zero.
 = 1(p Subshell) i.e. Value of m = –0 + 
 = 2(d Subshell) Case-I If = 0 then m = 0 it implies that s subshall has
 = 3(f Subshell) only one orbital called as s orbital.
Ex. If n = 1 then  = 0 1s i.e. in n =1
shell, only one subshell 's' is present. Shapes of s-orbitals:
If n = 2 then  = 0,1 2s,2p i.e. in n =2 shell, The s-orbitals are spherically symmetrical about
the nucleus, i.e., the probability of finding s elec-
two subshell 's' & 'p' are present.
tron is same in all directions from the nucleus.
If n = 3 then  = 0,1,2 3s, 3p, 3d i.e. in n =3
The size of the orbital depends on the value of
shell, three subshell 's' , 'p' & 'd' are present.
principal quantum number, there is one
If n = 4 then  = 0,1,2,3 4s,4p, 4d, 4f i.e. in n
spherically symmetrical orbital. The 1s orbital is
=4 shell, four subshell 's' , 'p' , 'd' & 'f' are present. smaller than 2s-orbital and 2s-orbital is smaller
 If the value of n is same then the order of energy than 3s, but all are spherical in shape as shown
of the various subshell will be in figure.
s<p<d<f
Ex. 4s < 4p < 4d < 4 f, 3s < 3p< 3d,2s <
2p
 If Value of  is same but value of n is different
then the order of energy will be.
Ex. 1s < 2s < 3s < 4s < 5s < 6s
3d < 4d < 5d < 6d
1S
4p < 5p <6p 2S Node 3S
Node
 ATOMIC STRUCTURE

Although the s-orbitals belonging to different shells These three p-orbitals are situated at right angle to
are spherically symmetrical, yet they differ in another and are directed along x, y and z axes
certain respects as explained below : (figure)
 The probability of 1s electron is found to be  Each p orbital has dumb bell shape ( 2 lobes which
maximum near the nucleus and decreases as are separated from each other by a point of zero
the distance from the nucleus increases. In case probability called nodal point or node or nucleus).
of 2s electrons, the probability is again maximum  The two lobes of each orbital are separated by a
near the nucleus and then decreases to zero and plane of zero electron density called nodal plane.
increases again and then decreases as the  Each p orbital of higher energy level are also dumb
distance from the nucleus from the nucleus bell shape but they have nodal surface.
increases. The intermediate region (a spherical
Nodal surface : Orbital Nodal surface
shell) where the probability is zero is called a
nodal surface of simply node. Thus, 2s-orbital 2 px 0
differs from 1s-orbital in having one node within 3 px 1
it. Similarly, 3s has two nodes. in general, any 4 px 2
ns orbital has (n -1) nodes. Nodal Plane : Orbital Nodal plane
 The size and energy of the s-orbital increases as px yz plane
the principal quantum number inreases, i.e., the py xz plane
size and energy of s-orbital increases in the order pz xy plane
1s < 2s < 3s .... npx (n – 2)
The s orbital of higher energy levels are also
Nodal y y
symmetrically spherical and can be represented surface
as follows: xz plane
x (Nodal plane)
y x
2Px
Nodal z 3Px
z
4Px
suface Py
Nodal point

(Nodal plane)
x y yz plane y

xy plane
(Nodal plane)

1s
z 2s x x
Case-II: 3s z Px
z
If = 1 ( p - subshell) Pz

Case III When  = 2, 'm' has five values – 2, –1, 0,

1 0 1 +1, +2. It implies that d subshell of any
then m=
px pz py energy shell has five orbitals. All the five
It implies that, p subshell have three orbitals orbitals are not identical in shape. Four of
called as px, py and pz. the d orbitals dxy, dyz, dxz, dx2  y2 contain four
Shape of p-orbitals : lobes while fifth orbital dZ2 consists of only
There are three p-orbitals, commonly two lobes. The lobes dxy orbital lie between
referred to as px, py and pz. These three p-orbitals, x and y axes. Similar is the case for dyz and
posses equivalent energy and therefore, have dxz. Four lobes of dx2  y2 orbital are lying
same relation with the nucleus. They, however, along x and y axes while the two lobes of
differ in their direction & distribution of the charge. dz2 orbital are lying along z axes and con-
y y y tain a ring of negative charge surrounding
z z z the nucleus in xy plane.Geometry of d orbital
is Double Dumb bell
x x x
2 1 0 1 2
px py pz m
d xy d yz d z2 d zx d x2  y2
 ATOMIC STRUCTURE

Shape of d–orbitals :  for clock wise spin/spin up() electron   12

 It implies that d subshell has 5 orbitals i.e. five
electron cloud and can be represented as follows.  for anticlock wise spin/spin down() electron   12
y z z y z
x
h
y x x y Spin angular momentum of an e –– = s s  1.
2

x
x
dx2–y2
dxy dyz dxz dz 2

or s s  1 
In d orbital:
 Each orbital can accomodate 2 electrons with
Nodal opposite spin or spin paired.
y
surface
—
Correct  Spin paired e
—
x Wrong  Spin parallel e
3P
3dxy
x
4P
4dxy
z 5P
x
5dxy
x

ELECTRONIC CONFIGURATION :
Number of nodal surface = n––1 Filling of electron in different energy subshell is
number of nodal plane =  electronic configuration.
total node = n –– 1 + 
= (n - 1) Rules for filling Subshell :
 Nodal plane: dxy  xz & yz nodal plane : 1. Aufbau Principle
dxz  xy & zy nodal plane : 2. (n +) rule
dzy  dzx & yx nodal plane :
dx2–y2  2, nodal plane : 3. Hund's maximum multiplicity principle
dz2  0, nodal plane : 4. Pauli's exclusion principle
Note: Orbitals of d subshell are Equivalent in
energy. 1. Aufbau Principle : Aufbau is a German word
f subshell When = 3 (f subshell)
and its meaning ' Building up'
Then
 Aufbau principle gives a sequence in which various
m  3 2 1 0 1 2 3
subshell are filled up depending on the relative order
fx3 fy3 fxyz fz3 fx(y2  z2 ) fy(z2  x2 ) fz(x2  y2 ) of the Energies of various subshell.

 The structure of f-orbital is very complex  Principle :The subshell with minimum energy is
 The number of values of f-orbitals in f subshell show filled up first when this subshell obtained maximum
that f subshell has 7 orbitals which are equivalent quota of electrons then the next subshell of higher
in Energy. energy starts filling.
Representation of the orbitals :  The sequence in which various subshell are filled
s are as follows.
s subshell  s

p 1s 2s 3s 4s 5s 6s 7s
p subshell  x y pz
p p Starting
point 2p 3p 4p 5p 6p 7p
d
d subshell  dxy dyz dz dxz dx – y 2 2 2

3d 4d 5d 6d
f
f subshell  fx fy fz fxyz fx(y–z )
3 3 3 2 2 fy(z–x )
2 2 fz(x–y )
2 2
4f 5f

4. Spin Quantam number (s) : Given by Gold

schmidt) 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2,
It represents the direction of electron Spin around 4f14, 5d10, 6p6,7s2, 5f14, 6d10 , ....
its own axis
 ATOMIC STRUCTURE

For Ex. 2. ( n+) rule : According to it the sequence in which

1
H  1s1 various subshell are filled up can also be
2
He  1s2 determined with the help of ( n +) value for a given
3
Li  1s2 , 2s1 subshell.
4
Be  1s2, 2s2 Principle of ( n+) rule : The subshell with lowest
5
B  1s2, 2s2, 2p1 (n+) value is filled up first, When two or more
6
C  1s2 , 2s 2,2p2 subshell have same (n+) value then the subshell
N  1s2 , 2s2 , 2p3
7 with lowest value of n is filled up first.
O  1s2 , 2s2 , 2p4
8 In case of H-atom : Energy only depends on
F  1s2 , 2s2 , 2p5
9 principle quantum number
Ne  1s2 , 2s2 , 2p6
10 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f <
Na  1s2 , 2s2 , 2p6 , 3s1
11 ...................
Mg  1s2 , 2s2 , 2p6 , 3s2
12
n  n +
A
Sub Shell
13
 1s2 , 2s2 , 2p6 , 3s2 , 3p1
14
Si  1s2 , 2s2 , 2p6 , 3s2 , 3p2 1s 1 0 1
p  1s2 , 2s2 , 2p6 , 3s2 , 3p3
15 2s 2 0 2
S  1s2 , 2s2 , 2p6 , 3s2 , 3p4
16
Cl  1s2 , 2s2 , 2p6 , 3s2 , 3p5 2p 2 1 3 (1)
3
17

18
Ar  1s2 , 2s2 , 2p6 , 3s2 , 3p6 3s 3 0 (2)
K  1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s1
19
3p 3 1 4 (1)
Ca  1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2
4 (2)
20

21
Sc  1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d1 4s 4 0
Ti  1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d2
22 3d 3 2 5 (1)
23
V  1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d3 
Cr  1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s1, 3d5 4p 4 1 5 (2)
24
[Exception] 5s 5 0 5 (3)
25
Mn  1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d5
Fe  1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d6 4d 4 2 6 (1)

26

27
Co  1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d7 5p 5 1 6 (2)
Ni  1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d8 6 (3)
28 6s 6 0
29
Cu  1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s1, 3d10
[Exception] Order : 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2,
4d10, 5p6, 6s2, 4f14, 5d10, 6p6,7s2, 5f14, 6d10 , ....
Zn
30
 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d10
Electronic configuration can be written by 3. Hund's Maximum Multiplicity Rule :
following different methods : (Multiplicity : Many of the same kind)
 According to Hund's rule electrons are distributed
 26Fe (1) 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d6
among the Orbitals of subshell in such a way as to
(2) 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 3d6 , 4s2
give maximum number of unpaired electron with
(3) 1s2 , 2s2p6 , 3s2p6d6 , 4s2 parallel spin.
(4) [Ar] 4s2 3d6  Thus the Orbital available in the subshell are first
2 2
1s  6
2s 2p 2 6
3s 3p 3d 
4s 6 2 filled singly with parallel spin electron before they
 Fe n begin to pair this means that pairing of electrons
26 (n–2) (n−1)
occurs with the introduction of second electron in
n  Outer most Shell 's' subshell, fourth electron in 'p' subshell, 6th
or Ultimate Shell or Valence Shell electron in 'd' Subshell & 8th e– in 'f' subshell.
In this Shell e– are Called as Valance electron or Ex. 5
B 1s2 2s2 2p1
this is called core charge
  
(n–1)  Penultimate Shell or core or
pre valence Shell 6
C 1s2 2s2 2p2
(n–2)  Pre Penultimate Shell    
 If we remove the last n Shell (ultimate Shell) then N 1s2 2s2 2p3
7
the remaining shell collectivly be called as Kernal.
    
