ATOMIC STRUCTURE

INTRODUCTION :
Structure of Atom

Rutherford's Model Bohr's Model Wave mechanical model

Dalton’s concept of the indivisibility of the atom was completely discredited by a series of experimental
evidences obtained by scientists. A number of new phenomenone were brought to light and man’s idea
about the natural world underwent a revolutionary change. The discovery of electricity and spectral
phenomena opened the door for radical changes in approaches to experimentation. It was concluded
that atoms are made of three particles : electrons, protons and neutrons. These particles are called the
fundamental particles of matter.

CATHODE RAYS (DISCOVERY OF ELECTRON)

In 1859 Julius Plucker started the study of conduction of electricity through gases at low pressure
(10–4atm) in a discharge tube When a high voltage of the order of 10,000 volts or more was impressed
across the electrodes, some sort of invisible rays moved from the negative electrode to the positive
electrode these rays are called as cathode rays.
PROPERTIES OF CATHODE RAYS :
(i) Path of travelling is straight from the cathode with a very high velocity as it produces shadow of
an object placed in its path
(ii) Cathode rays produce mechanical effects. If small paddle wheel is placed between the
electrodes, it rotates. This indicates that the cathode rays consist of material part.
(iii) When electric and magnetic fields are applied to the cathode rays in the discharge tube, the
rays are deflected thus establishing that they consist of charged particles. The direction of
deflection showed that cathode rays consist of negatively charged particles called electrons.
(iv) They produce a green glow when strike the glass wall beyond the anode. Light is emitted when
they strike the zinc sulphide screen.
(v) Cathode rays penetrate through thin sheets of aluminium and metals.
(vi) They affect the photographic plates
(vii) The ratio of charge(e) to mass(m) i.e. charge/mass is same for all cathode rays irrespective of
the gas used in the tube. e/m = 1.76 × 1011 Ckg–1
Thus, it can be concluded that electrons are basic constituent of all the atoms.
ANODE RAYS (DISCOVERY OF PROTON) :
Goldstein (1886) repeated the experiment with a discharge tube filled with a perforated cathode and
found that new type of rays came out through the hole in the cathode.

1
 When this experiment is conducted, a faint red glow is observed on the wall behind the cathode. Since
these rays originate from the anode, they are called anode rays.
PROPERTIES OF ANODE RAYS :
⚫ Anode rays travel along straight paths and hence they cast shadows of object placed in their
path.
⚫ They rotate a light paddle wheel placed in their path. This shows that anode rays are made
up of material particles.
⚫ They are deflected towards the negative plate of an electric field. This shows that these rays
are positively charged.
⚫ For different gases used in the discharge tube, the charge to mass ratio (e/m) of the positive
particles constituting the positive rays is different. When hydrogen gas is taken in the discharge
tube, the e/m value obtained for the positive rays is found to be maximum. Since the value of
charge (e) on the positive particle obtained from different gases is the same, the value of m
must be minimum for the positive particles obtained from hydrogen gas. Thus, the positive
particle obtained from hydrogen gas is the lightest among all the positive particles obtained
from different gases. This particle is called the proton.

DISCOVERY OF NEUTRON :
Later, a need was felt for the presence of electrically neutral particles as one of the constituent of atom.
These particles were discovered by James Chadwick in 1932 by bombarding a thin sheet of Beryllium
with -particles, when electrically neutral particles having a mass slightly greater than that of the
protons were emitted. He named these particles as neutrons.
4 Be + 2 He ⎯⎯→ C+
9 4 12 1
6 0 n

ATOMIC MODELS :
THOMSON’S MODEL OF THE ATOM :
An atom is electrically neutral. It contains positive charges as well as negative charges (due to the
presence of electrons). Hence, J J Thomson assumed that an atom is a uniform sphere of positive
charges with electrons embedded in it. The magnitude of total positive charge on sphere is equal to
total negative charge of embedded electrons.

RUTHERFORD’S EXPERIMENT :

2
 ⚫ Most of the -particles passed straight through the gold foil without suffering any deflection
from their original path.
⚫ A few of them were deflected through small angles, while a very few were deflected to a large
extent.
⚫ A very small percentage (1 in 100000) was deflected through angles ranging from 90° to 180°.

Rutherford’s nuclear concept of the atom.

⚫ The atom of an element consists of a small positively charged ‘nucleus’ which is situated at the
centre of the atom and which carries almost the entire mass of the atom.
⚫ The electrons are distributed in the empty space of the atom around the nucleus in different
concentric circular paths, called orbits.
⚫ The number of electrons in orbits is equal to the number of positive charges (protons) in the
nucleus. Hence, the atom is electrically neutral.
⚫ The volume of the nucleus is negligibly small as compared to the volume of the atom.
⚫ Most of the space in the atom is empty.

NUCLEUS :
Electrons, protons & neutrons are the fundamental particles present in all atoms,(Except hydrogen)

Particles Symbol Mass Charge Discoverer

–1e or  9.1096 x 10-31 kg – 1.602 x l0–19
Electron 0 J.J. Thompson
Coulombs Stoney Lorentz 1887
0.000548 amu – 4.803 × 10–10 esu

1
1H
Proton 1.6726 x 10–27 kg + 1.602 x 10–19 Goldstein
Coulombs Rutherford1907
1.00757 amu + 4.803 x 10–10 esu
1
Neutron 0n 1.6749 x 10–27 kg Neutral James Chadwick
1.00893 amu 0 1932
1 amu  1.66 × 10–27 kg

Uncommon Fundamental Particles :– Other less important fundamental particles of atom are
positrons, neutrinos and mesons.
0
(A) Positron (Positive electron, 1e ) :– It is the positive counterpart of electron, discovered by
Anderson in 1932. It is very unstable and combines with electron producing –rays (energy radiations).
0 +
Positron is symbolised as 1e or e

(B) Neutrino :– These are particles of small (  0) mass and zero charge. These were postulated by
Paulling in 1934.

3
 (C) Antineutrino :– These are particles of small (  0) mass and zero charge. These were postulated
by Fermi in 1934. "Identical to neutrino but opposite spin"
(D) Mesons :– (postulated by Yukawa in 1935) Mesons may be either positively or negatively charged
or neutral. Their mass is intermediate between that of electron and proton.
MESONS CHARGE MASS
 +/– 286 me

 +/–/0 270 me

k +/–/0 970 me

(E) Anti proton :– Discovered by Segre (1955) its mass equal to proton and charge equal to electron.
(F) Anti neutron :– Discovered by Cork (1956) its mass equal to neutron and charge equal to
zero."Identical to neutron but opposite spin"
ATOMIC NUMBER AND MASS NUMBER :
 Atomic number of an element
= Total number of protons present in the nucleus
= Total number of electrons present in the atom

• Atomic number is also known as proton number because the charge on the nucleus depends
upon the number of protons.

• Since the electrons have negligible mass, the entire mass of the atom is mainly due to protons
and neutrons only. Since these particles are present in the nucleus, therefore they are
collectively called nucleons.

• As each of these particles has one unit mass on the atomic mass scale, therefore the sum of
the number of protons and neutrons will be nearly equal to the mass of the atom.
 Mass number of an element = No. of protons + No. of neutrons.
• The mass number of an element is nearly equal to the atomic mass of that element. However,
the main difference between the two is that mass number is always a whole number whereas
atomic mass is usually not a whole number.
• The atomic number (Z) and mass number (A) of an element ‘X’ are usually represented
alongwith the symbol of the element as

23 35
e.g. 11 Na, 17 Cl and so on.

IMPORTANT DEFINATIONS :
1. ISOTOPES :
(i) First proposed by soddy.
(ii) The isotopes have same atomic number but different atomic weight.
(iii) They have same chemical properties because they have same atomic number.
(iv) They have different physical properties because they have different atomic masses.
16 17 18 12 13 14 14 15
eg. 8O , 8O , 8O ; 6C , 6C , 6C ; 7N , 7N
1 2 3
eg. 1H 1H 1H

Protium deuterium Tritium

4
 Z=1 1 1
A=1 2 3
2. ISOBAR :
The different atoms which have same atomic masses but different atomic number are called as Isobar.
40 40 40
eg. 18Ar 19K 20Ca

Atomic mass 40 40 40
Atomic number 18 19 20
3. ISOTONE :
Elements which contain same no. of neutron are called as Isotones.
30 31 32
eg. 14Si 15P 16S

Number of neutrons 16 16 16
4. ISOELECTRONIC :
Ion or atom or molecule or species which have the same number of electrons are called as
isoelectronic species.
– + +2
eg. 17Cl 18Ar 19K 20Ca

No. of electron 18 18 18 18

1. The triad of nuclei that is isotonic -

14 15 17 12 14 19 14 14 17 14 14 19
(1) 6C , 7N , 9F (2) 6C , 7N , 9F (3) 6C , 7N , 9F (4) 6C , 7N , 9F Ans. (1)

2. Complete the following table :

Particle Mass No. Atomic No. Protons Neutrons Electrons

Nitrogen atom – – – 7 7
Calcium ion – 20 – 20 –
Oxygen atom 16 8 – – –
Bromide ion – – – 45 36

Sol. For nitrogen atom.

No. of electron = 7 (given)
No. of neutrons = 7 (given)
 No. of protons = Z = 7 ( atom is electrically neutral)
Atomic number = Z = 7 Mass No. (A) = No. of protons + No. of neutrons
= 7 + 7 = 14
For calcium ion.
No. of neutrons = 20 (Given)
Atomic No. (Z) = 20 (Given)
 No. of protons = Z = 20 ;
No. of electrons in calcium atom = Z = 20

5
 But in the formation of calcium ion, two electrons are lost from the extranuclear part according to the
equation Ca → Ca2+ + 2e– but the composition of the nucleus remains unchanged.
 No. of electrons in calcium ion
= 20 – 2 = 18
Mass number (A) = No. of protons + No. of neutrons
= 20 + 20 = 40.
For oxygen atom.
Mass number (A) = No. of protons + No. of neutrons
= 16 (Given
Atomic No. (Z) = 8 (Given)
No. of protons = Z = 8,
No. of electrons = Z = 8
No. of neutrons = A – Z = 16 – 8 = 8

ELECTROMAGNETIC RADIATIONS :
The oscillating electrical/magnetic field are electromagnetic radiations. Experimentally, the direction of oscillations
of electrical and magnetic field are prependicular to each other.

→ →
E = Electric field, B = Magnetic field

Direction of wave propogation.

WAVE NATURE OF LIGHT :

 Wavelength of a wave is defined as the distance between any two consecutive crests or
troughs. It is represented by  (lambda) and is expressed in Å or m or cm or nm (nanometer) or
pm (picometer).
1 Å = 10– 8 cm = 10–10 m
1 nm = 10– 9 m, 1 pm = 10–12 m

 Frequency of a wave is defined as the number of waves passing through a point in one
second. It is represented by  (nu) and is expressed in Hertz (Hz) or cycles/sec or simply sec –1
or s–1.
1 Hz = 1 cycle/sec
 Velocity of a wave is defined as the linear distance travelled by the wave in one second. It is
represented by v and is expressed in cm/sec or m/sec (ms –1).

6
  Amplitude of a wave is the height of the crest or the depth of the trough. It is represented by ‘a’
and is expressed in the units of length.
 Wave number is defined as the number of waves present in 1 cm length. Evidently, it will be
equal to the reciprocal of the wavelength. It is represented by  (read as nu bar).
1
=

If  is expressed in cm, v will have the units cm–1.
Relationship between velocity, wavelength and frequency of a wave. As frequency is the number
of waves passing through a point per second and  is the length of each wave, hence their product will
give the velocity of the wave. Thus
c=×

Order of wavelength in electromagnetic spectrum

Cosmic rays <  – rays < X-rays < Ultraviolet rays < Visible < Infrared < Micro waves < Radio waves.