1s2 2s22p6 3s2 3p6 3d6 4s2
26 Fe →  O 1s2 2s2 2p4
Ex. 8
Kernal
    
 ATOMIC STRUCTURE

9
F 1s2 2s2 2p5 2. Which of the following statements is correct for
     an electron having azimuthal quantum number
Ne  1s2 2s2 2p6 l = 2?
10
(1) The electron may be in the lowest energy
     shell.
4. Pauli's Exclusion principle : In 1925 Pauli (2) The electron is in a spherical orbital.
stated that no two electron in an atom can have 1
same values of all four quantum numbers.i.e. An (3) The electron must have spin ms = +
2
orbital can accomodates maximum 2 electrons with (4) The electron may have a magnetic quantum
opposite spin. number = – 1
Ex.1. 6C12  1s2 2s2 2p2
3. Four electrons in an atom have the sets of quan-
    tum numbers as given below. Which electron in
px pz py at the highest energy level?
n 1 2 2 (1) n = 4, l = 0, ml = 0, ms = +1/2
 0 0 1 (2) n = 3, l = 1, ml = 0, ms = –1/2
m 0 0 + 1, – 1, 0 (3) n = 3, l = 2, ml = 0, ms = +1/2
s  12 ,  12  12 ,  12  12 ,  12 , (4) n = 4, l = 1, ml = –1, ms = –1/2

4. 2 (psi) the wave function represents the prob-

Ex.2
ability of finding electron, Its values depends :
Cl  1s2 2s2 2p6 3s2 3P5
17 (1) inside the nucleus
             (2) far from the nucleus
n = 1 2 2 3 3 (3) near the nucleus
 = 0 0 1 0 1 (4) upon the type of orbital
m= 0 0 +1, –1, 0 0 +1, –1, 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
5. Which set of quantum numbers is possible for
 ,  ,  ,  ,  ,  ,  ,  , 
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
the last electron of Mg+ ion ?
Exception of Aufbau principle : In some cases it is (1) n = 3, l = 2, m = 0, s = +1/2
seen that the electronic configuration is slightly (2) n = 2, l = 3, m = 0, s = +1/2
different from the arrangement given by Aufbau (3) n = 1, l = 0, m = 0, s = +1/2
principle. A simple region behind this is that half (4) n = 3, l = 0, m = 0, s = +1/2
filled & full filled subshell have got extra stability.
6. In any subshell, the maximum number of elec-
trons having same value of spin quantum num-
ber is :
(1) (  + 1) (2) l + 2
(3) 2l + 1 (4) 4l + 2
C
O
N
C
T
P 7.
E The electronic configuration of silver atom in ground
state is :
(1) [Ar]3d10, 4s1 (2) [Xr]4f14, 5d10,6s1
10 1
(3) [Kr]4d , 5s (4) [Kr]4d9, 5s2
APPLICATION EXERCISE-4 8. Which of the following statement is (are) correct?
1. Which of the following set of quantum number (1) The electronic configuration of Cr is
is impossible for an electron ? [Ar]3d5,4s1 (Atomic no. of Cr = 24)
1 (2) The magnetic quantum number may have a
(1) n = 1, l = 0, ml = 0, ms = + negative value
2
(3) In silver atom, 23 electrons have a spin of
1
(2) n = 9, l = 7, ml = –6, ms = – one type and 24 of the opposite type,
2
(Atomic no. of Ag =47)
1
(3) n = 2, l = 1, ml = 0, ms = + (4) All of the above
2
9. A subshell n = 5, l = 3 can accommodate :
1
(4) n = 3, l = 2, ml = –3, ms = + (1) 10 electrons (2) 14 electrons
2
(3) 18 electrons (4) None of these
 ATOMIC STRUCTURE

EXTRA STABILITY OF HALF

FILLED AND COMPLETELY
d5 (1)
FILLED ORBITALS
Half-filled and completely filled sub-shells have 4 exchanges by 1st e¯
extra stability due to the following reasons.
Symmetry of orbitals :
(2)
(a) It is a well known fact that symmetry leads
to stability.
3 exchanges by 2nd e¯
(b) Thus, if the shift of an electron from one orbital
to another orbital differing slightlyi n energy (3)
results in the symmetrical electronic
configuration. it becomes more stable
2 exchanges by 3rd e¯
(c) For example p3, d5, f7 confiogurations are more
stable than their near ones (D)
Exchange Energy
1 exchange by 4th e¯
(a) The e¯ in various subshells can exchange their
positions, since e¯ in the same subshell have Total number of possible exchanges
equal energies. = 4 + 3 + 2 + 1 = 10
(b) The energy is released during the exchange
process with in the same subshell. Probability distribution curve

(c) In case of half filled and completely filled orbitals, The orbital wave funtion doesnot have any physi-
the exchange energy is maximum and is greater cal significance but the square of () provides the
then the loss of orbital energy due to the transfer information regarding the probability of electron at
of electron from a higher to a lower sublevel e.g. a point in an atom.
from 4s to 3d orbitals in case of Cu and Cr
To draw the representation of variation of 2 in
(d) The greater the number of possible exchanges space we required following functions :
between the electrons of parallel spins present
(a) Radial wave function
in the degenerate orbitals, the higher would be
the amount of energy released and more will (b) Radial probability density (R2)
be the stability (c) Radial probability function (4r2R2)
(e) Let us count the number of exchange that are
possible in d4 and d5 configuration among
(a) Radial wave functions (R) : By drawing these
electrons with parallel spins :
curve we can find the node in 2s, radial function. At
node as figure indicates the value of radial function
d4 (1)
changes from positive to negative.
3 exchanges by 1st e¯ By drawing these we can get the information that how
the radial wave function changes with distance r.
(2)

2 Exhanges by 2nd e¯
1s 2s 2p

(3) R R R

only 1 exchange by 3rd e¯ Node

Total number of possible exchanges r r r
= 3 + 2 + 1 = 6
 ATOMIC STRUCTURE

(b) Radial probability density (R2) : The square of (c) Radial probability function (4r2R2) : The shape
radial wave function R2 for an orbital give the radial of an atom is assumed to be spherical so it better
to dicuss the probability of finding the electron in a
density & this radial density give the probability
speherical shell between the radius (r + dr) and r.
density of finding the electron. This probability which is independent of direction is
By drawing the curve we can obtain the useful in- called radial probability & equal to (4r2 dr, R2)
By drawing the curve we can obtain the informa-
formation about probability density or relative elec-
tion regarding the variation of radial probabiility func-
tron density at a point as a function of radius.
tion (4r2, R2) with distance r.
1s 2s 2p
1s 2s 2p

4r2R2

2
4r R

4r R
2 2 2
R R R

2
Node

r r r r r r

CONCEPT APPLICATION EXERCISE

ANSWER KEY
EXERCISE-1

1. (3) 2. (2) 3. (4) 4. (1) 5. (3) 6. (3) 7. (4)

EXERCISE-2

1. (1) 2. (1) 3. (3) 4. (3) 5. (1) 6. (3) 7. (2)

EXERCISE-3

1. (4) 2. (3) 3. (1) 4. (2) 5. (1) 6. (3) 7. (3)

8. (4)
EXERCISE-4

1. (4) 2. (4) 3. (4) 4. (4) 5. (4) 6. (3) 7. (3)

8. (4) 9. (2)
 ATOMIC STRUCTURE

EXERCISE-I
ATOMIC NUMBER, MASS NUMBER AND 6. Atomic weight of Ne is 20.2. Ne is mixture of Ne20
IMPORTANT DEFINITIONS and Ne22, Relative abundance of heavier isotope is :-

1. Which one of the following consitutes a group of (1) 90 (2) 20

the isoelectronic species? (3) 40 (4) 10
2- -
(1) C 2 ,O 2 ,CO,NO
7. Naturally occurring boron has two isotopes whose
+ 2- -
(2) N O ,C 2 ,C N ,N 2 atomic weights are 10.00 (I) and 11.00 (II). Atomic
weight of natural boron is 10.80. The percentage
- 2- 2-
(3) C N ,N 2 ,O 2 ,C 2 of isotopes (I) and (II) respectively is :-
- + (1) 20 and 80 (2) 10 and 20
(4) N 2 ,O 2 ,NO ,CO
(3) 15 and 75 (4) 30 and 70
2. (i) 26Fe54,26Fe56,26Fe57,26Fe28 (a) Isotopes
8. Let mass of electron is half, mass of proton is
(ii) 1H3, 2He3 (b) Isotones
two times and mass of neutron is three fourth of
(iii) 32Ge76, 33As77 (c) Isodiaphers orignal. The find out new atomic wt. of O16 atom:-
(iv) 92U235, 90Th231 (d) Isobars (1) increases by 37.5%
(v) 1H1, 1D2, 1T3 (2) Remain constant
Match the above correct terms:- (3) increases by 12.5%
(1) [(i), - a], [(ii) - d], [(iii) - b], [(iv) - c], [(v) - a] (4) decreases by 25%
(2) [(i) - a] [(ii) - d], [(iii) - d] [(iv) - c] [v - a]
(3) [v -a] [(iv) - c]. [(iii) - d] [(ii) - b] [(i) - a] 9. In 7N14 if mass attributed to electron were doubled
& the mass attributed to protons were halved,
(4) None of them
the atomic mass would become approximately:-
3. The atom A, B, C have the configuration (1) Halved
A  [Z(90) + n(146)], B  [Z(92) + n(146)], (2) Doubled
C  [Z(90) + n(148)] So that :- (3) Reduced by 25%
(a) A and C - Isotones (b) A and C - Isotopes (4) Remain same
(c) A and B - Isobars (d) B and C - Isobars
(e) B and C - Isotopes 10. A certain negative ion X–2 has in its nucleus
18 neutrons and 18 electrons in its extra nuclear
The wrong statement's are:-
structure. What is the mass number of the most
(1) a, b only (2) c, d, e only abundant isotope of 'X' :-
(3) a, c, d only (4) a, c, e only (1) 35.46 (2) 32
(3) 36 (4) 39
4. In an atom 13Al27. number of protons is (a) electron
is (b) and neutron is (c). Hence ratio will be 11. Isotopes of an element have
[in order c : b : a] (1) similar chemical properties but different
(1) 13 : 14 : 13 (2) 13 : 13 : 14 physical properties
(2) similar chemical and physical properties
(3) 14 : 13 : 13 (4) 14 : 13 : 14
(3) similar physical properties but different
chemical properties.
5. The relative abundance of two rubidium isotopes
(4) different chemical and physical properties.
of atomic weights 85 and 87 are 75% and 25%
respectively. The average atomic wt. of rubidium
12. The charge to mass ratio of protons is
is:–
(1) 9.55 × 10–4 C/g (2) 9.55 × 104 C/g
(1) 75.5 (2) 85.5 8
(3) 1.76 × 10 C/g (4) 1.76 × 10–8 m
(3) 86.5 (4) 87.5
 ATOMIC STRUCTURE