PARTICLE NATURE OF LIGHT (PLANK’S QUANTUM THEORY)

The smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation is called
as quantum of light.
According to Planck, the light energy coming out from any source is always an integral multiple of a smallest
energy value called quantum of light.
Let quantum of light be = E0(J), then total energy coming out is = nE0 (n = Integer)
Quantum of light = Photon ( Packet or bundle of energy)
Energy of one photon is given by
E0 = h (- Frequency of light)
h = 6.625 x 10–34 J-Sec (h - Planck const.)
hc
E0 = (c - speed of light)

( - wavelength)

10−34  108
Order of magnitude of Eo = = 10–16 J
10−10

Some of the experimental phenomenon such as diffraction and interference can be explained by the
wave nature of the electromagnetic radiation. However, following are some of the observations which
could not be explained
(i) the nature of emission of radiation from hot bodies (black - body radiation)
(ii) ejection of electrons from metal surface when radiation strikes it (photoelectric effect)

Photoelectric Effect :
When certain metals (for example Potassium, Rubidium, Caesium etc.) were exposed to a beam of light
electrons were ejected as shown in Fig.

7
 The phenomenon is called Photoelectric effect. The results observed in this experiment were :
(i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface, i.e.,
there is no time lag between the striking of light beam and the ejection of electrons from the metal
surface. This phenomenon is called SINGLE COLLISION event. The electron at the surface collides
with only one photon once.
(ii) The number of electrons ejected is proportional to the intensity or brightness of light.
(iii) For each metal, there is a characteristic minimum frequency, 0 (also known as threshold frequency)
below which photoelectric effect is not observed. At a frequency  > 0, the ejected electrons come out
with certain kinetic energy. The kinetic energies of these electrons increase with the increase of
frequency of the light used.
When a photon of sufficient energy strikes an electron in the atom of the metal, it transfers its energy
instantaneously to the electron during the collision and the electron is ejected without any time lag or
delay. Greater the energy possessed by the photon, greater will be transfer of energy to the electron
and greater the kinetic energy of the ejected electron. In other words, kinetic energy of the ejected
electron is proportional to the frequency of the electromagnetic radiation. Since the striking photon has
energy equal to h and the minimum energy required to eject the electron is h0 (is also called work
function, W0) then the difference in energy (h – h0) is transferred as the kinetic energy of the
photoelectron. Following the conservation of energy principle, the kinetic energy of the ejected electron
is given by the equation
1
h = h0 + m 2
2 e
where me is the mass of the electron and v is the velocity associated with the ejected electron.

Example 1.
The threshold frequency 0 for a metal is 6 × 1014 s–1. Calculate the kinetic energy of an electron emitted
when radiation of frequency  = 1.1 × 1015 s–1 hits the metal.
1
Sol. K.E. = m V2 = h ( – 0)
2 e
 K.E. = (6.626 × 10–34) (1.1 × 1015 – 6 × 1014)
 K.E. = (6.626 × 10–34) (5 × 1014)
= 3.313 × 10–19 J

Example 2.
When electromagnetic radiation of wavelength 310 nm fall on the surface of Sodium, electrons are
emitted with K.E. = 1.5 eV. Determine the work function (W0) of Sodium.
12400
Sol. h = = 4 eV
3100

8
 1
m V2 = 1.5 eV
2 e
1
 h0 = W0 = h – m V2 = 4 – 1.5 = 2.5 eV
2 e

SOME IMPORTANT FORMULAE :

1. A = Z + N (Number of neutrons)
2. dynamic mass of particle m = m0 / [1 – (v/c)2]1/2
3. Radius of nucleus R = R0 (A)1/3, R0 = 1.2 x 10–15 m

4. c = 
5. wave number  = 1 / 
6. E = h  = hc /  = hc 
q1 q2 1
7. F=K ;K= = 9.0 × 109 Nm2/C2
r2 40

8. E = h = E2 – E1

Ex-3. Calculate number of photon coming out per sec. from the bulb of 100 watt. If it is 50% efficient and
wavelength coming out is 600 nm.
Sol. Energy = 100J
hc 6.625  10−34  3  108 6.625
Energy of one photon = = = × 10–19
 600  10−9 2
100
no. of photon = × 1019 = 15.09 × 1019
6.625

Ex-4. Certain sun glasses having small of AgCl incorporated in the lenses, on exposer to light of appropriate
wavelength turns to gray colour to reduce the glare following the reactions:
hv →
AgCl ⎯⎯⎯ Ag(Gray) + Cl
If the heat of reaction for the decomposition of AgCl is 248 kJ mol–1, what maximum wavelength is
needed to induce the desired process?
Sol. Energy needed to change = 248 × 103 J/mol
If photon is used for this purpose, then according to Einstein law one molecule absorbs one photon.
hc
Therefore,  NA . = 248 × 103

6.625  10−34  3.0  108  6.023  1023
= = 4.83 × 10–7 m
248  103

Ex-5. Sodium street lamp gives off a characteristic yellow light of wavelength 588 nm. Calculate the energy
mole (in kJ/mol) of these photons.
Sol.  = 588 nm = 588  10–9 m
c = 3  108 ms–1
c
E = N0h = N0 h


9
 6.02  1023  6.63  10−34  3  108
=
588  10−9
= 2.04  105 J mol–1 = 2.04  102 kJ mol–1

Ex-6. Find the wavelength of 100 g particle moving with velocity 100 ms –1
h 6.626  10−34 kg m2s−1
Sol. = =
mv 0.1 kg 100 ms−1
= 6.626  10–35 m.

Ex-7. A bulb emits light of wavelength 4500 Å. The bulb is rated as 150 watt and 8% of the energy is emitted
as light. How many photons are emitted by the bulb per second.
Sol. A 150 watt bulb emits 150 J of energy per second.
The energy emitted by the bulb as light
8
= 150  = 12 J
100
suppose the bulb emits n photons per second
nh
E = nh =

E 12J  4500  10−10 m
n= = = 2.715  1019
c  h 3  108 ms−1  6.63  10−34 Js

BLACK BODY RADIATION :

When solids are heated they emit radiation over a wide range of wavelengths.
The ideal body, which emits and absorbs all frequencies, is called a black body and the radiation
emitted by such a body is called black body radiation. The exact frequency distribution of the emitted
radiation (i.e., intensity versus frequency curve of the radiation) from a black body depends only on its
temperature.

The above experimental results cannot be explained satisfactorily on the basis of the wave theory of
light. Planck suggested that atoms and molecules could emit (or absorb) energy only in discrete
quantities and not in a continuous manner.
BOHR’S ATOMIC MODEL : It is based on quantum theory of light.
It is applicable only for Hydrogen like species, for example He+, Li2+, Be3+, etc.
Assumptions of Bohr’s model :
⚫ There are certain orbits around the nucleus such that if electron will be revolving in these orbit,
then it does not emit any electromagnetic radiation. These are called stationary orbit for the e –.
The neccessary centripetal force is produced by attraction forces of nucleus.

mv 2 Ke2Z
=
r r2

10
 h
⚫ Angular momentum of the electron in these stationary orbit is always an integral multiple of
2
nh
mvr = 2

⚫ Electron can make jump from one stationary orbit to another stationary orbit by absorbing or
emitting a photon of energy equal to difference in the energies of the stationary orbit i.e. energy
change does not take place in continuous manner.

hc
=E E – difference in the energy of orbit


E
= This is Bohr’s frequency rule.
h

Mathematical forms of Bohr’s Postulates :

Calculation of the radius of the Bohr’s orbit : Suppose that an electron having mass ‘m’ and charge
‘e’ revolving around the nucleus of charge ‘Ze’ (Z is atomic number & e = charge) with a
tangential/linear velocity of ‘v’. Further consider that ‘r’ is the radius of the orbit in which electron is
revolving.
According to Coulomb’s law, the electrostatic force of attraction (F) between the moving electron and
nucleus is :
KZe2 1
F= 2
where : K = constant = = 9 x 109 Nm2/C2
r 40
mv 2
and the centrifugal force F =
r
For the stable orbit of an electron both the forces are balanced.
mv 2 KZe2
i.e =
r r2
KZe 2
then v2 = ......... (i)
mr
From the postulate of Bohr,
V

e
r
+Ze

nh nh
mvr =  v=
2 2mr

n2h2
On squaring v2 = ........ (ii)
42m2r 2
From equation (i) and (ii)
KZe 2 n2h2
=
mr 42m2r 2
On solving, we will get
n2h2
r=
42mKZe2
On putting the value of e , h , m, the radius of nth Bohr orbit is given by :

11
 n2
rn = 0.529 x Å
Z

Ex-8. Calculate radius ratio for 2nd orbit of He+ ion & 3rd orbit of Be+++ ion.
 22 
Sol. r1 (radius of 2nd orbit of He+ ion) = 0.529   Å
 2
 
 32 
r2 (radius of 3rd orbit of Be+++ ion) = 0.529   Å
 4
 
r1 0.529  22 / 2 8
Therefore = =
r2 0.529  32 / 4 9

Calculation of velocity of an electron in Bohr’s orbit :

Angular momentum of the revolving electron in nth orbit is given by
nh
mvr =
2
nh
v= ......... (iii)
2mr
put the value of ‘r’ in the equation (iii)
nh  42mZe2K
then, v=
2mn2h2

Ze2K
v=
nh
on putting the values of , e-, h and K
Z 1
velocity of electron in nth orbit vn = 2.18 x 106 x m/sec ; v Z ; v
n n
2r
T, Time period of revolution of an electron in its orbit =
v
v
f, Frequency of revolution of an electron in its orbit =
2r
Calculation of energy of an electron :
The total energy of an electron revolving in a particular orbit is
T.E. = K.E. + P.E.
where : P.E. = Potential energy , K.E. = Kinetic energy , T.E. = Total energy
1
The K.E. of an electron = mv2
2
KZe 2
and the P.E. of an electron = –
r

1 KZe 2
Hence, T.E. = mv2 –
2 r
mv 2 KZe2 KZe 2
we know that, = 2 or mv2 =
r r r
substituting the value of mv2 in the above equation :

12
 KZe 2 KZe 2 KZe 2
T.E. = – =–
2r r 2r
KZe 2
So, T.E. = –
2r
substituting the value of ‘r’ in the equation of T.E.
KZe 2 42 Ze2m 22Z2e4m K 2
Then T.E. = – x =–
2 2
nh 2
n2h2
Thus, the total energy of an electron in nth orbit is given by

22 me4 k 2  z2 
T.E. = En = –  2  ... (iv)
h2 n 
Putting the value of m,e,h and  we get the expression of total energy
Z2
En = – 13.6 eV / atom n  T.E.  ; Z  T.E. 
n2
Z2
En = – 2.18 × 10–18 J/atom
n2
1
T.E. = P.E.
2
T.E. = – K.E.

Note : - The P.E. at the infinite = 0

The K.E. at the infinite = 0

Conclusion from equation of energy :

(a) The negative sign of energy indicates that there is attraction between the negatively charged
electron and positively charged nucleus.
(b) All the quantities on R.H.S. in the energy equation [Eq. iv] are constant for an element having atomic
number Z except ‘n’ which is an integer such as 1,2,3, etc . i.e. the energy of an electron is constant as
long as the value of ‘n’ is kept constant.
(c) The energy of an electron is inversely proportional to the square of ‘n’ with negative sign.
(d) Negative charge of the energy of e– in the atom indictes that the energy of e– in the atom is at lower
energy than the energy of a free e– at rest (which is taken to be zero).