13. The charge to mass ratio of -particles is 21. The ratio of the energy of a photon of 2000 Å
approximately ..... the charge to mass ratio of wavelength radiation to that of 4000 Å radiation
is
protons
(1) 1 / 4 (2) 4
(1) twice (2) half
(3) 1 / 2 (4) 2
(3) four times (4) Six times

ELECTROMAGNETIC WAVES AND BOHR’S ATOMIC MODEL

PLANCK’S QUANTUM THEORY
22. The expression for calculation of velocity is
14. The energy of a photon of radiation having
wavelength 3000Å is nearly  KZe2 
(1) 6.63 × 10 J –19
(2) 6.63 × 10 J –18
(1) v   
(3) 6.63 × 10–16 J (4) 6.63 × 10–49 J
 mr 

15. Energy of a photon having wave number 2KZe2

(2) v =
1.00 cm–1 is nh
(1) 6.62 × 10–34 J (2) 1.99 × 10–23 J nh
(3) v =
(3) 6.62 × 10–32 J (4) 6.62 × 10–36 J 2KZe 2
(4) Both (1) & (2)
16. The frequency of wave is 6 × 1015 s–1. Its wave
number would be
23. If the velocities of first, second, third and fourth
(1) 105 cm–1 (2) 2 × 10–5 cm–1
orbits of hydrogen atom are v1, v2, v3 and v4
(3) 2 × 10–7 cm (4) 2 ×105 cm–1 respectively, then which of the following should
be their increasing order?
17. The momentum of a photon of frequency
(1) v1 > v2 > v3 > v4 (2) v4 < v3 < v1 < v2
5 × 1017 s–1 is nearly:
(3) v1 > v2 < v3 > v4 (4) Equal for all
(1) 1.1 × 10–24 kg ms–1
(2) 3.33 × 10–43 kg ms–1
24. If v1, v2, v3 and v4 are velocities of the electron
(3) 2.27 × 10–40 kg ms–1 present in the first orbit of H, He+, Li+2 and Be+3,
(4) 2.27 × 10–38 kg ms–1 then which of the following should be their
increasing order?
18. The number of photons of light having wavelength (1) v1 < v2 < v3 < v4 (2) v4 = v3 = v2 = v1
100 nm which can provide 1.00 J energy is nearly
(3) v1 < v2 < v3 > v4 (4) v1 > v2 < v3 > v4
(1) 107 photons (2) 5 × 1018 photons
(3) 5 × 1017 photons (4) 5 × 107 photons 25. Which orbits of H, He+ and Li+2 have identical
energies?
19. How many times does light travel faster in (1) Second orbits of all the three
vacuum than an electron in Bohr first orbit of
(2) First orbit of H, second orbit of He+ and third
hydrogen atom?
orbit of Li+2
(1) 13.7 times (2) 67 times
(3) Third orbit of all the three
(3) 137 times (4) 97 times
(4) Fourth orbit of H, First orbit of He+ and Fifth
orbit of Li+2
20. The value of Planck’s constant is 6.63 × 10–34 Js.
The velocity of light is 3.0 × 108 ms–1, Which
value is closest to the wavelength in nanometers 26. What should be the correct order of energies of
of a quantum of light with frequency of the first orbits of H, He+ and Li+2, if these are
8 × 1015 s –1? represented as E1, E2 and E3 respectively?

(1) 5 × 10 –18 (2) 4 × 101 (1) E1 < E2 < E3 (2) E3 < E2 < E1

(3) 3 × 107 (4) 2 × 10–25 (3) E1 < E2 > E3 (4) E1 = E2 = E3

 ATOMIC STRUCTURE

27. What should be the ratio of energies of the fifth 35. The ratio of the radii of first orbits of H, He+ and
orbits of Li+2 and He+? Li+2 is :
(1) 4 : 9 (2) 9 : 4 (1) 1 : 2 : 3 (2) 6 : 3 : 1
(3) 12 : 16 (4) 7 : 2 (3) 9 : 4 : 1 (4) 6 : 3 : 2

28. The ratio of radii of the third and fifth orbits of Li+2 36. For the energy levels in an atom, which one of
will be the fallowing statement is correct?

(1) 9 : 25 (2) 25 : 9 (1) There are seven principal electron energy

levels
(3) 2 : 3 (4) 4 : 5
(2) the second principal energy level has four
orbitals and contains a maximum of eight
29. What should be the circumference of the second
electrons
orbit of hydrogen atom?
(3) The M energy level can have a maximum of
(1) 13.3 Å (2) 3.7 Å 32 electrons
(3) 26.6 Å (4) 3.4 Å (4) The 4s sub-energy level has higher energy
that 3d sub-energy level.
30. The diameter of the second orbit of hydrogen atom
should be: 37. The approximate quantum number of a circular
(1) 2.12 Å (2) 4.23 Å orbit of diameter 20.6 nm of the hydrogen atom
(3) 2.10 Å (4) 4.01 Å according to Bohr’s theory is :
(1) 10 (2) 14
31. What should be the ratio of the radii of the orbits (3) 12 (4) 16
of electron in Na+10 and hydrogen atom?
38. The ratio of the radius of the atom to the radius of
(1) 11:1 (2) 1 : 11
the nucleus is of the order of
(3) 1 : 1 (4) 1 : 2
(1) 105 (2) 106
(3) 10–5 (4) 10–8
32. What should be the velocity of the electron
present in the fourth orbit of hydrogen atom, if 39. The first use of quantum theory to explain the
the velocity of the electron present in its first orbit structure of atom was made by
is 2.188 × 10+8 cm per second?
(1) Heisenberg (2) Bohr
(1) 1.094 × 108 cm per second
(3) Planck (4) Einstien
(2) 5.47 × 107 cm per second
(3) 4.376 × 108 cm per second 40. The energy of an electron in the first Bohr orbit of
(4) 2.188 × 106 cm per second H atom is–13.6 eV. The possible energy value(s)
of the excited state for electrons in Bohr orbits of
Li2+ is
33. What should be the velocity of the electrons
present in the first, second and third orbits of H, (1) –61.2 eV (2) –13.6 eV
He+ and Li+2, respectively? (3) –122.4 eV (4) +6.8 eV
(1) 2.188 × 108 cm per second
41. In a hydrogen atom, if the energy of electron in
(2) 5.47 × 107 cm per second
the ground state is –13.6 eV, then that in the 2nd
(3) 4.376 × 108 cm per second excited state is:
(4) 2.188 × 106 cm per second (1) –1.51 eV (2) – 3.4 eV
(3) –6.04 eV (4) –13.6 eV
34. What should be the ratio of energies of the third
and fifth orbits of He+?
42. In hydrogen atom energy of first excited state is
(1) 25 : 9 –3.4 eV. Then find out the K.E. of the electron in
(2) 5 : 3 the same orbit of hydrogen atom:
(3) 16 : 9 (1) 3.4 eV (2) + 6.8 eV
(4) None of these (3) –13.6 eV (4) +13.6 eV
 ATOMIC STRUCTURE

43. According to Bohr ’s theory, the angular 51. Potential energy is – 27.2 eV in second orbit of
momentum of an electron in 5th orbital is: He+ then calculate, double of total energy in first
excited state of hydrogen atom :
(1) 25 h/ (2) 1.0 h/
(1) – 13.6 eV (2) – 54.4 eV
(3) 10 h/ (4) 2.5 h/
(3) – 6.8 eV (4) – 27.2 eV

44. Angular momentum for P-shell electron is : 52. The graphical representation of energy of e– and
3h atomic number is :
(1) (2) Zero
π

2h
(3) (4) None
2π (1) (2)

45. Multiplication of electron velocity and radius for a

orbit in an atom is :
(1) Proportional to mass of electron
2
(2) Proportional to square of mass of electron Z
(3) Inversely proportional to mass of electron (3) (4) E
(4) Does not depend upon mass of electron

46. The radius of a shell for H-atom is 4.761Å. The

value of n is : 53. Maximum frequency of emission is obtained for
(1) 3 (2) 9 the transition :
(3) 5 (4) 4 (1) n = 2 to n = 1 (2) n = 6 to n = 2
(3) n = 1 to n = 2 (4) n = 2 to n = 6
47. The ratio of the radii of two Bohr orbits of H-atoms
is 4 : 1, what would be their nomenclature: 54. If the ionization energy of hydrogen is 313.8 K
cal per mole, then the enrgy of the electron in 2nd
(1) K & L (2) L & K
excited state will be :
(3) N & L (4) 2 & 3 both
(1) – 113.2 Kcal/mole (2) – 78.45 Kcal/mole
48. The velocity of electron in third excited state of (3) – 313.8 Kcal/mole (4) – 35 Kcal/mole
Be3+ ion will be :
55. Which of the following electron transition will
3 require the largest amount of energy in a hydrogen
(1) (2.188 × 108)ms–1 atom :
4
(1) From n = 1 to n = 2
3 (2) From n = 2 to n = 3
(2) (2.188 × 106)ms–1
4
(3) From n =  to n = 1
(3) (2.188 × 106) Kms–1 (4) From n = 3 to n = 5
(4) (2.188 × 103) Kms–1
56. A single electron orbits a stationary nucleus
49. The energy of H-atom in nth orbit is En then energy (z = 5). The energy required to excite the electron
in nth orbit of singly ionised helium atom will be : from third to fourth Bohr orbit will be :
(1) 4En (2) En/4 (1) 4.5 eV (2) 8.53 eV
(3) 2En (4) En/2 (3) 25 eV (4) 16.53 eV

50. The energy of second Bohr orbit of the hydrogen 57. En = – 313.6/n2. If the value of En = – 34.84 then
atom is – 328 kJ/mol. Hence the energy of fourth to which of the following values does ‘n’
Bohr orbit should be : correspond :
(1) 1 (2) 2
(1) – 41 KJ/mol (2) – 1312 KJ/mol
(3) – 164 KJ/mol (4) – 82 KJ/mol (3) 3 (4) 4
 ATOMIC STRUCTURE

58. Which of the following is a correct relationship : 64. The radius of which of the following orbit is same
(1) E1 of H = 1/2 E2 of He+ = 1/3m, E3 of Li+2 = 1/4, as that of the first Bohr’s orbit of hydrogen atom?
E4 of Be+3 (1) He+ (n = 2) (2) Li2+ (n = 2)
(3) Li2+ (n = 3) (4) Be3+ (n = 2)
(2) E1(H) = E2(He+) = E3(Li+2) = E4(Be+3)
(3) E1(H) = 2E2(He+) = 3E3(LI+2) = 4E4(Be+3)
(4) No relation SPECTRUM AND SPECTRUM LINE

59. Energy required to remove an e– from M shell of 65. What should be the correct order of energies, if
H-atom is 1.51 eV, then energy of first excited E1, E2, E3 and E4 are the energies of Lyman,
state will be : Balmer, Paschen and Brackett series,
respectively.
(1) – 1.51 eV (2) +1.51 eV
(1) E1 > E2 > E3 > E4 (2) E4 > E3 > E2 > E1
(3) – 3.4 eV (4) – 13.6 eV
(3) E1 < E3 < E2 < E4 (4) E4 < E2 < E3 < E1

60. The ionisation energy for excited hydrogen atom

66. In Bohr series of lines of hydrogen spectrum, the
in eV will be :
third line from the red end corresponds to which
(1) 13.6 (2) Less than 13.6 one of the following inter-orbit jumps of the
(3) Greater than 13.6 (4) 3.4 or less electron for Bohr’ orbits in an atom of hydrogen?
(1) 2  5 (2) 3  2
61. The energy required to excite an electron of H-
atom from first orbit to second orbit is : (3) 5  2 (4) 4  1

3
(1) of its ionisation energy 67. The spectrum of He+ is expected to be similar to
4
that of
1 (1) Li+ (2) H (3) Na (4) He
(2) of its ionisation energy
2

1 68. Which of the following should be the expression

(3) of its ionisation energy for the last line of Paschen series?
4
1 1 1  1 1 1
(4) None
(1)  R  2  (2)  R  
 9    4 9
62. Which of the following is a correct graph :

1 1 1  1  1 1
(3)  R   (4)  R  
K.E.   9 16    16  
(1) (2)
69. Which of the following should be the correct value
n of the wave number of first line in Balmer series
of hydrogen atom?
(1) 5R/36 (2) –5R/36
K.E. (3) R/9 (4) –R/9

(3) (4)
70. What should be the frequency of radiation of the
n emission spectrum when the electron present in
63. The energy of an electron in the first Bohr orbit of hydrogen atom undergoes transition from n = 3
energy level to the ground state?
H atom is  13.6 eV . The possible energy
value(s) of the excited state(s) for electrons in (1) 3 × 1015 second–1
Bohr orbits of hydrogen is/are : (2) 3 × 105 second–1
(1)  3.4 eV (2)  4.2 eV (3) 3 × 1010 second–1
(3)  6.8 eV (4) + 6.8 eV (4) 3 × 108 second–1
 ATOMIC STRUCTURE

71. What should be the quantum number of the 79. If H-atom is supplied with 12.1 eV energy and
highest energy state when an electron falls from electron returns to the ground state after excitation
the highest energy state to Lyman series and for the number of spectral lines in Balmer series
this transition the wave number will be would be :
= 97492.2cm–1? (use energy of ground state of H-atom = – 13.6 eV)
(1) 2 (2) 3 (3) 4 (4) 5 (1) 1 (2) 2
(3) 3 (4) 4
72. The ratio of minimum frequency of Lyman &
Balmer series will be :
80. If the shortest wavelength of Lyman series of H
(1) 1.25 (2) 0.25 atom is x, then the wave length of first line of
(3) 5.4 (4) 10 Balmer series of H atom will be :

73. The wavelength of the radiation emitted, when in 9x 36x

(1) (2)
a hydrogen atom electron falls from infinity to 5 5
stationary state 1 would be
5x 5x
(Rydberg constant = 1.097 × 107 m–1) (3) (4)
9 36
(1) 91 nm (2) 192 nm
(3) 406 nm (4) 9.1 × 10–8 nm 81. What transition in He+ will have the same  as
the I line in Lyman series of H - atom :
74. The ratio of minimum wavelengths of Lyman &
(1) 5  3 (2) 3  2
Balmer series will be :
(3) 6  4 (4) 4  2
(1) 1.25 (2) 0.25
(3) 5 (4) 10 82. In H-atom, electron transits from 6th orbit to 2nd
orbit in multi step. Then total spectral lines
75. Which transition emits photon of maximum (without Balmer series) will be :
frequency :
(1) 6 (2) 10
(1) second spectral line of Balmer series
(3) 4 (4) 0
(2) second spectral line of Paschen series
(3) fifth spectral line of Humphery series 83. An atom has x energy level, then total number of
(4) first spectral line of Lyman series lines in its spectrum are :
(1) 1 + 2 + 3 ..... (x + 1)
76. The wavelength of photon obtained by electron (2) 1 + 2 + 3 ... (x2)
transition between two levels in H-atom and singly
(3) 1 + 2 + 3 .......(x – 1)
ionised He are 1 and 2 respectively, then :
(4) (x + 1)(x + 2)(x + 3)
(1) 2 = 1 (2) 2 = 21
(3) 2 = 1/2 (4) 2 =1/4 84. The figure indicates the energy level diagram for
the origin of six spectral lines in emission
77. Find out ratio of following for photon spectrum(e.g line no. 5 arises from the transition
(vmax.)Lyman : (vmax)Bracket from level B to X) which of the following spectral
(1) 1 : 16 (2) 16 : 1 lines will not occur in the absorption spectrum :
(3) 4 : 1 (4) 1 : 4 C
B
78. The limiting line in Balmer series will have a
X
frequency of :
A
(1) 3.65 × 1014 sec–1 1 2 3 4 5 6
15 –1
(2) 3.29 × 10 sec (1) 1, 2, 3 (2) 3, 2
(3) 8.22 × 1014 sec–1 (3) 4, 5, 6 (4) 3, 2, 1
(4) – 8.22 × 1014 sec–1
 ATOMIC STRUCTURE

85. A certain electronic transition from an excited 93. If the radius of first Bohr orbit of hydrogen atom
state to ground state of the H-atom in one or is ‘x’ then de Broglie wavelength of electron in
more step gives rise to three lines in the ultra 3rd orbit is nearly:
violet region of the spectrum. How many lines (1) 2x (2) 6x
does this transition produce in the infrared region x
of the spectrum : (3) 9x (4)
3
(1) 1 (2) 2
(3) 3 (4) 4 94. What is the ratio of the De-Broglie wave lengths
for electrons accelerated through 200 volts and
86. Which electronic level would allow the hydrogen 50 volts :-
atom to absorb a photon but not to emit a photon
(1) 3s (2) 2p (1) 1 : 2 (2) 2 : 1
(3) 2s (4) 1s (3) 3 : 10 (4) 10 : 3

87. An electron in a hydrogen atom in its ground state 95. A particle X moving with a certain velocity has
absorbs energy equal to the ionisation energy of a debroglie wavelength of 1Å. If particle Y has
Li+2. The wavelength of the emitted electron is: a mass of 25% that of X and velocity 75% that
(1) 3.32 ×10–10 m (2) 1.17 Å
of X, debroglies wavelength of Y will be :-
(3) 2.32 × 10–9 nm (4) 3.33 pm
(1) 3Å (2) 5.33 Å
88. The energy photon emitted corresponding to (3) 6.88 Å (4) 48Å
transition n = 3 to n = 1 is :
[h = 6 ×10–34 J-sec.]
–18 96. The wavelength associated with a golf ball
(1) 1.76 ×10 J (2) 1.98 ×10–18 J
–17 weighing 200 g and moving at a speed of 5 m/h
(3) 1.76 ×10 J (4) None of these
is of the order
(1) 10–10m (2) 10–20m
89. The difference in the wavelength of the 1st line of
–30
Lyman series and 2nd line of Balmer series in a (3) 10 m (4) 10–40 m
hydrogen atom is
9 4 97. The de-Broglie wavelength of a tennis ball of mass
(1) (2) 60 g moving with a velocity of 10 metres per
2R R
second is approximately
88 [Planck’s constant, h = 6.63 × 10–34 Js]
(3) (4) None
15R (1) 10–25 metres (2) 10–33 metres
(3) 10–31 metres (4) 10–16 metre
90. The wave number of electromagnetic radiation
emitted during the transition of electron in
between two levels of Li2+ ion whose principal 98. A ball of 100 g mass is thrown with a velocity of
quantum numbers sum is 4 and difference is 2 is 100 ms–1. The wavelength of the de Broglie wave
(1) 3.5 R (2) 4 R associated with the ball is about
(1) 6.63 × 10–35 m (2) 6.63 × 10–30 m
8
(3) 8 R (4) R (3) 6.63 × 10–35 cm (4) 6.63 × 10–33 m
9
DE-BROGLIE CONCEPT 99. An electron has kinetic energy 2.8 × 10–23 J.
de-Broglie wavelength will be nearly :
91. No. of wave in fourth orbit :– (me = 9.1 × 10–31 kg)
(1) 4 (2) 5 (3) 0 (4) 1 (1) 9.28 × 10–24 m (2) 9.28 × 10–7 m
(3) 9.28 × 10–8 m (4) 9.28 × 10–10 m
92. (n +1) is the principal quantum number of the
energy state for an atom. What are the number
100. Which of the following has least de-Broglie ?
of elliptical orbits associated with it :–
(1) (n – 1) (2) (n + 1) (1) e– (2) p
(3) (n – 2) (4) n (3) CO2 (4) SO2
 ATOMIC STRUCTURE