Ex-9 What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n
= 2 state in the hydrogen atom ?
Sol. Since ni = 5 and nf = 2, this transition gives rise to a spectral line in the visible region of the Balmer
series.
1 1
E = 2.18 × 10–18 J  2 – 2  = – 4.58 × 10–19 J
5 2 
It is an emission energy
The frequency of the photon (taking energy in terms of magnitude) is given by

13
 E 4.58  10–19 J
= = = 6.91 × 1014 Hz
h 6.626  10–34 Js
c 3.0  108 ms–1
= = = 434 nm
 6.91 1014 Hz

Ex-10. If the energy of an electron in 3rd Bohr orbit is –E, what is the energy of the electron of the electron in
(i) 1st Bohr orbit (ii) 2nd Bohr orbit ?
Sol. According to Bohr’s model,
E1
En =
n2
E1
E3 = = –E (given)
(3)2
 E1 = – 9E
E1 9E
E2 = 2
=– = 2.25 E
(2) 4

Ex-11. The ratio of (E2 – E1) to (E4 – E3) for He+ ion is approximately equal to (where En is the energy of nth
orbit) (1) 10 (2) 15 (3) 17 (4) 12
 1 1 
13.6 (2)2  2 − 2 
 (1) (2) 
Sol. = 15 Ans. (2)
 1 1 
13.6 (2)  2 −
2

 (3) (4)2 

Failures / limitations of Bohr’s theory:

(a) He could not explain the line spectra of atoms containing more than one electron.
(b) He also could not explain the presence of doublet i.e. 2 closely spaced lines.
(c) He was unable to explain the splitting of spectral lines in magnetic field (Zeeman effect) and in
electric field (Stark effect)
(d) No conclusion was given for the principle of quantisation of angular momentum.
(e) He was unable to explain the de-Broglie’s concept of dual nature of matter.
(f) He could not explain Heisenberg’s uncertainty principle.
(g) Could’t explain the ability of atoms to form molecules by chemical bonds.

Energy Level Diagram :

(i) Orbit of lowest energy is placed at the bottom, and all other orbits are placed above this.
(ii) The gap between two orbits is proportional to the energy difference of the orbits.

0 eV n= 
-0.85 eV n=4
-1.51eV n=3
-3.4eV n=2

12.1eV
10.2eV
-13.6eV n=1
Energy level diagram of H-atom

DEFINITION VALID FOR SINGLE ELECTRON SYSTEM :

14
 (i) Ground state :
Lowest energy state of any atom or ion is called ground state of the atom. For it n = 1.
Ground state energy of H–atom = – 13.6 ev
Ground state energy of He+ on = – 54.4 ev
(ii) Excited State :
States of atom other than the ground state are called excited states :
n=2 first excited state
n=3 second excited state
n=4 third excited state
n=n+1 nth excited state
(iii) Ionisation energy (E) :
Minimum energy required to move an electron from ground state to
n =  is called ionisation energy of the atom or ion.
onisation energy of H–atom = 13.6 ev
onisation energy of He+ ion = 54.4 ev
onisation energy of Li+2 ion = 122.4 ev

(iv) Ionisation Potential (.P.) :

Potential difference through which a free electron must be accelerated from rest, such that its kinetic
energy becomes equal to ionisation energy of the atom is called ionisation potential of the atom.
.P. of H atom = 13.6 V, .P. of He+ on= 54.4 V
(v) Excitation Energy :
Energy required to move an electron from ground state of the atom to any other state of the atom is
called excitation energy of that state.
Excitation energy of 2nd state = excitation energy of 1st excited state = 1st excitation energy = 10.2 ev.
(vi) Excitation Potential :
Potential difference through which an electron must be accelerated from rest to so that its kinetic
energy become equal to excitation energy of any state is called excitation potential of that state.
Excitation potential of third state = excitation potential of second exicited state = second excitation
potential = 12.09 V.
(vii) Binding Energy ‘or’ Seperation Energy :
Energy required to move an electron from any state to n =  is called binding energy of that state.
Binding energy of ground state = .E. of atom or on.

Ex-12 A single electron system has ionization energy 11180 kJ mol –1 . Find the number of protons in the
nucleus of the system.
Z2
Sol. .E. = × 21.69 × 10–19 J
n2
11180  103 Z2
= × 21.69 × 10–19 Ans. Z = 3
6.023  1023 12

Ex-13. The ionization energy of He+ is 19.6  10–18 J/atom. Calculate the energy of the first stationary state of
Li2+.
Sol. For H–like species
I.E. = I.E.(H)  Z2
 I.E.(He+ ) = Z2  I.E.(H)

19.6  10–18 J = (2)2  I.E.(H)

15
 19.6  10 −18
I.E.(H) = J
4
I.E.(Li+2 ) = Z2  I.E.(H)

(3)2  19.6  19 −18

=
4
= 4.41  10–17 J/atom
I.E.(Li+2 ) = E  – E1

4.41  10–17 = 0 – E1
E1 = – 4.41  10–17 J/atom.

Ex-14. If the binding energy of 2nd excited state of a hydrogen like sample is 24 eV approximately, then the
ionisation energy of the sample is approximately
(1) 54.4 eV (2) 24 eV (3) 122.4 eV (4) 216 eV
13.6(Z)2
Sol. = 24
(3)2
I.E. = 13.6(Z)2 = (24 × 9) = 216 ev Ans. (4)
Ex-15. The ionisation energy of H atom is 21.79 × 10 –19 J . Then the value of binding energy of second excited
state of Li2+ ion
(1) 32 × 21.7 × 10–19 J (2) 21.79 × 10–19 J
1 1
(3) × 21.79 × 10–19 J (4) × 21.79 × 10–19 J
3 32
21.79  10−19 (3)2
Sol. B.E. = = 21.79 × 10–19 J Ans. (2)
(3)2

HYDROGEN SPECTRUM :
Study of Emission and Absorption Spectra :
An instrument used to separate the radiation of different wavelengths (or frequencies) is called
spectroscope or a spectrograph. Photograph (or the pattern) of the emergent radiation recorded on the
film is called a spectrogram or simply a spectrum of the given radiation.
The branch or science dealing with the study of spectra is called spectroscopy.
Spectrum

Based on Based on
Nature origin

Continuous Discrete Absorption Emission

spectrum spectrum
Band Line
Spectrum Spectrum
Emission spectra :
When the radiation emitted from some source e.g. from the sun or by passing electric discharge
through a gas at low pressure or by heating some substance to high temperature etc, is passed directly
through the prism and then received on the photographic plate, the spectrum obtained is called
‘Emission spectrum’.
Depending upon the source of radiation, the emission spectra are mainly of two type :
(a) Continuous spectra :

16
 When white light from any source such as sun, a bulb or any hot glowing body is analysed by passing
through a prism it is observed that it splits up into seven different wide band of colours from violet to
red. These colours are so continuous that each of them merges into the next. Hence the spectrum is
called continuous spectrum.

(b) Discrete spectra : It is of two type

(i) Band spectrum
Band spectrum contains colourful continuous bands sepearted by some dark space.
Generally, molecular spectrum are band spectrum.
(ii) Line Spectrum :
This is the ordered arrangement of lines of particular wavelength seperated by dark space eg. hydrogen
spectrum.
Line spectrum can be obtained from atoms.
Absorption spectra :
When white light from any source is first passed through the solution or vapours of a chemical
substance and then analysed by the spectroscope, it is observed that some dark lines are obtained in
the continuous spectrum. These dark lines are supposed to result from the fact that when white light
(containing radiations of many wavelengths) is passed through the chemical substance, radiations of
certain wavelengths are absorbed, depending upon the nature of the element.

EMISSION SPECTRUM OF HYDROGEN :

When hydrogen gas at low pressure is taken in the discharge tube and the light emitted on passing
electric discharge is examined with a spectroscope, the spectrum obtained is called the emission
spectrum of hydrogen.

Line Spectrum of Hydrogen :

17
Line spectrum of hydrogen is observed due to excitation or de-excitation of electron from one stationary
orbit to another stationary orbit
Let electron make transition from n2 to n1 (n2 > n1) in a H-like sample
– 13.6 Z
2

2 eV n2
n 2

– 13.6 Z2 n1
2
eV
n1

photon

−13.6Z2  −13.6Z2   1 1 
Energy of emitted photon = (E)n2 → n1 = – 
 n2  = 13.6Z2  2 − 2
n22 n n2 
 1   1
Wavelength of emitted photon
hc
=
( E)n2 →n1

hc
=
 1 1 
13.6Z 2  2 − 2 
n 
 1 n2 

1 (13.6)z2  1 1 
=  2 − 2
 
hc  n1 n2 

1  1 1 
Wave number, =  = RZ2  2 − 2 
  
 n1 n2 
13.6eV
R = Rydberg constant = 1.09678 × 107m–1 ; R 1.1 × 107 m–1 ; R = ; R ch = 13.6 eV
hc

18
Ex-16 Calculate the wavelength of a photon emitted when an electron in H- atom maker a transition from n = 2
to n = 1
1  1 1 
Sol. = RZ2  2 − 2
  n1 n2 
1 1 1
 = R(1)2  2 − 2 
 1 2 
1 3R 4
 = or  =
 4 3R

SPECTRA LINES OF HYDROGEN ATOM :

LYMAN SERIES
* It is first spectral series of H.
* It was found to be in ultraviolet region by Lyman in 1898.
* For it value of n1 = 1 and n2 = 2,3,4 where ‘n1’ is ground state and ‘n2’ is called excited state of electron
present in a H - atom.
1 1 1
* = RH  2 − 2  where n2 > 1 always.
  1 n2 
n12 1
* The wavelength of marginal line (i.e. n2 = ) = for all series. So for lyman series  = .
RH RH
* st line of lyman series  2 → 1
nd line of lyman series = 3 → 1
Last line of lyman series =  → 1
[10.2 eV  (E) lyman  13.6 eV]
12400 12400
  lyman  Aº
13.6 10.2
12400
* Longest line : longest wavelength line  longest or  max. =
( E)min
12400
* Shortest line : shortest wavelength line  shortest or  min =
( E)max
* First line of any spectral series is the longest ( max) line.
* Last line of any spectral series is the shortest ( min) line.
Series limit :
t is the last line of any spectral series.
1 1 1 
Wave no of st line of Lyman series = =  = R × 12  2 − 2 
 1 2 
 4 − 1
 = R × 12  
 4 
R3 3 R
 = =
4 4
 4 
  = 3R 
 
Wave no of last line of Lyman series
1 1 
 = R × 12  2 − 2 
1  
 =R

19
 For Lyman series,
12400 12400
 longest = , shortest =
( E)2−1 ( E ) → 1
BALMER SERIES :
* It is the second series of H-spectrum.
* It was found to be in visible region by Balmer in 1892.
* For it value of n1 = 2 and n2 = 3,4,5,.............
n12 22 4
* The wavelength of marginal line of Balmer series = = =
RH RH RH

1  1 1
* = RH  2 − 2  where n2 > 2 always.
 2 n2 

1.9  (E)balmer  3.4 eV.
All the lines of balmer series in H spectrum are not in the visible range. nfact only st 4
lines belongs to visible range.
12400 12400
Aº   balmer  Å
3.4 1.9
3648 Å   balmer  6536 Å
Lines of balmer series (for H atom) lies in the visible range.
Ist line of balmer series = 3 → 2
last line of balmer series =  → 2
 1 1  5R
(  ) 1st line = R ×1  2 − 2  =
 2 3  36
 1 1  R
(  ) last line = R  2 − 2  =
 2   4
PASCHEN SERIES :
(a) It is the third series of H - spectrum.
(b) It was found to be in infrared region by Paschen.
(c) For it value of n1 = 3 and n2 = 4,5,6 ........
n12 32 9
(d) The wavelength of marginal line of Paschen series = = = .
RH RH RH

1 1 1
(e) = RH  2 − 2  where n2 > 3 always.
  3 n2 

BRACKETT SERIES :
(a) It is fourth series of H - spectrum.
(b) It was found to be in infrared region by Brackett.
(c) For it value of n1 = 4 and n2 = 5,6,7 ..............
n12 42 16
(d) The wavelength of marginal line of brackett series = = =
RH RH RH

1  1 1
(e) = RH  2 − 2  where n2 > 4 always.
  4 n2 

PFUND SERIES :
(a) It is fifth series of H- spectrum.
(b) It was found to be in infrared region by Pfund.
(c) For it value of n1 = 5 and n2 = 6,7,8 ............... where n1 is ground state and n2 is excited state.