101. If the de-Broglie wavelength of the fourth Bohr QUANTUM NUMBERS

orbit of hydrogen atom is 4Å, the circumference
of the orbit will be : 109. In an atom, for how many electrons, the quantum
(1) 4Å (2) 4 nm 1
(3) 16 Å (4) 16 nm numbers will be , n = 3, l = 2, m = + 2, s = + :-
2
(1) 18 (2) 6
102. What is the de-Broglie wavelength associated
(3) 24 (4) 1
with the hydrogen electron in its third orbit :
(1) 9.96 × 10–10 cm (2) 9.96 × 10–8 cm 110. For the azimuthal quantum number (l), the total
(3) 9.96 × 104 cm (4) 9.96 × 108 cm number of magnetic quantum number is given
by:-
103. The correct order of wavelength of Hydrogen (1H1), (m  1) (m  1)
Deuterium (1H2) and Tritium (1H3) moving with (1) l = (2) l =
2 2
same kinetic energy is
(1) H > D > T (2) H = D = T (2m  1) (2m  1)
(3) H < D < T (4) H < D > T (3) l = (4) l =
2 2

HEISENBERG UNCERTAINITY PRINCIPLE 111. Spin angular momentum for electron :-

h h
104. For the electron if the uncertainty in velocity is (1) s(s  1) (2) 2s(s  1)
2 2
v, the uncertainty in its position (x) is given
below: h
(3) s(s  2) (4) None
h 2 2
(1) mv (2)
2 hmv
h 2m 112. If l = 3 then type and number of orbital is :–
(3) (4) (1) 3p, 3 (2) 4f, 14
4mv hv
(3) 5f, 7 (4) 3d, 5
105. The uncertainty in momentum of moving particle 113. The total value of m for the electrons (n = 4) is -
is 1.0 × 10–15 kg m s–1, the minimum uncertainty
(1) 4 (2) 8 (3) 16 (4) 32
in its position would be
(1) 5.28 × 10–20 m (2) 5.28 × 10–49 m 114. An electron has magnetic quantum number as -
(3) 6.63 × 10–49 m (4) 6.63 × 10–22 m 3, what is its principal quantum number :-
(1) 1 (2) 2 (3) 3 (4) 4
106. Heisenberg Uncertainity principle is not valid for
(1) Moving electron (2) Motor car 115. The total spin resulting from a d7 configuration
(3) Stationary particles (4) 2 & 3 both is:-

107. The uncertainty in the position of an electron 1 3

(1) (2) 2 (3) 1 (4)
(mass 9.1 × 10–28gm) moving with a velocity of 2 2
3 × 104 cm sec –1, uncertainity in velocity is
0.011% will be:- 116. No. of all subshells of n +  = 7 is:-
(1) 1.92 cm (2) 7.68 cm (1) 4 (2) 5 (3) 6 (4) 7
(3) 0.175 cm (4) 3.84 cm
117. The quantum number of 20th electron of
108. The uncertainity in position of an electron & helium Fe(Z = 26) ion would be :–
atom are same. If the uncertainity in momentum (1) 3, 2, – 2, – ½ (2) 3, 2, 0, ½
for the electron is 32 × 105 , then the uncertainity
(3) 4, 0, 0, + ½ (4) 4, 1, – 1, + ½
in momentum of helium atom will be
(1) 32 × 105 (2) 16 × 105 118. Which orbitlal has two angular nodal planes :-
5
(3) 8 × 10 (4) None (1) s (2) p (3) d (4) f
 ATOMIC STRUCTURE

119. In an atom having 2K, 8L, 8M and 2N electrons, 128. Which atom has as many as s - electron as
p - electron ?
1
the number of electrons with m = 0; S =  are (1) H (2) Mg (3) N (4) Na
2
(1) 6 (2) 2 (3) 8 (4) 16 129. Correct set of four quantum number for valency
(outermost) electron of rubidium (Z = 37) is :
120. Orbital angular momentum of a 3d electron is:-
(1) 5 , 0 , 0 , + 1/2 (2) 5 , 1 , 0 , + 1/2
h h (3) 5 , 1 , 1 , + 1/2 (4) 6 , 0 , 0 , + 1/2
(1) 2 2 (2) 6 2
130. The probability of finding an electron residing
h h in a px orbital is not zero :
(3) (4)
2 4 (1) In the yz plane (2) In the xy plane
(3) In the y direction (4) In the z direction
121. An orbital with = 0 is symmetrical about the:-
(1) x-axis only (2) y-axis only 131.  2 (psi) the wave function represents the
(3) z-axis only (4) The nucleus probability of finding electron . Its value
depends :
122. In n &  are principal and azimuthal quantum no. (1) inside the nucleus
respectively then the expression for calculating (2) far from the nucleus
the total no. of electron in any energy level is:- (3) near the nucleus
 n  n1 (4) upon the type of orbital
(1)  2 ( 2 1) (2)  2 ( 2 1)
0  1 132. The number of nodal planes in a px orbital is :
(1) 1 (2) 2 (3) 3 (4) zero
 n1  n1

(3)  2 ( 2 1) (4)  2 ( 2 1) 133. The orbital angular momentum of an electron
0 0 in p-orbital is:
123. For azimuthal quantum number l = 3, the h
maximum number of electrons will be : (1) zero (2)
2
(1) 2 (2) 6 (3) zero (4) 14
h 1 h
124. Which d - orbital has different shape from rest (3) (4)
of all d - orbitals ? 2 2 2

(1) d (2) d z2 134. The value of Azimuthal quantum number for all
x 2  y2
d xy electrons present in 5p orbitals is
(3) (4) dxz
(1) 4 (2) 5 (3) 2 (4) 1
125. The total number of orbitals in a shell with
principal quantum number ‘n’ is : 135. Which of the following sets of quantum number
(1) 2 n (2) 2 n2 is not possible?
(3) n 2
(4) n + 1 1
(1) n = 3; l = 0 , ml = 0, ms = +
2
126. What is the correct orbital designation for the
1
electron with the quantum numbers , (2) n = 3; l = 0 , ml = 0, ms = –
n = 4 , l = 3 , m = – 2 , s = 1/2
2
(1) 3 s (2) 4 f (3) 5 p (4) 6 s 1
(3) n = 3; l = 0, ml = –1, ms = +
2
127. The total number of electrons that can be 1
accommodated in all the orbitals having (4) n = 3; l = 1, ml = 0, ms = –
principal quantum number 2 and azimuthal
2
quantum number 1 is : 136. Which quantum number is sufficient to determine
(1) 2 (2) 4 (3) 6 (4) 8 the energy of the electron in hydrogen atom?
(1) l (2) n (3) ms (4) ml
 ATOMIC STRUCTURE

1 1 143. For principal quantum number n = 4, the total

137. The quantum numbers + and – for the number of orbitals having l = 3 is
2 2
electron spin represent (1) 3 (2) 7 (3) 5 (4) 9
(1) rotation of the electron in clockwise and
anticlockwise direction respectively 144. Which of the following sets of quantum numbers
(2) rotation of the electron in anticlockwise and represents the highest energy of an atom?
clockwise direction respectively 1
(3) magnetic moment of the electron pointing up (1) n = 3, l = 1, m = 1, s = +
and down respectively 2
(4) two quantum mechanical spin states which 1
have no classical analogue. (2) n = 3, l = 2 m = 1 s = +
2
138. The number of radial nodes in 3s and 2p
respectively are:
1
(3) n = 4, l = 0, m = 0, s = +
(1) 2 and 0 (2) 1 and 2 2
(3) 0 and 2 (4) 2 and 1
1
(4) n = 3, l = 0, m = 0, s = +
139. The following quantum number are possible for 2
how many orbitals?
n = 3, l = 2, m = +2 145 The orbital angular momentum of an electron in
(1) 3 (2) 2 (3) 1 (4) 4 2s orbital is:
140. The 19th electron of chromium has which of the 1 h
following sets of quantum numbers? (1) + . (2) Zero
2 2π
n l m s
1 h h
(1) 3 0 0 (2) (4) 2
2 2π 2π
1
(2) 3 2 –2 ELECTRONIC CONFIGURATION
2
1 146. A transition metal 'X' has a configuration [Ar] 3d5 in
(3) 4 0 0
2 its + 3 oxidation state. Its atomic number is:–
1 (1) 22 (2) 26 (3) 28 (4) 19
(4) 4 1 –1 147. 2
4s is the configuration of the outermost orbit of
2
an element. Its atomic number would be :–
141. The orbital angular momentum for an electron
(1) 29 (2) 24 (3) 30 (4) 19
h
revolving in an orbit is given by l (l  1) . . 148. Aneutral atom of anelementhas 2K,8L,11 M and2N
2 electrons. The number of s-electron in the atom are
This momentum for an s-electron will be given by
h 1 h (1) 2 (2) 8 (3) 10 (4) 6
(1) 2. (2) . 149. The explaination for the presence of three unpaired
2 2 2
electrons in the nitrogen atom can be given by:-
h
(3) zero (4) (1) Pauli's exclusion principle
2
(2) Hund's rule
142. Which of the following sets of quantum numbers
is correct for an electron in 4f-orbital? (3) Aufbau's principle
1 (4) Uncertainty principle
(1) n = 4, l = 3, m = 4, s = +
2 150. n and  values of an orbital 'A' are 3 and 2, of
1 another orbital 'B' are 5 and 0. The energy of
(2) n = 4, l = 4, m = –4, s = – (1) B is more than A
2
1 (2) A is more than B
(3) n = 4, l = 3, m = +1, s = + (3) A and B are of same energy
2
1 (4) None
(4) n = 3, l = 2, m = –2, s = +
2
 ATOMIC STRUCTURE