20
 n12 52 25
(d) The wavelength of marginal line of Pfund series = = =
RH RH RH

1 1 1
(e) =  2 − 2  RH where n2 > 5 always.
  5 n2 

HUMPHRY SERIES :
(a) It is the sixth series of H-spectrum.
(b) It was found to be in infrared region by Humphry.
(c) For it value of n1 = 6 and n2 = 7 , 8 , 9 ...................
n12 62 36
(d) The wavelength of marginal line of Humphry series = = =
RH RH RH

1  1 1
(e) = RH  2 − 2  where n2 > 6.
  6 n2 

Ex-17 Electron in hydrogen atom is in fourth stationary state. What is the maximum number of spectral lines
that can be observed as it de-excites.
n ( n − 1)
Sol. Maximum number of spectral lines from nth state =
2
n=4
n ( n − 1) 4 ( 4 − 1)
= =6
2 2
Ans. 6

Ex-18 Calculate wavelength for 2nd line of Balmer series of He+ ion
1 1 1
Sol. = R(2)2  2 − 2 
  n1 n2 
n1 = 2 n2 = 4
1 1 1
= R(22 )  2 − 2 
 2 4 
1 3 R 4
= = Ans.
 4 3 R

Ex-19. calculate the wavelength and energy of radiation emitted for the electronic transition from infinity to
stationary state one of the hydrogen atom.
1 1 1
Sol. = 109678  2 − 2
  n1 n2 
 1 1
= 109678  2 −  = 109678 cm–1
 (1) 
1
= cm = 9.1  10–6 cm
109678
= 9.1  10–8 m = 91 nm.
hc 6.63  10−34 J.s 3  108 ms−1
E= =
 9.1 10−8 m
= 2.186  10–18 J

21
Ex-20. Calculate the wavelength of the radiation emitted, producing a line in the Lyman series when an
electron falls from fourth stationary state in hydrogen atom (RH = 1.1  107 m–1)
Sol. For Lyman series n1 = 1

1 1 1  1 1 
= RH  2 − 2  = 1.1  107 m–1  2 − 
  n1 n2   (1) (4)2 
15
= 1.1  107 
16
 = 9.7  10 m = 97 nm
–8

Ex-21. Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic
hydrogen.
Sol. The shortest wavelength transition corresponds to n2 = to n1 = 2 transition
1 1
v = 109678  2 − 2  cm–1
 n1 n2 
 1 1
= 109678  2 − 
 (2)  
109678
= = 27419.5 cm–1
4

Ex-22. The energy of the electron in the second and third Bohr orbits of hydrogen atom is – 5.42  10–12 ergs
and – 2.41  10–12 ergs respectively. Calculate the wavelength of the emitted ratiation when electron
drops from third to second orbits.
Sol. E = E3 – E2
= –2.41  10–12 – (–5.42  10–12) ergs
= – 2.41  10–12 + 5.42  10–12 ergs
= 3.01  10–12 ergs
Now h = 6.63  10–27 ergs s,
c = 3  1010 cm/s
hc
E =

hc 6.63  10−27 ergs.s 3  1010 cms−1
= = = 6.6  10–5 cm = 6.6  103 Å
E 3.01 10−12 ergs

Ex-23. Calculate the energy emitted when electrons of 1.0 g atom of hydrogen undergo transition giving the
spectral line of lowest energy in the visible region of its atomic spectrum. (RH = 1.1  107 m–1)

1 1 1
Sol. = RH  2 − 2
  n1 n2 
 1 1 
= 1.1  107 m–1  2 − 
 (2) (3)2 
5
= 1.1  107  m–1
36
 = 6.545  10–7 m.
hc
Energy emitted per atom, E =


22
 6.62  10−34 J.s 3  108  ms−1
=
6.545  10−7 m
= 3.03  10–19 J
1.0 g atom hydrogen contains 1 mole hydrogen atoms
Energy emitted by 1.0 g atom of hydrogen
NA  3.03  10–19 J
= 6.02  1023  3.03  10–19 J
= 1.82  105 J

Ex-24. The wave number of the first line in the Balmer series of hydrogen is 15200cm−1. What would be the
wavenumber of the first line in the Lyman series of the Be3+ ion?
(1) 2.4 x 105cm−1 (2) 24.3 x 105cm−1 (3) 6.08 x 105cm−1 (4) 1.313 x 106 cm−1
 1 1 
Sol. Given 15200 = R(1)2  2 −  ..... (1)
 (2) (3)2 
 1 1 
Then  = R(4)2  2 −  ..... (2)
 (1) (2)2 
from (1) and (2) equation
 = 1.313 × 106 cm–1 Ans. (4)

Ex-25. What would be the maximum number of emission lines for atomic hydrogen that you would expect to
see with the naked eye if the only electronic energy levels involved are those as shown in the Figure?

Hint : Balmer series lines lies in visible region.

(1) 4 (2) 6 (3) 5 (4) 15
Sol. Only four lines are present in visible region, 6 → 2, 5 → 2, 4 → 2 & 3 → 2.

No. of photons emitted by a sample of H atom :

If an electron is in any higher state n = n and makes a transition to ground state, then total no. of
n  (n − 1)
different photons emitted is equal to .
2
If an electron is in any higher state n = n2 and makes a transition to another excited state n = n 1, then
n ( n + 1)
total no. of different photons emitted is equal to , where n = n2 – n1
2
Note : In case of single isolated atom if electron make transition from n th state to the ground state then
max. number of spectral lines observed = (n-1)

23
Ex-26 If electron make transition from 7th excited state to 2nd state in H atom sample find the max. number of
spectral lines observed.
Sol. n = 8 – 2 = 6
 6 + 1 7
spectral lines = 6   = 6× = 21
 2  2

Dual nature of electron (de-Broglie Hypothesis):

(a) Einstein had suggested that light can behave as a wave as well as like a particle i.e. it has dual
character.
(b) In 1924, de-Broglie proposed that an electron behaves both as a material particle and as a wave.
(c) This proposed a new theory, the wave mechanical theory of matter. According to this theory, the
electrons protons and even atom when in motion possess wave properties.
(d) According to de-Broglie, the wavelength associated with a particle of mass m, moving with velocity v
is given by the relation,
h
=
mv
where h is Planck’s constant

(e) This can be derived as follows according to Planck’s equation.

h.c
E = h =

Energy of photon on the basis of Einstein’s mass energy relationship
h
E = mc2 or =
mc
Equating both we get
h.c
= mc2

h
or =
mc
Which is same as de - Broglie relation.
h
=
p

h h h
= = =
mv 2mKE 2mqV

* If an electron is accelerated through a potential difference of ‘V’ volt from rest then :
h
 =
2me (eV)

 150 
 =  V Å (on putting values of h, me and e)
 
12.3
= Å (V in volt)
V

24
 h
* mvr = n ×
2
h
=
mv
h nh
mv = putting this in mvr =
 2
h nh  2r 
 r   = = de Broglie wavelength
 2  n 

Ex-27 What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s –1 ?
Sol. According to de Broglie equation
h (6.626  10 –34 Js)
= = = 6.626 × 10–34 m (J = kg m2 s–2).
mv (0.1 kg) (10 m s –1)

Ex-28. A moving electron has 5  10–25 J of kinetic energy. What is the de-broglie wavelength ?
Sol. Mass of the electron m = 9.1  10–31 kg
1
K.E. = mv2 = 5  10–25 J
2
h
=
2m  K.E.
6.626  10 −34
=
2  9.1 10 −31  5  10 −25
= 6.95  10–7 m.

Ex-29. The de Brogile wavelength of an electron moving in a circular orbit is . The minimum radius of orbit is
   
(1) (2) (3) (4)
 2 4 3
Sol. We know 2r = n
For minimum radius n = 1

2rmin =  ; rmin = Ans. (2)
2
Ex-30. An electron, practically at rest, is initially accelerated through a potential difference of 100 volts. It then
has a de Broglie wavelength = 1 Å. It then get retarded through 19 volts and then has a wavelength 2
3 −  2
Å. A further retardation through 32 volts changes the wavelength to 3, What is ?
1
20 10 20 10
(1) (2) (3) (4)
41 63 63 41
150 150
Sol. 1 = Å .... (1) 2 = Å .... (2)
100 81
150
3 = Å .... (3)
49
From (1), (2) and (3)

25
 3 −  2 20
= Ans. (3)
1 63

HEISENBERG’S UNCERTAINITY PRINCIPLE :

It is impossible to measure simultaneously both the position and velocity (or momentum) of a
microscopic particle with absolute accuracy or certainity.
h h h
x.p  or m x.v  or x.v 
4 4 4m
where, x = uncertainity in position
p = uncertainity in momentum
h = Plank’s constant
m = mass of the particle
v = uncertainity in velocity
 In terms of uncertainity in energy E, and uncertainity in time t, this principle is written as,
h
E.t  .
4
 Heisenberg replaced the concept of definite orbits by the concept of probability.

Ex-31. Alveoli are the tiny sacs of air in the lungs whose average diameter is 50 pm. Consider an oxygen
molecule trapped within a sac. Calculate uncertainty in the velocity of oxygen molecule?
(1) 1.98 × 10–2 ms–1 (2*) 19.8 ms–1 (3) 198 × 10–4 ms–1 (4) 19.8 × 10–6ms–1
Sol. x = 50 pm
h 6 . 625  10−34  6.022  1023
So v = = m/sec
4 . m x 4  3.14  32  10 −3  50  10 −12
= 0.0019853 × 104 m/sec
= 19.853 m/sec
Ex-32. Determine the de-Broglie wavelength associated with an electron in the 3rd Bohr's orbit of He+ ion?
(1) 10 Å (2) 2 A (3*) 5 Å (4) 1 Å
Sol. n = 2r
2r 2 9
so = = × (53 pm) × 5Å
3 3 2
Ex-33. If the light of wavelength 5.77 × 10–10 cm is used to detect an electron then the uncertainly in velocity
(approximately) will be (h = 6.6 × 10–34 Js, me = 9.1 × 10–31 kg).
(1) 106 m/s (2) 106 m/s (3*) 107 m/s (4) None of these
−34
h 6.6  10
Sol. V = = = 107 m/s
4  m x 4  3.14  9.1 10−31  5.77  10−12

Ex-34. The uncertainty in position and velocity of the particle are 0.1 nm and 5.27×10 –24 ms–1 respectively then
the mass of the particle is : (h = 6.625 × 10–34Js)
(1) 200 g (2) 300 g (3*) 100 g (4) 1000 g
h 6.6  10−34
Sol. m= =
4x 4  3.14  10−10  5.27  10−24
100 gm.
3.3
Ex-35. Uncertainty in position of a hypothetical subatomic particle is 1Å and uncertainty in velocity is × 105
4
m/s then the mass of the particle is approximately [h = 6.6 × 10–34 Js]

26
 (1) 2 × 10–28 kg (2) 2 × 10–27 kg (3) 2 × 10–29 kg (4) 4 × 10–29 kg
Sol. x × m × v  h/4
3.3 6.6  10 −34
1 × 10-10 × m × × 105  m = 2 × 10– 29 kg Ans. (3)
4 4

THE SCHRODINGER EQUATION :

This equation determines the probability of position and energy of electron which is moving around the
nucleus.
2  2  2  8 2 m 82m
+ + + or  y +
2
(E – V) y = 0
x 2
y 2
z 2
h 2
h2
Where = Laplacian operator.
E = Total energy of the electron.
V = Potential energy of the electron
y = Wave function or wave amplitude
2
y = Probability density (probability of finding an electron in an atom)
Important features of the quantum mechanical model of atom
Quantum mechanical model of atom is the picture of the structure of the atom, which emerges from the
application of the Schrodinger equation to atoms. The following are the important features of the
quantum mechanical model of atom :
1. The energy of electrons in atoms is quantized (i.e. can only have certain specific values), for example
when electron are bound to the nucleus in atoms.
2. The existence of quantized electronic energy levels is a direct result of the wave like properties of
electrons and are allowed solutions of schrdinger wave equation.
3. Both the exact position and exact velocity of an electron in an atom cannot be determined
simultaneously (Heisenberg uncertainty principle). The path of an electron in an atom therefore, can
never be determined or known accurately. That is why, as you shall see later on, one talks of only
probability of finding the electron at different points in an atom .