151. Sum of the paired electrons present in the orbital 160. The maximum number of unpaired electrons are in
with  = 2 in all the species Fe2+, Co2+ and Ni+2 (1) Fe+2 (2) Fe+3 (3) Fe4+ (4) Fe
are:–
161. The number of d-electrons retained in
(1) 9 (2) 12 (3) 6 (4) 15 Fe2+ (At. no. of Fe = 26) ion is :
(1) 6 (2) 3 (3) 4 (4) 5
152. The number of electrons in the M-shell of the
element with atomic number 24 is :- 162. Which of the following ions has the maximum
(1) 24 (2) 12 (3) 8 (4) 13 value of magnetic moment?
(1) Cu+ (2) Cu2+ (3) Fe2+ (4) Fe3+
153. An improbable configuration is :
(1) [Ar] 3 d4 , 4 s 2 (2) [Ar] 3 d5 , 4 s 1 163. The atomic number of an element is 17, the
(3) [Ar] 3 d6 , 4 s 2 (4) [Ar] 3 d10 , 4 s 1 number of orbitals containing electron pairs in
the valency shell is:-
154. Krypton (36Kr) has the electronic configuration
(1) 8 (2) 2 (3) 3 (4) 6
(18Ar) 4 s2 3 d10 4 p6, the 37th electron will go into
which of the following subshells . 164. Which of the following transition neither shows
(1) 4 f (2) 4 d (3) 3 p (4) 5 s
absorption nor emission of energy in case of
155. The electronic configuration of the element Hydrogen atom :–
which is just above the element with atomic (1) 3px 3s (2) 3dxy 3dyz
number 43 in the same particle group is :
(3) 3s 3dxy (4) All the above
(1) 1 s 2 , 2 s 2 2 p6 , 3 s 2 3 p6 3 d10 , 4 s 1 4 p6
(2) 1 s 2 , 2 s 2 2 p6 , 3 s 2 3 p6 3 d5 , 4 s 2 165. In potassium the order of energy level for 19th
electron is :
(3) 1 s 2 , 2 s 2 2 p6 , 3 s 2 3 p6 3 d6 , 4 s 1
(4) 1 s 2 , 2 s 2 2 p6 , 3 s 2 3 p6 3 d10 , 4 s 2 4 p5 (1) 3 s > 3 d (2) 4 s < 3 d
(3) 4 s > 3 p (4) 4 s = 3 d
156. The number of vacant d - orbitals in completely
excited Cl atom is: 166. Remaining part of atom except outer orbit is
(1) 2 (2) 3 (3) 1 (4) 4 called:-
157. Which among the following is correct of 5B in (1) Kernel (2) Core
normal state? (3) Empty space (4) None of these
2s 2p
(1) :
167. For H atom, the energy required for the removal of
Against Hund’s rule electron from various sub-shells is given as under:-

(2) : 3s 3p 3d
E1 n=
Against Aufbau principle as well as Hunds rule 0
E2
0
(3) : E3
0
Violation of Pauli’s exclusion principle and not The order of the energies would be :-
Hund’s rule (1) E1 > E2 > E3 (2) E3 > E2 > E1
(4) : (3) E1 = E2 = E3 (4) None of these
Against Aufbau principle 168. The maximum number of electrons in a sub-shell
158. The atomic orbitals are progressively filled in order is given by the expression
of increasing energy. This principle is called. (1) 4l – 2 (2) 4l + 2
(1) Hund’s rule (2) Aufbau principle (3) 2l + 1 (4) 2n2
(3) Exclusion principle (4) de- Broglie rule.
159. For a given principal level n = 4, the energy of its 169. Total number of nodal planes, angular and
subshells is in the order spherical nodes in 3s-subshell are respectively :
(1) s < p < d < f (2) s > p > d > f (1) zero, zero, 2 (2) 2, 2, 2
(3) s < p < f < d (4) f < p < d < s. (3) zero, zero, zero (4) zero, 2, 2
 ATOMIC STRUCTURE

170. The maximum number of electrons with clockwise 175. Consider the ground state of Cr atom (z = 24).
spin that can be accommodated in a f-sub-shell The numbers of electrons with the azimuthal
is quantum numbers, l = 1 and 2 are, respectively.
(1) 14 (2) 7 (3) 5 (4) 10 (1) 12 and 4 (2) 12 and 5
(3) 16 and 4 (4) 16 and 5
171. In magnesium atom, in ground state, the number
of electrons with m = 0 is: 176. Which of the following statements in relation to
the hydrogen atom is correct?
(1) 4 (2) 6 (3) 2 (4) 8
(1) 3s, 3p and 3d orbitals all have the same
172. In chromium atom, in ground state, the number energy.
of occupied orbital is (2) 3s and 3p orbitals are of lower energy than
(1) 14 (2) 15 (3) 7 (4) 12 3d orbital.
(3) 3p orbital is lower in energy than 3d orbital.
173. A sub-shell with n = 6, l = 2 can accommodate a
(4) 3s orbital is lower is energy than 3p orbital.
maximum of
(1) 12 electrons (2) 36 electrons
(3) 10 electrons (4) 72 electrons

174. How many electrons in argon have m = 0?

(1) 6 (2) 8 (3) 10 (4) 12.

ANSWER KEY [EXERCISE-I]

Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans . 2 1 4 3 2 4 1 1 3 2 1 2 2 1 2
Que . 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Ans . 4 1 3 3 2 4 2 1 1 2 2 2 1 1 2
Que . 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
Ans . 2 2 1 1 4 2 2 1 2 2 1 1 4 1 3
Que . 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans . 1 4 4 1 4 3 4 1 4 1 4 3 2 3 4
Que . 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
Ans . 1 3 1 4 1 3 2 1 1 1 2 3 1 2 4
Que . 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
Ans . 4 2 3 1 2 4 1 3 3 1 4 2 1 2 3
Que . 91 92 93 94 95 96 97 98 99 10 0 10 1 10 2 10 3 10 4 10 5
Ans . 1 4 2 1 2 3 2 1 3 4 3 2 1 3 1
Que . 106 107 108 109 110 111 112 113 11 4 11 5 11 6 11 7 11 8 11 9 12 0
Ans . 4 3 1 4 2 1 3 3 4 4 1 3 3 1 2
Que . 121 122 123 124 125 126 127 128 12 9 13 0 13 1 13 2 13 3 13 4 13 5
Ans . 4 4 4 2 3 2 3 2 1 2 4 1 2 4 3
Que . 136 137 138 139 140 141 142 143 14 4 14 5 14 6 14 7 14 8 14 9 15 0
Ans . 2 4 1 3 3 3 3 2 2 2 2 3 2 2 1
Que . 151 152 153 154 155 156 157 158 15 9 16 0 16 1 16 2 16 3 16 4 16 5
Ans . 2 4 1 4 2 1 3 2 1 2 1 4 3 4 2
Que . 166 167 168 169 170 171 172 173 17 4 17 5 17 6
Ans . 1 3 2 1 2 4 2 3 3 2 1
 ATOMIC STRUCTURE

EXERCISE-II (AIPMT, AIIMS & NEET (2008-2023))

AIPMT 2008 7. If n = 6, the correct sequence for filling of electrons
will be :
(1) ns (n–2)f (n–1)d  np
1. If uncertainty in position and momentum are
(2) ns (n–1)d (n–2)f  np
equal, then uncertainty in velocity is ?
(3) ns (n–2)f np (n–1)d
h 1 h (4) ns np (n–1)d (n–2)f
(1) (2)
 2m 
AIPMT MAINS 2011
h 1 h
(3) (4)
2 m  8. According to the Bohr Theory, which of the
following transitions in the hydrogen atom will give
rise to the least energetic photon?
2. The measurement of the electron position is (1) n = 5 to n = 3 (2) n = 6 to n = 1
associated with an uncertainty in momentum, (3) n = 5 to n = 4 (4) n = 6 to n = 5
which is equal to 1× 10 –18 g cm s –1 . The
uncertainty in electron velocity is :
AIMS 2011
(mass of electron = 9×10–28g)
(1) 1 × 1011 cm s–1 (2) 1 × 109 cm s–1 9. Smallest wavlength occurs for
(3) 1 × 106 cm s–1 (4) 1 × 105 cm s–1 (1) Lyman series (2) Balmar series
(3) Paschen series (4) Brackett series
AIPMT 2009
10. Which of the following is wrong for Bohr model
(1) It establishes stability of atom
3. Maximum number of electrons in a subshell of an (2) It is contradicted with Heisenber uncertainity
atom is determined by the following :- principle
(1) 2n
2
(2) 4 + 2 (3) 2 + 1 (4) 4 – 2 (3) It explain the concept of spectral lines
(4) e– behaves as particle & wave

4. W hich of the following is not permissible

arrangement of electrons in an atom ? AIPMT PRE 2012
(1) n = 3,  = 2, m = –2, s = –1/2
11. Maximum number of electrons in a subshell with
(2) n = 4,  = 0, m = 0, s = –1/2
 = 3 and n = 4 is:
(3) n = 5,  = 3, m = 0, s = +1/2
(1) 10 (2) 12
(4) n = 3,  = 2, m = –3, s = –1/2 (3) 14 (4) 16

AIPMT 2010 12. The correct set of four quantum numbers for the
valence electron of rubidium atom (Z = 37) is:-
(1) 5, 0, 0, + ½ (2) 5, 1, 0, + ½
5. A 0.66 kg ball is moving with a speed of 100 m/s.
(3) 5, 1, 1, + ½ (4) 6, 0, 0 + ½
The associated wavelength will be
–34
(h = 6.6 × 10 Js) :-
–34 –35 AIPMT MAINS 2012
(1) 6.6 × 10 m (2) 1.0 × 10 m
–32 –32
(3) 1.0 × 10 m (4) 6.6 × 10 m
13. The orbital angular momentum of a p-electron is
given as :-
AIPMT PRE 2011
3 h h
(1) (2) 6.
2 π 2π
6. The total number of atomic orbitals in fourth energy
level of an atom is :- h
(3) (4) h
(1) 8 (2) 16 (3) 32 (4) 4 3
2π 2π
 ATOMIC STRUCTURE

AIIMS 2012 AIIMS 2013

14. Threshold frequency of a metal is 5×1013sec–1 upon 19. A particle is moving with 3 times faster than speed
which 1 × 10+14 sec–1 frequency light is focused of e–. Ratio of wavelength of particle & electron
then maximum kinetic energy of emitted is 1.8 × 10–4 then particle L is:-
electron:- (1) Neutron (2) - particle
(1) 3.3 × 10–21 (2) 3.3 × 10–20 (3) Deutron (4) Tritium
(3) 6.6 × 10–21 (4) 6.6 × 10–20
AIPMT 2014
nh
15. In Bohr's orbit indicates :- 20. What is the maximum number of orbitals that can
2
be indetinfied witht he following quantum
(1) Momentum
(2) Kinetic energy numbers? n = 3,  = 1, m = 0

(3) Potential energy (1) 1 (2) 2 (3) 3 (4) 4

(4) Angular momentum
21. Calculate the energy in joule corresponding to light
of wavelength 45 nm :
NEET UG 2013 (Planck’s constant h = 6.63 × 10–34 Js ; speed
of light c = 3 × 108 ms–1)
(1) 6.67 × 1015 (2) 6.67 × 1011
16. The value of Planck’s constant is 6.63 × 10–34 Js. –15
(3) 4.42 × 10 (4) 4.42 × 10–18
The speed of light is 3 × 1017 nms–1. Which value
is closest to the wavelength in nanometer of a
22. Magnetic moment 2.83 BM is given by which of
quantum of light with frequency of 6 × 1015 s–1?
the following ions?
(1) 75 (2) 10 (3) 25 (4) 50
(At. nos. Ti = 22, Cr = 24, Mn = 25, Ni = 28) :-
(1) Ti3+ (2) Ni2+ (3) Cr3+ (4) Mn2+

 Z2 
17. Based on equation E = –2.178 × 10–1Js  n2  AIIMS 2014

certain conclusions are written. Which of them 23. The energy of an electron of 2py orbital is
is not correct? (1) greater than 2p, orbital
(1) For n = 1, the electron has a more negative (2) Less than 2pz orbital
energy than it does for n = 6 which means (3) same as that of 2px and 2pz orbital
that the electron is more lossely bound int (4) Equal to 2s orbital
he smallest allowed orbit.
(2) The negative sign in equation simply means AIPMT 2015
that the energy of electron bound to the
nucleus is lower than it would be if the
24. W hich of the following pairs of ions are
electrons were at the infintite distance form
isoelectronic and isostructural?
the nucleus.
(3) Larger the value of n, the larger is the orbit (1) CIO3− ,CO32− (2) SO32− ,NO3−
radius (3) ClO3− ,SO32− (4) CO32− ,SO32−
(4) Equation can be used to calculate the change
in energy when the electron change orbit. 25. The number of d-electrons in Fe2+(Z = 26) is not
equal to the number of electrons in which one of
18. What is the maximum number of electrons that the following?
can be associated with the following set of (1) p-electrons in Cl (Z=17)
quantum numbers? n = 3 ; l = 1 and m = –1 (2) d-electrons in Fe (Z=26)
(1) 2 (2) 10 (3) 6 (4) 4 (3) p-electrons in Ne (Z=10)
(4) s-electrons in Mg (Z=12)
 ATOMIC STRUCTURE

26. Magnetic moment 2.84 B.M. is given by :– NEET 2017

(At. no.), Ni = 28, Ti = 22, Cr = 24, Co = 27
34. Which one is the wrong statement? :
(1) Ti3+ (2) Cr2+ (3) Co2+ (4) Ni2+
h
(1) de–Broglie's wavelength is given by λ = ;
27. The angular momentum of electron in ‘d’ orbital mv
is equal to :_ where m = mass of the particle, v = group
velocity of the particle.
(1) 2 (2) 2 3 (3) 0 (4) 6
h
(2) The uncertainty principle is ∆E × ∆t ≥ .
4π
RE-AIPMT 2015
(3) Half filled and fully filled orbitals have greater
28. Which is the correct order of increasing energy stability due to greater exchange energy,
of the listed orbitals in the atom of titanium? greater symmetry and more balanced
(At. no. Z = 22) arrangement.
(1) 3s 3p 3d 4s (2) 3s 3p 4s 3d (4) The energy of 2s orbital is less than the
energy of 2p orbital in case of Hydrogen like
(3) 3s 4s 3p 3d (4) 4s 3s 3p 3d
atoms.

AIIMS 2015 NEET 2018

29. In which tranistion of hydrogen atom have same

35. Which one is a wrong statement ?
wavlength as in Balmer series transition of He+
ion(n =4 to n = 2) (1) The electronic configuration of N atom is
(1) 4 to 2 (2) 3 to 2 (3) 2 to 1 (4) 4 to 1

NEET(I) 2016
(2) An orbital is designated by three quantum
30. The electronic configurations of Eu(Atomic No. 63), numbers while an electron in an atom is
Gd(Atomic No 64) and Tb (Atomic No. 65) are designated by four quantum numbers.
(1) [Xe]4f76s2, [Xe]4f86s2 and [Xe]4f85d16s2
(3) Total orbital angular momentum of electron in
(2) [Xe]4f75d16s2, [Xe]4f75d16s2 and [Xe]4f96s2
's' orbital is equal to zero.
(3) [Xe]4f65d16s2, [Xe]4f75d16s2 and [Xe]4f85d16s2
(4) [Xe]4f76s2, [Xe]4f75d16s2 and [Xe]4f96s2 (4) The value of m for d 2 is zero.
z

36. Match the metal ions given in Column I with the

31. Two electrons occupying the same orbital are
spin magnetic moments of the ions given in
distinguished by
column II and assign the correct code :
(1) Principal quantum number
Column I Column II
(2) Magnetic quantum number
(3) Azimuthal quantum number a. Co3+ i. 8 B.M.
(4) Spin quantum number
b. Cr3+ ii. 35 B.M.
NEET(II) 2016
32. Which of the following pairs of d-orbitals will have c. Fe3+ iii. 3 B.M.
electron density along the axis?
d. Ni2+ iv. 24 B.M.
(1) dz2 ,dx2 − y2 (2) dxy ,dx2 − y2
v. 15 B.M.
(3) dz2 ,dxz (4) dxz ,dyz
a b c d
33. How many electrons can fit in the orbital for which (1) iv i ii iii
n = 3 and l = 1? (2) i ii iii iv
(1) 10 (2) 14 (3) 2 (4) 6 (3) iv v ii i
(4) iii v i ii
 ATOMIC STRUCTURE

AIIMS 2018 43. Which of the following series of transitions in the

37. Wave length of particular transition for H atom is spectrum of hydrogen atom falls in visible region ?

400 nm. What can be wavelength of He+ for same (1) Paschen series (2) Brackett series

transition : (3) Lyman series (4) Balmer series

(1) 400 nm (2) 100 nm NEET 2020

(3) 1600 nm (4) 200 nm
44. The number of protons, neutrons and electrons
175
38. A gas metal in bivalent state have approximately in 71 Lu , respectively, are
23e– what is spin magnetic moment in elemental
(1) 71, 104 and 71 (2) 104, 71 and 71
state :
(3) 71, 71 and 104 (4) 175, 104 and 71
(1) 2.87 (2) 5.5 (3) 5.9 (4) 4.9
NEET 2021

39. Whatis maximum wavelengthof line of Balmer series

45. From the following pairs of ions which one is not
of Hydrogen spectrum (R = 1.09 × 107 m–1) :
an iso-electronic pair?
(1) 400 nm (2) 654 nm
(1) Fe2+, Mn2+ (2) O2–, F–
(3) 486 nm (4) 434 nm
(3) Na+, Mg2+ (4) Mn2+, Fe3+

40. In second orbit of H atom what is velocity of e–

46. A particular station of All India Radio, New Delhi,
(1) 2.18 × 106m/sec (2) 3.27 × 106m/sec
broadcasts on a frequency of 1,368 kHz (kilohertz).
(3) 10.9 × 105m/sec (4) 21.8 × 106m/sec
The wavelength of the electromagnetic radiation
emitted by the transmitter is (speed of light,
41. When on metal sheet fall 1 light will eject
c = 3.0 × 108 ms–1 )
electron with v1 velocity and with 2 light eject
(1) 21.92 cm (2) 219.3 m
electron of v2 velocity, what is v 22  v12 value
(3) 219.2 m (4) 2192 m

2hc  1 1 hc  1 1 NEET 2022

(1) m      (2) m     
 2 1  2 1

47. If radius of second Bohr orbit of the He+ ion is

105.8 pm, what is the radius of third Bohr orbit of
2hc  1 1 m  1 1
(3) m      (4) 2hc      Li2+ ion?
 1 2   2 1
(1) 158.7 pm (2) 15.87 pm
(3) 1.587 pm (4) 158.7 Å
NEET 2019

48. Identify the incorrect statement from the following.

42. 4d, 5p, 5f and 6p orbitals are arranged in the or-
(1) All the five 5d orbitals are different in size when
der of decreasing energy. The correct option is :
compared to the respective 4d orbitals.
(1) 6p > 5f > 4d > 5p (2) 5f > 6p > 4d > 5p
(2) All the five 4d orbitals have shapes similar to
(3) 5f > 6p > 5p > 4d (4) 6p > 5f > 5p > 4d
the respective 3d orbitals.
 ATOMIC STRUCTURE

(3) In an atom, all the five 3d orbitals are equal in

energy in free state.

(4) The shapes of dxy, dyz and dzx orbitals similar

to each other ; and dx2 − y2 and dz2 are similar

to each other.
NEET 2023

49. Select the correct Statements from the following

:

A. Atoms of all elements are composed of two

fundamental particles.

B. The mass of the electron is 9.10939×10–31

kg.

C. All the isotopes of a given elements show

same chemical properties.

D. Protons and electrons are collectively known

as nucleons.

E. Dalton’s atomic theory, regarded the atom as

an ultimate particle of matter.

Choose the correct answer from the options

given below.

(1) C,D and E only (2) A and E only

(3) B,C and E only (4) A,B and C only

ANSWER KEY [EXERCISE-II]

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 2 2 2 4 2 2 1 4 1 4 3 1 3 2 4
Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Ans. 4 1 1 1 1 4 2 3 3 1 4 4 2 3 4
Que. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
Ans. 4 1 3 4 1 3 2 3 2 3 1 3 4 1 1
Que. 46 47 48 49
Ans. 2 1 4 3
 ATOMIC STRUCTURE

EXERCISE-III
1. Radius of H-atom in its ground state is 6. The ionization energy of the ground state hydrogen
5.3 × 10–11 m. After collision with an electron it atom is 2.18 × 10-18 J. The energy of an electron
is found to have a radius of 21.2 × 10–11m. What in second orbit of He+ will be
is the principal quantum no. 'n' of the final state (1) –1.09 × 10-18 J (2) – 4.36 × 10-18 J
of the atom:-
(3) – 2.18 × 10-18 J (4) –2.18 × 10-18 J
(1) n = 2 (2) n = 3
(3) n = 4 (4) n = 16
7. If kinetic energy of a proton is increased nine
times the wavelength of the de-Broglie wave
2. What should be the wavelength and energy
associated with it would become
respectively of the emitted light when the electron
of hydrogen atom undergoes transition from first (1) 3 times (2) 9 times
excited state to ground state? (3) 1/3 times (4) 1/9 times
(1) 1215 Å and 21.8 × 10–12 erg
(2) 6.560 Å and 16.35 × 10–12 erg 8. An ion Mna+ has the magnetic moment equal
(3) 1215 Å and 16.35 × 10–12 erg to 4.9 B.M. The value of ‘a’ is :
(4) 6560 Å and 21.8 × 10–12 erg
(1) 3 (2) 4 (3) 2 (4) 5