4. An atomic orbital is the wave function  for an electron in an atom. Whenever an electron is
described by a wave function, we say that the electron occupies that orbial
5. The probability of finding an electron at a point within an atom is porportional to the square of the
2 2
orbital wave function i.e. || at that point. || is known as probability density and is always positive.
Hydrogen atom and the schrodinger Equation :
When Schrodinger equation is solved for hydrogen atom. The solution gives the possible energy levels
the electron can occupy and the corresponding wave function(s) () of the electron associated with
each energy level. These quantized energy states and corresponding wave functions which are
characterized by a set of three quantum numbers (principal quantum number n, azimuthal quantum
number  and magnetic quantum number m) arise as a natural consequence in the solution of the
Schrodinger equation.
Application of Schrodinger equation to multi-electron atoms presents a difficulty : the Schrodinger
equation cannot be solved exactly for a multi electron system. This difficulty can be overcome by using
approximate methodes such calculation with the aid of modern computers show that orbitals in atoms
other than hydrogen do not differ in any radical way from the hydrogen orbitals discussed above. The
principle difference lies in the consequence of increased nuclear charge. Because of this all the orbitals
are somewhat contracted. Unlike orbitals of hydrogen or hydrogen like species, whose energies

27
 depends only on the quantum number n, the energies of the orbitals in multi-electron atoms depends on
quantum numbers n and .

QUANTUM NUMBERS :
The set of four numbers required to define an electron completely in an atom are called quantum
numbers. The first three have been derived from Schrodinger wave equation.
(i) Principal quantum number (n) : (Proposed by Bohr)
It describes the size of the electron wave and the total energy of the electron. It has integral values 1, 2,
3, 4 ...., etc., and is denoted by K, L, M, N. ..., etc.

* Number of subshell present in nth shell = n

n subshell
1 s
2 s, p
3 s, p, d
4 s, p, d, f
* Number of orbitals present in nth shell = n2 .
* The maximum number of electrons which can be present in a principal energy shell is equal to 2n2.
No energy shell in the atoms of known elements possesses more than 32 electrons.
nh
* Angular momentum of any orbit =
2
(ii) Azimuthal quantum number (l) : (Proposed by Sommerfield)
It describes the shape of electron cloud and the number of subshells in a shell.
* It can have values from 0 to (n – 1)

* value of  subshell
0 s
1 p
2 d
3 f

* Number of orbitals in a subshell = 2 + 1

* Maximum number of electrons in particular subshell = 2 × (2 + 1)

h  h 
* Orbital angular momentum L = ( + 1) =  ( + 1)  = 2 
2  
h
i.e. Orbital angular momentum of s orbital = 0, Orbital angular momentum of p orbital = 2 ,
2
h
Orbital angular momentum of d orbital = 3
2
(iii) Magnetic quantum number (m) : (Proposed by Linde)
It describes the orientations of the subshells. It can have values from – to +  including zero, i.e., total
(2 + 1) values. Each value corresponds to an orbital. s-subshell has one orbital, p-subshell three
orbitals (px, py and pz), d-subshell five orbitals (dxy ,dyz , dzx , dx2 − y2 , dz2 ) and f-subshell has seven
orbitals. The total number of orbitals present in a main energy level is ‘n2’.
(iv) Spin quantum number (s) : (Proposed by Goldschmidt & Uhlenbeck)

28
 It describes the spin of the electron. It has values +1/2 and –1/2. (+) signifies clockwise spinning and (–)
signifies anticlockwise spinning.
eh
* Spin magnetic moment s = s(s + 1) or  = n(n + 2) B.M. (n = no. of unpaired
2 mc
electrons)
h
* It represents the value of spin angular momentum which is equal to s(s + 1)
2
1
* Maximum spin of atom = × No. of unpaired electron.
2
⚫ Total no. of e– in main energy shell = 2n2
Total no. of e– in a subshell = 2 (2 + 1)
Maximum no. of e– in an orbital = 2
Total no. of orbitals in a subshell = (2 + 1)
No. of subshells in main energy shell = n
No. of orbitals in a main energy shell = n2
h
⚫ Orbital angular momentum L = ( + 1)
2
h 1
⚫ Spin angular momentum S = S (S + 1) ; S =
2 2
= n (n + 2) B.M. (n = no. of unpaired electrons)
eh
⚫ Orbital magnetic moment s = × ( + 1)
4 me

ELECTRONIC CONFIGURATION
Pauli’s exclusion principle :
No two electrons in an atom can have the same set of all the four quantum numbers, i.e., an orbital
cannot have more than 2 electrons because three quantum numbers (principal, azimuthal and
magnetic) at the most may be same but the fourth must be different, i.e., spins must be in opposite
directions.

Aufbau principle :
Aufbau is a German word meaning builiding up. The electrons are filled in various orbitals in order of
their increasing energies. An orbital of lowest energy is filled first. The sequence of orbitals in order of
their increasing energy is : 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, ....
The energy of the orbitals is governed by (n + ) rule.
(n + l) Rule :
This rule states that electrons are filled in orbitals according to their (n + ) values.Electrons are filled in

increasing order of their (n + ) values. When (n + ) is same for sub energy levels, the electrons first
occupy the sublevels with lowest "n" value.

Energy level diagram :

The representation of relative energy levels of various atomic orbital is made in terms of energy level
diagrams.

29
 7s
6p 5d
6s 4f

n=1 1s 5p
4d
2s 5s
n=2 2p
4p

Increasing energy
3s 3p 3d
n=3 3d 4s

n=4 4s 4p
4d 3p
4f
5s 5p
n=5 5d 3s
5f
n=6 6s 6p
6d 2p
7s 7p
n=7 2s

1s

Fig. Sequence of filling of electrons in orbitals belonging to different energy levels

Hund’s rule :
No electron pairing takes place in the orbitals in a subshell until each orbital is occupied by one electron
with parallel spin. Exactly half filled and fully filled orbitals make the atoms more stable, i.e., p 3, p6, d5,
d10, f7 and f14 configuration are most stable.

Ex-36. Calculate total spin, magnetic moment for the atoms having at. no. 7, 24 and 36.
Sol. The electronic configuration are
7N : 1s2, 2s2 2p3 unpaired electron = 3
24Cr : 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1 unpaired electron = 6
36Kr : 1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p6 unpaired electron = 0
 Total spin for an atom = ± 1/2 × no. of unpaired electron
For 7N, it is = ± 3/2 ; For 24Cr, it is = ± 3 ; For 36Kr, it is = 0
Also magnetic moment = n (n + 2)
For 7N, it is = 15 ; For 24Cr, it is = 48 ; For 36Kr, it is = 0

Ex-37. Write down the four quantum numbers for fifth and sixth electrons of carbon atom.
Sol. 2 2 2
6C : 1s , 2s 2p

1 1
fifth electron : n=2 =1 m = – 1 or +1 s = + or –
2 2
1 1
sixth electron : n=2 =1 m=0 s=+ or –
2 2
Ex-38. Given below are the sets of quantum numbers for given orbitals. Name these orbitals.
(a) n = 3 (b) n = 5 (c) n = 4 (d) n = 2 (e) n = 4
=1 = 2 = 1 =0 =2
Ans. 3p, 5d, 4p, 2s, 4d

30
Sol. (a) n = 3 ,  = 1  3p (b) n = 5 ,  = 2  5d (c) n = 4 ,  = 1  4p

(d) n = 2 ,  = 0  2s (e) n = 4 ,  = 2  4d

Ex-39. Which of the following set of quantum numbers is not valid.

1 1
(1) n = 3, = 2, m = 2, s = + (2) n = 2, = 0, m = 0, s = –
2 2
1 1
(3) n = 4, = 2, m = –1, s = + (4) n = 4, = 3, m = 4, s = –
2 2
Sol. Not valid Ans. (4)
Ex-40. What is the total spin value in case of 26Fe3+ ion
(1) +1 or –1 (2) +2 or –2 (3) + 2.5 or – 2.5 (4) +3 or –3
 1
Sol. Total spin = no. of unpaired e– ×   
 2
 1 5
= 5 ×   =± Ans. (3)
 2 2

Shape of the orbitals :

Shape of the orbitals are related to the solutions of Schrodinger wave equation and gives the space in
which the probability of finding an electron is maximum.
s- orbital : Shape → spherical

s- orbital is non directional and it is closest to the nucleus, having lowest energy.
s-orbital can accomodate maximum no. of two electrons.

p-orbital : Shape → dumb bell

Dumb bell shape consists of two loops which are separated by a region of zero probability called node.

p - orbital can accomodate maximum no. of six electrons.

d - Orbital : Shape double → dumb bell

31
 d - orbital can accomodate maximum no. of 10 electrons.

f - orbital : Shape leaf like

f - orbital can accomodate maximum no. of 14 electrons.

Nodes in orbitals
(i) total nodes = n – 1,
(ii) angular nodes = ,
(iii) radial nodes = n – – 1.

Nuclear Chemistry
Spontaneous disintegration of nuclei due to emission of radiations like , ,  is called radioactivity.
Radioactivity is a nuclei phenomenon.
Radioactivity is not dependent on external conditions like temperature, pressure etc.
Radioactivity of a substance is independent to its physical state.
x(s), x(l), x(g), (x)+(g), (x)–(g) in all form, x is radioactive.
14
CO2, 146C(s), 146C(g) is radioactive.
Radiations :
 : 2He4 ( 24He2+) (nucleus of He-atom)
 or – : –1e0 (fast moving electron emitted from nucleus)
 : 00 (electromagnetic radiation (waves) of high frequency)
speed : >>
penetrating power : >>
ionisation power : >>
Emission of rays Usual condition Effect Process representation / example
n
1.  Z > 83 ratio increases Z
XA → Z– 2X’A – 4 + 2He4
p
92
U238 → 90Th234 + 2He4
n n
2.  If ratio is high. ratio decreases Z
YA → Z+ 1Y’A – 4 + –1e0
p p
n 6
eg. 6
C12 (stable) = 6
C14 → 7N14 + –1e0
p 6
n 8 n 8 n 7
6
C14 (radioactive) = (high) = =
p 6 p 6 p 7

32
 n 13
eg. 11
Na24 (radioactive) = (high) 0
n1 → 1p1 + –1e0 (from nucleus)
p 11
n 12
11
Na23 (stable) =
p 11
n 11 n
11
Na22 = ( ratio low)
p 11 p
3.  If nucleus energy nucleus energy 43
Tc99 → 43Tc99 + 
level is high level decreases high low
nucleus nucleus
energy energy
(metastable)
n n
4. (a) Positron emission If ratio is low ratio increases Z
YA → Z – 1Y´A + +1e0
p p
(+1e0) 11
Na22 → 10Ne22 + +1e0
1
p1 → 0n1 + +1e0 (from nucleus)
n n
−1e →
0
(b) Electron capture If ratio is low ratio increases Z
X´A + Z–1
X´´A
p p K −shell

(EC) or K-shell 80
Hg197 + –1e0 → 79Au197

Electron capture

1p
1
+ −1 e0 ⎯⎯→ 0 x1
(K − shell)

Nuclear stability belt :

200
(1.5:1 ratio) 80 Hg
B
Belt of
stability
Number of neutrons

(1.40:1 ratio) 168

70 Yb

(1.25:1 ratio) C
90
40 Zr

1 :1 neutron to
proton ratio

A
Number of protons
-emission
0
n1 → 1p1 + –1e0
* Z upto 20 : nuclei stable with n/p ratio nearly 1 : 1
* Z > 20 : n/p ratio increases with Z in stable nuclie region.

33
 * More number of neutrons are required to reduce repulsion between protons.
* 83
Bi209 : Stable with largest n/p ratio
n 1.52
=
p 1

Even - odd rule :

no. of n no. of p no. of stable nuclie
even even 155 (max)
even odd 55
odd even 50
odd odd 5 (min)
* Expected pairing of nucleus
Magic Numbers :
Nuclei in which nucleons have magic no. (2, 8, 20, 28, 50 ....) are more stable.
e.g. 2
He4 , 8O16

Group displacement law : (Given by Soddy and Fajan)

* When 1 emission takes place from a nuclie, new formed nuclie occupy two position left in periodic
table.