3. The correct Schrodinger’s wave equation for

9. For an electron in a hydrogen atom , the wave
an electron with E as total energy and V as r / a0
potential energy is : function ,  is proportional to exp , where
a0 is the Bohr’s radius . What is the ratio of
2  2  2  8 2 the probability of finding the electron at the
(1) + + +
 x2  y2  z2 m h2 nucleus to the probability of finding it at a0 .
(E – V) =0 (1) e (2) e2 (3) 1/e2 (4) zero

2  2  2  8m
(2) 2 + 2 + 2 +
10. The maximum probability of finding electron in
x y z h2 the dxy orbital is :
(E – V)  = 0 (1) along the x-axis
 
2
 
2
 
2
8 m
2
(2) along the y-axis
(3) 2 + 2 + 2 +
x y z h2 (3) at an angle of 45º from the x & y-axis
(E – V) =0 (4) at an angle of 90º from the x & y-axis

2  2  2  8  m2
(4) 11. Maximum value (n + l + m) for unpaired
2 + 2 + 2 +
x y z h electrons in second excited state of chlorine
(E – V) =0 17
Cl is:
(1) 28 (2) 25
4. What should be the ratio of energies of the (3) 20 (4) none of these
electrons of the first orbits of Na+10 and H?
(1) 11:1 (2) 121 : 1 12. The distance between 3rd and 2nd Bohr orbits
(3) 1 : 121 (4) 1 : 11 of hydrogen atom is
(1) 0.529 × 10–8 cm (2) 2.645 × 10–8 cm
5. What should be the frequency (in cycles per (3) 2.116 × 10–8 cm (4) 1.058 × 10–8 cm
second) of the emitted radiation when an electron
undergoes transition from M energy level to K
energy level, if the value of R is 105 cm–1. 13. The ratio of (E 2– E 1) to (E 4 – E 3) for the
(1) 3/2 × 1015 (2) 8/3 × 1015 hydrogen atom is approximately equal to:
15
(3) 8/5 × 10 (4) 9/4 × 1015 (1) 10 (2) 15 (3) 17 (4) 12
 ATOMIC STRUCTURE

14. Which of the following graphs correspond to 18. The ratio of specific charge of an electron to that
one node? of a proton is
(1) 1 : 1 (2) 1837 : 1
(3) 1 : 1837 (4) 2 : 1
(1)
19. The electrons, identified by quantum numbers n
and l, (i) n = 4, l = 1 (ii) n= 4, l = 0 (iii) n = 3,
l = 2 (iv) n = 3, l = 1 can be placed in order of
a0 increasing energy, from the lowest to highest, as
(1) (iv) < (ii) < (iii) < (i) (2) (ii) < (iv) < (i) < (iii)

(2) (3) (i) < (iii) < (ii) < (iv) (4) (iii) < (i) < (iv) < (ii)

20. Uncertainty in the position of an electron

(mass = 9.1 × 10–31 kg) moving with a velocity
300 ms–1, accurate upto 0.001%, will be:
a0
(1) 19.2 × 10–2 m (2) 5.76 × 10–2 m
(3) 1.92 × 10–2 m (4) 3.84 × 10–2 m
(h = 6.63 × 10–34Js)

21. The ionization enthalpy of hydrogen atom is

(3)
1.312 × 106 J mol–1. The energy required to excite
a0 the electron in the atom from n = 1 to n = 2 is
(1) 8.51 × 105 J mol–1 (2) 6.56 × 105 J mol–1
(3) 7.56 × 105 J mol–1 (4) 9.84 × 105 J mol–1

22. Which of the following has maximum energy?

(4) 3s 3p 3d
(1)
a0
3s 3p 3d
(2)
15. The number of waves made by a Bohr electron in
3s 3p 3d
an orbit of maximum magnetic quantum number 3
is: (3)

(1) 3 (2) 4 (3) 2 (4) 1 3s 3p 3d

(4)

16. A compound of vanadium has a magnetic

m om ent of 1.73 BM. T he elec tr onic 23. The frequency of radiation emitted when the
configuration of vanadium ion in the compound electron falls from n = 4 to n = 1 in a hydrogen
is: atom will be (Given ionization energy of
(1) [Ar]3d2 (2) [Ar]3d14s0 H = 2.18 × 10–18 J atom–1)
(3) [Ar]3d3 (4) 3d04s1 (1) 1.03 × 1015 s–1 (2) 3.08 × 1015 s–1
15 –1
(3) 2.00 × 10 s (4) 1.54 × 1015 s–1
17. The wave number of the limiting line in Lyman
series of hydrogen is 109678 cm–1. The wave 24. What is the maximum number of electrons which
number of the limiting line in Balmer series of can be accommodated in an atom in which the
He+ would be: highest principal quantum number value is 4?
(1) 54839 cm–1 (2) 219356 cm–1 (1) 10 (2) 18
(3) 109678 cm –1
(4) 438712 cm –1 (3) 36 (4) 54
 ATOMIC STRUCTURE

25. The wavelength of radiation emitted when an 32. A single electron orbits a stationary nucleus of
electron in a hydrogen atom makes a transition charge +Ze, where Z is a constant, It requires
from an energy level with n = 3 to a level with 47.2 eV to excite electron from second Bohr orbit
to third Bohr orbit, find the value of Z :
−1312
n = 2 is : [Given that En = kJmol–1] (1) 1 (2) 3
n2
(1) 6.56 × 10–7 m (2) 65.6 nm (3) 5 (4) 4
(3) 65.6 × 10–7 m (4) any of the above
33. A photon of energy 12.75 eV is completely
26. The energy required to escape the electron from absorbed by a hydrogen atom initially in ground
ground state of H is 13.6 eV then the same for Ist state. The principle quantum number of the
excited state of H atom : excited state is :
(1) 3.4 (2) 13.6 (1) 1 (2) 3
(3) 27.2 (4) can’t say anything (3) 4 (4) 

27. A gas absorbs a photon of 355 nm and emits at 34. An hydrogen atom (ionisation energy 13.6eV)
two wavelengths. If one of the emissions is at jumps from third excited state to first excited
680 nm, the other is at : state. The energy of photon emitted in the
(1) 743 nm (2) 518 nm process is :
(3) 1035 nm (4) 325 nm (1) 1.89 eV (2) 2.55 eV
(3) 12.09 eV (4) 12.75 eV
28. The frequency of light emitted for the transition
n = 4 to n = 2 of He+ is equal to the transition in
H atom corresponding to which of the following : 35. If a photon of energy 14 eV. is incident on an H-
atom, what is true :
(1) n = 3 to n = 1 (2) n = 2 to n = 1
(1) Atom will be ionised and electron will have a
(3) n = 3 to n = 2 (4) n = 4 to n = 3
kinetic energy of 14 eV
(2) Atom will be ionised and electron will have a
29. Energy of an electron is given by
kinetic energy of 0.4 eV
 Z2  (3) Photon passes through atom without
–18
E= – 2.178 ×10 J  2  . Wavelength of light
n  interacting with it
required to excite an electron in an hydrogen atom (4) More than one electrons will make transitions
from level n = 1 to n = 2 will be :
(h = 6.62 × 10–34 Js and c = 3.0 × 108 ms–1) 36. An electron of energy 10.8 eV is incident on an
(1) 1.214 × 10–7 m (2) 2.816 × 10–7 m H-atom then :
(3) 6.500 × 10–7 m (4) 8.500 × 10–7 m (1) The electron will come out with 10.8 eV energy
(2) The electron will be completely absorbed
30. Supposing the I.P. of hydrogen atom is 960 eV.
(3) 10.2 eV. of the electron would be absorbed
Find out the value of principal quantum number
by H atom and it would come out with 0.6
having the energy equal to – 60 eV :
eV energy.
(1) n = 2 (2) n = 3
(4) None
(3) n = 4 (4) n = 5

37. The ratio of the difference in energy between the

31. If the ionisation potential of an atom is 20V, its first and second Bohr orbit to that between second
first excitation potential will be : and third Bohr orbit in H-atom is :
(1) 5 V (2) 10 V (1) 4/9 (2) 1/3
(3) 15 V (4) 20 V
(3) 27/5 (4) 1/2
 ATOMIC STRUCTURE

38. Match the following : 41. When a hydrogen sample in ground state is
bombarded then what potential is required to
(A) Energy of ground (i) + 6.04 eV accelerate electron so that first Paschen line is
+
State of He emitted :
(B) Potential energy of (ii) –27.2 eV
I orbit of H-atom (1) 2.55 V (2) 0.65 V
–18
(C) Kinetic energy of (iii) 8.72 × 10 J (3) 12.09 V (4) 12.75 V
II excited state of
+
He
(D) Ionisation potential (iv) –54.4 eV
of He
+ 42. The binding energy of e– in ground state of
hydrogen atom is 13.6 eV. The energies required
(1) A-(i), B-(ii), C-(iii), D-(iv) to eject out an electron from three lowest states
of He+ atom will be : (in eV)
(2) A-(iv), B-(iii), C-(ii), D-(i)
(1) 13.6, 10.2, 3.4 (2) 13.6, 3.4, 1.5
(3) A-(iv), B-(ii), C-(i), D-(iii)
(3) 13.6, 27.2, 40.8 (4) 54.4, 13.6, 6
(4) A-(ii), B-(iii), C-(i), D-(iv)

43. Given that in the H-atom the transition energy for

39. In the following transition which statement is
n = 1 to n = 2, Rydberg states is 10.2 eV. The
correct :
energy for the same transition in Be3+ is :
E3 (1) 20.4 eV (2) 163.2 eV
E2 (3) 30.6 eV (4) 40.8 eV
E1

(1) E3 – 1 = E3-2 – E2-1

44. The wavelength of first line of the Lyman series
(2) 3 = 1 + 2 for hydrogen is 1216Å. The wavelength for the
first line of this series for a 10 time ionised sodium
(3) 3 = 2 + 1
atom (z = 11) will be :
(4) All of these
(1) 1000 Å (2) 100 Å
(3) 10 Å (4) 1 Å
40. In which transition, one quantum of energy is
emmited :
(1) n = 4  n = 2 (2) n = 3  n = 1
(3) n = 4  n = 1 (4) All of them

ANSWER KEY [EXERCISE-III]

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 1 3 3 2 2 3 3 1 4 3 2 2 2 2 2
Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Ans. 2 3 2 1 3 4 2 2 3 1 1 1 2 1 3
Que. 31 32 33 34 35 36 37 38 39 40 41 42 43 44
Ans. 3 3 3 2 2 3 3 3 3 4 4 4 2 3