* When 1 emission takes place from a nuclie, new formed nuclie occupy one position right in periodic
table.

Due to emission of 1 particle; isobars are formed.

Due to emission of 1 particle; isodiaphers are formed.

Due to emission of 1 and 2 ; isotopes are formed.

Isotopes : same number of proton eg. 6C14 and 6C12
Isobars : same mass number eg. 6C14 and 7N14
Isotones : same number of neutron eg. 2He4 and 1H3

Isodiaphers : Same (n – p) difference

e.g. 9F19 and 19K39 ; (n – p) = 10

Isosters : Same number of atoms and electrons

e.g. N2 and CO
N2O and CO2

Artifical nuclear reaction :

* specific nuclei + stricking particle —→ New nuclei + emitted particle

eg. 1. (, p type) 7
N14 + 2
He4 —→ 8
O17 + 1
p1 (or 1H1)
(s.p.) (e.p.)
2. (n,  type) 11
N 23
+ 1
n0 —→ 11
Na 24
+ 
3. (D, p type) 13
Al 27
+ 1
H 2
—→ 13
Al
28
+ 1
H1
4. (p,  type) 3
Li7 + 1
H1 —→ 2
He4 + 2
He4

Nuclear fission and nuclear fusion :

34
 In both processes, large amount of heat evolved due to conversion of some mass into energy.
Nuclear fission : Is a process where heavy nuclei splits into large nuclei.

92
U235 + 0n1 —→ 92
U*236

eg. atom bomb is based on fission.

Nuclear fusion :
Is a process where light nuclei fused together to form heavy nuclei.
1
H2 + 1H3 —→ 2He4 + 0n1
1
H2 + 1H2 —→ 2
He4
Hydrogen bomb is based on fusion. Very high temperature is required in this process.

24
Ex-41. Na is the most stable isotope of Na. Find out the process by which 11 Na can undergo radioactive
23

decay.
Sol. n/p ratio of 24Na is 13/11 and thus greater than one. It will therefore decay following -emission.
24 24 0
11 Na ⎯⎯→ 12 Mg + −1e

Ex-42. The number of -particle emitted during the change c

aX
b
⎯⎯→ d Y is :

a−b a−b c −b a−b

(A) (B) d +   +c (C) d +   –a (D) d +  –c
4  2   2   2 
c b 4 0
Sol. aX ⎯⎯→ d Y + m 2 He + n −1e

 c = b + 4m ......(i)
and a = d + 2m – n ......(ii)
by (i) & (ii)

c −b
n = d +  – a. Ans. (C)
 2 

13
Ex-43. The decay product of 7 N is :
13 0 13 0
(A) 8 O + −1e (B) 6 C + +1e

13
(C) 6 C + K electron capture (D) 95 Be + 24 He

13 n
Sol. 7 N is positron emitter ; ratio is low. Ans. (B)
p

Ex-44. A radioactive element X has an atomic numbers of 100. It decays directly into an element Y which
decays directly into an element Z. In both processes a charged particle is emitted. Which of the
following statement would be true?
(A) Y has an atomic number of 102. (B) Z has an atomic number of 101.
(C) Z has an atomic number of 97. (D) Z has an atomic number of 99.
Sol. X and Y can decay one  each or one  each or X-decays, 1 , Y-decays 1  or X-decays 1  or Y-
decays 1 . In either case (A), (B) and (C) cannot be true. Ans. (D)

35
 ONLY ONE OPTION CORRECT TYPE
Section (A) : Cathode rays, Anode rays, Basic definations and Rutherford model

1. Cathode ray are :

(1) stream of electrons (2) stream of -particles
(3) radiation (4) stream of cations

2. Which of the following statement is wrong about anode rays :

(1) They travel in straight line (2) They produce heating effect
(3) They carry positive charge (4) None of these

3. Which of the following is/are affected by electric field :

(1) Anode rays (2) Cathode rays (3) Both (1) and (2) (4) None of these

4. The e/m ratio for Anode rays :

(1) varies with the element forming the anode in the discharge tube.
(2) varies with the gas in the discharge tube.
(3) is constant.
(4) Both (1) & (2).

5. Millikan's oil drop expriments is used to find -

(1) e/m ratio of an electron (2) Charge of an electron
(3) Mass of an electron (4) Velocity of an electron

6. Match the following :

Column-I Column-II
Sub-atomic particles Persons responsible for discovery
(1) Electron (p) James Chadwick
(2) Proton (q) J.J. Thomson
(3) Neutron (r) Rutherford
(4) Nucleus (s) Goldstein
(1) (1 - q, 2 - s, 3 - r, 4 - p) (2) (1 - p, 2 - p, 3 - q, 4 - s)
(3) (1 - r, 2 - s, 3 - p, 4 - q) (4) (1 - q, 2 - s, 3 - p, 4 - r)

7. An element having atomic number 25 and atomic weight 55 will have –

(1) 25 protons and 30 neutrons (2) 25 neutrons and 30 protons
(3) 55 protons (4) 55 neutrons

8. Which of the following is isoelectronic with N2O :

(1) NO (2) N2O5 (3) CO2 (4) CO

9. The charge on the atom having 17 protons, 18 neutrons and 18 electrons is

(1) + 1 (2) – 1 (3) – 2 (4) zero

281
10. Number of protons, neutrons and electrons in the element 89 Ac are respectively :
(1) 89, 231, 89 (2) 89, 89, 242 (3) 89, 142, 89 (4) 89, 192, 89
11. An isotone of 16
8 O is :

36
 17 12 14 32
(i) 8 O (ii) 6 C (iii) 6 C (iv) 16 S
(1) (ii & iii) (2) (i & ii) (3) (iii) (4) (ii & iii & iv)

12. Which of the following are isoelectronic with one another ?

(1) Na+ and Ne (2) K+ and O (3) Ne and O (4) Na+ and K+

13. Rutherford's experiment on scattering of particles showed for the first time that the atom has
(1) Electrons (2) Protons (3) Nucleus (4) Neutrons

14. When atoms are bombarded with alpha particles, only a few in million suffer deflection, others pass out
undeflected. This is because
(1) The force of repulsion on the moving alpha particle is small
(2) The force of attraction on the alpha particle to the oppositely charged electrons is very small
(3) There is only one nucleus and large number of electrons
(4) The nucleus occupies much smaller volume compared to the volume of the atom

15. Which of the following elements has maximum density of nucleus :

(1) 14Si30 (2) 15P31 (3) 8O16 (4) All have same density

Section (B) : Nature of Light and photoelectric effects

1. A photon in X region is more energetic than in the visible region ; X is :

(1) IR (2) UV (3) Microwave (4) Radio wave

2. Photon of which light has maximum energy :

(1) red (2) blue (3) violet (4) green

3. The frequency of yellow light having wavelength 600 nm is :

(1) 5.0 × 1014 Hz (2) 2.5 × 107 Hz (3) 5.0 × 107 Hz (4) 2.5 × 1014 Hz

4. Wave number of radiations having frequency of 4 × 10 4 Hz will be :

(1) 1.33 × 10–6 cm–1 (2) 1.33 × 10–7 cm–1 (3) 9 × 10–11 cm–1 (4) 4 × 10–5 cm–1

5. A wavelength of 400 nm of an electromagnetic radiation is not correspond to :

(1) frequency = 7.5 × 1014 Hz (2) wave number = 2.5 × 106 m–1.
(3) velocity = 3 × 10 ms
8 –1
(4)  = 40 Å

6. Which one of the following is not the characteristic of Planck's quantum theory of radiation-
(1) The energy is not absorbed or emitted in whole number multiple of quantum.
(2) Radiation is associated with energy.
(3) Radiation energy is not emitted or absorbed continously but in the form of small packets called
quanta.
(4) This magnitude of energy associated with a quantum is proportional to the frequency.

7. Wavelength of a photon having an energy of 2 eV. will be :

(1) 6.2  10–7 m (2) 6.2  10–6 m (3) 6.2  10–9 m (4) 6.2  10–8 m

8. One quantum is absorbed per gaseous molecule of X 2 for converting it into X atoms. If light absorbed
has wave length 1240Å, then bond energy of X2 will be :
(1) 10 eV/molecule (2) 20 J/mole (3) 48 eV/molecule (4) 184 J/mol

9. The work function for a metal is 4 eV. To eject a photoelectron of zero velocity from the surface of the
metal, the wavelength of incident light should be above :
(1) 310 Å (2) 1550 Å (3) 155 Å (4) 3100 Å

37
10. The energy required to remove an electron from a metal X is 3.31 × 10 –20 J. Wavelength/s of light that
can not photoeject an electron from metal X is/are
(1) 4 m (2) 6 m (3) 7 m (4) 5 m

11. Light of wavelength  falls on metal having work function hc/0. Photoelectric effect will take place only
if :
(1)   0 (2)   20 (3)   0 (4)   0/2

12. Cu metal (work function = 4.8 eV) can show photoelectric effect if wavelength of photon is :
(1)  = 5000 Å (2)  = 6000 Å (3)  = 2000 Å (4)  = 4000 Å

13. Maximum kinetic energy of photoelectron using photon of wavelength 2000Å at Cu metal will be (If work
function of Cu is 4.8 eV.) :
(1) 2.4 ev (2) 1.4 ev (3) 1.9 ev (4) 3.4 ev

14. When the frequency of light incident on a metallic plate is doubled, the maximum KE of the emitted
photoelectrons will be :
(1) Doubled
(2) Halved
(3) Increased but more than double of the initial maximum KE
(4) Unchanged

15. The number of photoelectrons emitted depends upon :

(1) The intensity of the incident radiation
(2) The frequency of the incident radiation
(3) The product of intensity and frequency of incident radiation
(4) None of these

Section (C) : Bohr's Model

1. The expression for Bohr’s radius of an atom is
n2h2 n2h2 n2h2 n2h2
(1) r = 2 4 2 (2) r = 2 2 (3) r = 2 2 2 (4) r =
4 me z 4 me z 4 me z 4 m 2 e 2 z 2
2

2. Ratio of radii of second and first Bohr orbits of H atom is :

(1) 2 (2) 4 (3) 3 (4) 5

3. The ratio of radii of second orbits of He+ , Li2+ and Be3+ is :

(1) 1 : 2 : 3 (2) 6 : 4 : 3 (3) 3 : 4 : 6 (4) none of these

4. If r is the radius of first orbit, the radius of nth orbit of H atom is given by -
(1) r n (2) r n2 (3) r/n (4) r2 n2

5. If the speed of electron in second orbit of He+ is "v", then the speed of electron in first Bohr orbit of
hydrogen atom will be :
(1) v/2 (2) 2v (3) v (4) 4v

6. What is the ratio of speeds of electrons in Ist orbit of H-atom to IVth orbit of He+ ion .
(1) 2 : 1 (2) 8 : 3 (3) 3 : 2 (4) 27 : 5

7. If the radius of Ist orbit of hydrogen atom is 0.53 Å then radius of Ist orbit of He+ is :

38
 (1) 1.27 Å (2) 0.265 Å (3) 1.59 Å (4) 0.132 Å

8. Which state of the triply ionized Beryllium (Be3+) has the same orbit radius as that of the ground state of
hydrogen atom ?
(1) 1 (2) 2 (3) 3 (4) 4

9. If the velocity of the electron in first orbit of H atom is 2.18 × 106 m/s, what is its value in third orbit ?
(1) 7.27 × 105 m/s (2) 4.36 × 106 m/s (3) 1.24 × 105 m/s (4) 1.09 × 106 m/s

10. The difference in angular momentum associated with the electron in two successive orbits of hydrogen
atom is :
(1) h/ (2) h/2 (3) h/2 (4) (n – 1)h/2

11. The angular momentum of an electron in a given orbit is J, Its kinetic energy will be :
1 J2 Jv J2 J2
(1) (2) (3) (4)
2 mr 2 r 2m 2

12. Angular momentum in 2nd Bohr orbit of H-atom is x. Then angular momentum of electron in Ist excited
state of Li+2 is :
x
(1) 3x (2) 9x (3) (4) x
2

14. When an electron drops from a higher energy level to a low energy level, then :
(1) energy is absorbed (2) energy is emitted
(3) atomic number increases (4) atomic number decreases

14. The maximum energy of an electron in an atom will be at :

(1) Nucleus (2) Ground state
(3) First excited state (4) Infinite distance from the nucleus

15. The ratio of potential energy and total energy of an electron in a Bohr orbit of hydrogen like species is :
(1) 2 (2) –2 (3) 1 (4) –1

16. The ratio of energies of hydrogen atom for first and second excited state is :
(1) 4/1 (2) 1/4 (3) 4/9 (4) 9/4

17. In hydrogen atom, energy of first excited state is – 3.4 eV. Then, KE of same orbit of hydrogen atom is:
(1) + 3.4 eV (2) + 6.8 eV (3) – 13.6 eV (4) + 13.6 eV

18. Potential energy of electron is – 27.2 eV in 2nd orbit of He+. Then total energy of electron in first excited
state of Hydrogen atom will be :
(1) –3.4 eV (2) –13.6 eV (3) 3.4 eV (4) 13.6 eV

19. If the potential energy of electron in hydrogen atom is –3.02eV then in which of the following excited
level is electron present :
(1) 1st (2) 2nd (3) 3rd (4) 4th

20. The ionisation energy for the H-atom is 13.6 eV. Then the required energy in eV to excite the electron it
from the ground state to next higher state will be : (in eV)
(1) 3.4 (2) 10.2 (3) 12.1 (4) 1.5

39
21. The energy of an electron in an excited H-atom is –1.51 eV. Angular momentum of electron in the given
orbit will be
(1) 3h/ (2) 3h/2 (3) 2h/ (4) h/

22. The ionization energy of H-atom is 13.6 eV. The ionization energy of Li+2 ion will be :
(1) 54.4 eV (2) 122.4 eV (3) 13.6 eV (4) 27.2 eV

23. If the binding energy of 3rd orbit of a H-like species is 13.6 eV, then the species must be :
(1) Be3+ (2) Li+ (3) He+ (4) None of these

24. In which excited state that a hydrogen atom sample already in ground state can reach when it is
bombarded with photons of energy 12.75 eV, will be :
(1) 4 (2) 2 (3) 3 (4) No transition will occur.

25. If the binding energy of first excited state of a hydrogen like sample is 54.4 eV, then determine the
atomic number Z of the H-like species :
(1) 1 (2) 2 (3) 3 (4) 4

26. The excitation energy of first excited state of a hydrogen like atom is 40.8 eV. The energy needed to
remove the electron from ground state of the ion is :
(1) 54.4 eV (2) 122.4 eV (3) 40.8 eV (4) 13.6 eV

27. Match the following

(1) Energy of ground state of He+ (i) + 6.04 eV
(2) Potential energy of  orbit of H-atom (ii) –27.2 eV
(3) Kinetic energy of  excited state of He+ (iii) 54.4 V
(4) Ionisation potential of He+ (iv) – 54.4 eV
(1) A – (i), B – (ii), C – (iii), D – (iv) (2) A – (iv), B – (iii), C – (ii), D – (i)
(3) A – (iv), B – (ii), C – (i), D – (iii) (4) A – (ii), B – (iii), C – (i), D – (iv)

Section (D) : Atomic Spectrum

1. The wavelength of a spectral line for an electronic transition is inversly related to :
(1) No. of electrons undergoing transition
(2) The nuclear charge of the atom
(3) The velocity of an electron undergoing transition
(4) The difference in the energy levels involved in the transition

2. Different lines in Lyman series of hydrogen spectrum lie in :

(1) UV (2) Visible (3) IR (4) None of these

3. The spectral lines corresponding to the radiation emitted by an electron jumping from 6th, 5th and 4th
orbits to second orbit belong to :
(1) Lyman series (2) Balmer series (3) Paschen series (4) Pfund series

4. The transition of electron in H-atom that will emit maximum energy is :

(1) n3 ࢮ n2 (2) n4 ࢮ n3 (3) n5 ࢮ n4 (4) All have same energy

5. Which transition emits photon of maximum frequency in hydrogen like species :

(1) 2nd spectral line of Balmer series (2) 2nd spectral line of Paschen series

40
 (3) 5th spectral line of Humphery series (4) 5th spectral line of Lymen series

6. The shortest  for the Brakett series for H is : (Given RH = 109678 cm–1)
(1) 1459 Å (2) 4052 Å (3) 4052 nm (4) 1459 nm

7. Wavelength of 1st line of Balmer series in hydrogen spectrum is :

(1) 6656 Å (2) 6266 Å (3) 6626 Å (4) 6566 Å

8. When an electron in an excited hydrogen atom jumps 5th orbit to 3rd orbit the spectral line is observed
in the ..........region and in .............series of the hydrogen spectrum.
(1) Visible, Balmer (2) Visible, Lyman (3) Infrared, Paschen (4) Infrared, Balmer

9. No. of visible lines when an electron returns from 5th orbit to ground state in H spectrum -
(1) 5 (2) 4 (3) 3 (4) 10

10. Maximum number of spectral lines in Lyman series will be if electron makes transistion from n th orbit :
(1) n (2) n – 1 (3) n – 2 (4) n (n + 1)

11. In a isolated H-atom, electron transits from 6th orbit to 2nd orbit maximum number of spectral lines will be
(1) 6 (2) 10 (3) 4 (4) 0

12. The wave number of first line of Balmer series of hydrogen atom is 15200 cm –1. What is the wave
number of first line of Balmer series of Li2+ ion:
(1) 15200 cm–1 (2) 13680000 m–1 (3) 76000 cm–1 (4) 13680 cm–1

13. Calculate number of possible spectral lines which may be emitted in bracket series in H atom, if
electrons present in 9th excited level returns to ground level.
(1) 4 (2) 5 (3) 6 (4) 7

14. Ratio of wavelength of second line of Lyman series to that of series limit of Paschen series of H-atom.
(1) 1/8 (2) 1/9 (3) 8/9 (4) 9/8

Section (E) : de-Broglie concept & Heisenbergs uncertainity principle

1. The de Broglie equation suggests that an electron has

(1) Particle nature (2) Wave nature
(3) Both Particle & wave nature (4) Radiation behaviour

2. The wavelength of a charged particle ________the square root of the potential difference through which
it is accelerated :
(1) is inversely proportional to (2) is directly proportional to
(3) is independent of (4) is unrelated with

3. If the kinetic energy of an electron is increased 4 times, the wavelength of the de-Broglie wave
associated with it would become :
(1) four times (2) two times (3) half times (4) one fourth times
4. A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be : (h = 6.6 × 10–34 Js)
(1) 6.6 × 10–32 m (2) 6.6 × 10–34 m (3) 1.0 × 10–35 m (4) 1.0 × 10–32 m

5. The de-broglie wavelength associated with a ball of mass 1 kg having kinetic energy 0.5 J is.
(1) 6.626  10–34 m (2) 13.20  10–34m (3) 10.38  10–21 m (4) 6.626  10–34 Å

41
6. The speed of a proton is one hundredth of the speed of light in vacuum. What is its de-Broglie of proton
wavelength? Assume that one mole of protons has a mass equal to one gram [h = 6.626 × 10–27 erg
sec] :
(1) 13.31 × 10–3 Å (2) 1.33 × 10–3 Å (3) 13.13 × 10–2 Å (4) 1.31 × 10–2 Å

7. What possibly can be the ratio of the de Broglie wavelengths for two electrons each having zero initial
energy and accelerated through 50 volts and 200 volts ?
(1) 3 : 10 (2) 10 : 3 (3) 1 : 2 (4) 2 : 1

8. A helium molecule is moving with a velocity of 2.40 x 102 ms–1 at 300K. The de-Broglie wave length is
about
(1) 0.416 nm (2) 0.83 nm (3) 803 Å (4) 8000 Å

9. If wavelength is equal to the distance travelled by the electron in one second, then -
p h h h
(1)  = (2)  = (3)  = (4)  =
h m p m

10. de-Broglie wavelength of electron in second orbit of Li2+ ion will be equal to de-Broglie of wavelength of
electron in
(1) n = 3 of H-atom (2) n = 4 of C5+ ion (3) n = 6 of Be3+ ion (4) n = 3 of He+ ion

11. What is the de-Broglie wavelength associated with the electron in 3rd orbit of hydrogen :
(1) 9.96  10−10 cm (2) 9.96  10−8 cm (3) 9.96  104 cm (4) 9.96  108 cm

12. Select the incorrect relationship among the following :

h h h h
(1) x  p  (2) x p  (3) x  V  (4) E t 
4  4  m 4  m 4 

13. It the uncertainity in position of a moving particle is 0 then uncertainity in momentum will be :
(1) 0 (2) 1 (3)  (4) Can not predict

14. The Uncertainity in the momentum of an electron is 1.0 × 10 –10 kg m s–1 . The Uncertainity in its
position will be: (h = 6.62 × 10–34 Js)
(1) 1.05 × 10–28 m (2) 1.05 × 10–26 m (3) 5.27 × 10–25 m (4) 5.25 × 10–28 m

15. What is the minimum uncertainty in position of a proton whose velocity is given by 1.5 × 10 6  1500 m/s
(1) 21 m (2) 21 cm (3) 21 m (4) 21 pm

Section (F) : Quantum Numbers and Electronic configuration

1. Magnetic quantum number specifies -

(1) Size of orbitals (2) Shape of orbitals
(3) Orientation of orbitals (4) Nuclear stability

2. A p-orbital can accommodate

(1) 4 electrons (2) 6 electrons
(3) 2 electrons with parallel spins (4) 2 electrons with opposite spins

3. A given orbital is labeled by the magnetic quantum number m = –1. This could not be
(1) s - orbital (2) p-orbital (3) d-orbital (4) f-orbital

4. Which of the following represents the correct set of quantum numbers of a 4d electron ?

42
 1 1 1
(1) 4, 3, 2, + (2) 4, 2, 1, 0 (3) 4, 3, – 2, + (4) 4, 2, 1, –
2 2 2
1
5. An orbital containing electron having quantum number n = 4, l = 3, m = 0 and s = – is called
2
(1) 3s orbital (2) 3p orbital (3) 4d orbital (4) 4f orbital

6. The maximum number of electrons in a subshell is given by the expression

(1) 4l – 2 (2) 4l + 2 (3) 2l + 2 (4) 2n2

7. The electrons present in K-shell of the atom will differ in

(1) principal quantum number (2) azimuthal quantum number
(3) magnetic quantum number (4) spin quantum number

1
8. The maximum number of 3d-electrons that can have s = – , are
2
(1) 10 (2) 3 (3) 5 (4) 7

9. If n and  are respectively the principal and azimuthal quantum numbers, then the expression for
calculating the total number of electrons in any orbit is -
= n = n −1 = n +1 = n −1
(1) 2=1
(2 + 1) (2) =1
2 (2 + 1) (3) = 0
2(2 + 1) (4) = 0
2(2 + 1)

10. Number of electrons having  = 1 and m=0 in P-atom in its ground state :
(1) 3 (2) 1 (3) 2 (4) 0

11. Maximum number of electrons that can have n = 3,  = 2, m = +2, s = +½ in an atom are :
(1) 18 (2) 6 (3) 24 (4) 1

12. A correct set of four quantum numbers for unpaired electron in Cl-atom :
n l m s
(1) 3 2 0 +½
(2) 3 1 0 +½
(3) 3 1 +1 0
(4) 3 0 –1 +½

13. Spin angular momentum for an electron is given as :

h h h
(1) s ( s + 1) (2) 2s ( s + 1) (3) s ( s + 2) (4) None
2 2 2

14. The orbital angular momentum of an electron in 2s-orbital is -

h h h
(1) (2) zero (3) (4) 2
4 2 2

15. Which of the following set of quantum numbers is permitted

(1) n = 3, l = 2, m = – 2, s = +1/2 (2) n = 3, l = 2, m = – 1, s = 0
(3) n = 2, l = 2, m = +1, s = – 1/2 (4) n = 2, l = 2, m = +1, s = – 1/2

16. Which of the following principles/rules limits the maximum number of electrons in an orbital to two
(1) Aufbau principle (2) Pauli's exclusion principle
(3) Hund's rule of maximum multiplicity (4) Heisenberg's uncertainty principle

43
17. Which is not correct for an electron having n = 5, m = 2 :
(1)  = 4 (2)  = 0, 1, 2, 3 (3)  = 3 (4)  = 2, 3, 4

18. Which of the following orbital quantum number value is not possible for an electron present in 4d
subshell:
(1) n = 4 (2)  = 1 (3) m = 1 (4) m = 2

19. The atomic number of an element is 17. The number of orbitals containing electron pairs in the valency
shell in its ground state are is :
(1) 8 (2) 2 (3) 3 (4) 6

20. Nitrogen has the electronic configuration 1s2 ,2s2 2p1x 2p1y 2p1z and not 1s2 ,2s2 2p2x 2p1y 2p0z which is
determined by
(1) Aufbau's principle (2) Pauli's exclusion principle
(3) Hund's rule (4) Uncertainty principle

21. For sodium atom the number of electrons with m = 0 will be :

(1) 2 (2) 7 (3) 9 (4) 8

22. Which of the following ions has the maximum number of unpaired d-electrons?
(1) Zn2+ (2) Fe2+ (3) Ni3+ (4) Cu+

23. The total spin resulting from a d7 configuration is :

(1) 1 (2) 2 (3) 5/2 (4) 3/2

24. In hydrogen atom, which is incorrect order of their energies.

(1) 1s < 2p (2) 2p = 2s (3) 2p > 2s (4) 2p < 3s

25. For a given subshell let maximum number of electrons with same spin be x and number of possible m
values be y.
x y
(1) x = 2y (2) x = y (3) =y (4) x =
2 2

26. Ratio of number of unpaired electrons in Fe2+ to that of Ti is

(1) 1.2 (2) 3 (3) 2 (4) 4

27. Orbital angular momentum of 2s orbital is

h 2h 3h
(1) (2) (3) (4) Zero
2 2 2

Section (G) : Shape of orbitals

1. Which orbital is non-directional

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

2. An orbital with  = 0 is symmetrical about the :

(1) x-axis only (2) y-axis only (3) z-axis only (4) All

44
3. Which orbital has two angular nodal planes.
(1) s (2) p (3) d (4) f

4. Which d-orbital does not have four lobes

(1) dx2 − y2 (2) dxy (3) dz2 (4) dxz

5. The number of radial nodes of 5s atomic orbital are

(1) 1 (2) 2 (3) 3 (4) 4

6. The number of angular nodes of 3dyz atomic orbital are

(1) 1 (2) 2 (3) 3 (4) 4

7. The sum of angular nodes and radial nodes of 4dxy atomic orbital are
(1) 1 (2) 2 (3) 3 (4) 4

8. The number of angular nodes of 3p atomic orbital are

(1) 1 (2) 2 (3) 3 (4) 4

9. 3py orbital has..........nodal plane :

(1) XY (2) YZ (3) ZX (4) All of these

10. A 3p-orbital has

(1) Two non-spherical nodes (2) Two spherical nodes
(3) One spherical and one non spherical nodes (4) One spherical and two non spherical nodes

11. Which of the following d-orbitals has dough-nut shape ?

(1) dxy (2) dyz (3) dx2 − y2 (4) dz2

12. The permissible solution to the schrodinger wave equation gave an ideal of ........... quantum number
(1) 4 (2) 3 (3) 2 (4) 1

45
1. Which is not true with respect to cathode rays :
(1) Cathode rays consist of fast moving electrons.
(2) For production of cathode rays in a discharge tube, the gas filled should be at a low pressure.
(3) For production of cathode rays in a discharge tube, the voltage applied across the electrodes should
be high.
(4) None of these
54 56
2. (i) 26 Fe , 26 Fe , 57
26 Fe ,
58
26 Fe (a) Isotopes
3 3
(ii) H , He
1 2 (b) Isotones
76
(iii) 32 Ge , 77
33 As (c) Isodiaphers
235 231
(iv) 92 U, 90 Th (d) Isobars
1 2 3
(v) H , D , T
1 1 1

Match the above correct terms :

(1) (i – a), (ii – d), (iii – b), (iv – c), (v – a) (2) (i – a), (ii – c), (iii – b), (iv – d), (v – a)
(3) (i – a), (ii – d), (iii – b), (iv – c), (v – b) (4) (i – a), (ii – b), (iii – d), (iv – c), (v – a)

3. The ratio of the "e/m" (specific charge) values of a electron and an -particle is -
(1) 2 : 1 (2) 1 : 1 (3) 1 : 2 (4) None of these

4. An oil drop has 6.4 × 10–19 C charge. The number of electrons in this oil drop are :
(1) 2 (2) 4 (3) 6 (4) 8

5. Atomic radius is of the order of 10–8 cm and nuclear radius is of the order of 10 –13 cm. Calculate what
fraction of atom is occupied by nucleus ?
(1) 10–20 (2) 10–15 (3) 10–12 (4) None
27
6. The radius of nucleus of 13 Al is approximetely:
–15
(1) 1.2 × 10 m (2) 2.7 × 10–15 m (3) 10.8 × 10–15 m (4) 4 × 10–15 m

7 If the ratio of radius of two different nuclear are in the ratio 2 : 3, then ratio of their mass numbers will be
8 2 4 4
(1) (2) (3) (4)
27 3 9 27

8. For a broadcasted electromagnetic wave having frequency of 1200 KHz, number of waves that will be
formed in 2 km distance are :
(1) 8 (2) 80 (3) 12 (4) 120

9. If 10–17J of light energy is needed by the interior of human eye to see an object. The number of
photons of green light ( = 550 nm) needed to see the object are :
(1) 27 (2) 28 (3) 29 (4) 30

10. How many photons of light having a wavelength of 5000 Å are necessary to provide 1 joule of energy.
(1) 2.8  1018 photons (2) 2.5  1017 photons (3) 2.5  1018 photons (4) 2.6  1014 photons

11. A light source of wavelength  illuminates a metal and ejects photo-electrons with (K.E.)max = 1 eV

46
 
Another light source of wavelength , ejects photo-electrons from same metal with (K.E.)max = 4eV
3
Find the value of work function ?
(1) 1 eV (2) 2 eV (3) 0.5 eV (4) None of these

12. A photon of 300 nm is absorbed by a gas and then emits two photons. One photon has a wavelength
400 nm then the wavelength of second photon in nm :
(1) 1200 (2) 1600 (3) 800 (4) 400

13. The ratio of radius of two different orbits in a H-atom is 4 : 9. Then, the ratio of the frequency of
revolution of electron in these orbits is :
(1) 2 : 3 (2) 27 : 8 (3) 3 : 2 (4) 8 : 27

14. The ratio of the time period of electrons in 2nd Bohr orbit of He+ and 4th Bohr orbit of Li2+ is.
9 3 9 27
(1) (2) (3) (4)
32 8 8 32

15. If the ionisation potential of a hydrogen like species is 20V, its Ist excitation potential will be :
(1) 5 V (2) 10 V (3) 15 V (4) 20 V

16. The amount of energy required to excite the electron from 3 rd to 4th Bohr's orbit in B4+ will be :
(1) 4.5 eV (2) 8.53 eV (3) 25 eV (4) 16.53 eV

17. A single electron orbits a stationary nucleus of charge +Ze, where Z is a constant. It requires 47.2 eV to
excite electron from second Bohr orbit to third Bohr orbit, find the value of Z :
(1) 1 (2) 3 (3) 5 (4) 4

18. The ionization energy of He+ is 19.6 × 10–18 J atom–1. The energy of the first stationary state of Li+2 will
be:
(1) 84.2 × 10–18 J/atom (2) 44.10 × 10–18 J/atom
–18
(3) 63.2 × 10 J/atom (4) 21.2 × 10–18 J/atom

19. Energy required to pull out an electron from 1st orbit of hydrogen atom to infinity is 100 units. The
amount of energy needed to pull out the electron from 2nd orbit to infinity is :
(1) 50 units (2) 100 units (3) 25 units (4) Zero

20. If the I excitation potential of a hypothetical H-like atom is 10.2x V, then the value of II excitation energy
is about :
(1) 13.6x eV (2) 12.09x eV (3) -10.2x eV (4) 40.5x eV

21. The kinetic energy of the electron present in the ground state of Li2+ ion is represented by :
3e2 3e2 3e2 3e2
(1) (2) − (3) (4) −
8 0 r 8 0 r 4 0 r 4 0 r

22. If the energy of an electron in hydrogen atom is given by expression, –1312/n² kJ mol–1, then the
energy required to excite the electron from ground state to second orbit is
(1) 328 kJ/mol (2) 656 kJ/mol (3) 984 kJ/mol (4) 1312 kJ/mol

23. The frequency of radiation emitted when the electron falls from n = 2 to n = 1 in a hydrogen atom will
be ( h = 6.625  10−34 Js )
(1) 2.46  1015 s−1 (2) 2.00  1015 s−1 (3) 1.54  1015 s−1 (4) 1.03  1015 s−1

24. The kinetic energy of an electron in the 2nd orbit of hydrogen atom is : (a0 is Bohr radius of 1st orbit)

47
 h2 h2 h2 h2
(1) (2) (3) (4)
4ma02 162ma02 322ma02 642ma02

25. If the series limit of wavelength of the Lyman series for the hydrogen atoms is 912Å, then the series
limit of wavelength for the Balmer series of the hydrogen atom is :
(1) 912 Å (2) 912 × 2 Å (3) 912 × 4 Å (4) 912/2 Å

26. The difference between the wave number of 1st line of Balmer series and last line of paschen series for
Li2+ ion is
R 5R R
(1) (2) (3) 4R (4)
36 36 4

27. If the shortest wave length of Lyman series of H atom is x, then the wave length of the first line of
Balmer series of H atom will be -
(1) 9x/5 (2) 36x/5 (3) 5x/9 (4) 5x/36

28. Suppose that a hypothetical atom gives a red, green, blue and violet line spectrum . Which jump
according to figure would give off the red spectral line.

(1) 3 → 1 (2) 2 → 1 (3) 4 → 1 (4) 3 → 2

29. What electron transition in the He+ spectrum would have the same wavelength as the first Lyman
transition of hydrogen.
(1) n = 4 to n = 2 (2) n = 6 to n = 4 (3) n = 6 to n = 2 (4) n = 4 to n = 1

30. What will be the KE of photoelectrons ejected, when photon of 13eV is absorbed by H-atom in 3rd
excited state.
(1) 12.15 eV (2) 11.49 eV (3) 12.46 eV (4) 12.63 eV

31. What is the ratio of the wave lengths of last lines of Balmer and Lyman series for any hydrogen like
species.
(1) 2 (2) 3 (3) 4 (4) 5

32. An atom has n energy level then total number of lines in its spectrum are :
(1) 1 + 2 + 3 ................ (n + 1) (2) 1 + 2 + 3 ................. (n)2
(3) 1 + 2 + 3 ................. (n – 1) (4) (n + 1) (n + 2) (n + 4)

33. A certain electronic transition from an excited state to ground state in a sample of H-atoms gives rise to
maximum three lines in the ultra violet region of the spectrum. How many lines does this transition
produce in the infrared region of the spectrum :
(1) 1 (2) 2 (3) 3 (4) 4

34. What minimum number of atoms/ions should be present in a sample of H-like species, so that a
maximum of 6 spectral lines can be produced of electronic transition from fifth excited state upto n = 2 ?
(1) 1 (2) 2 (3) 3 (4) 4

35. A particle X moving with a certain velocity has a de Brogile wavelength of 1Å. If particle Y has a mass
of 25% that of X and velocity 75% that of X, deBroglies wavelength of Y will be :
(1) 3Å (2) 5.33 Å (3) 6.88 Å (4) 48 Å
36. Number of de Broglie waves formed in the 4th orbit of H are :
(1) 4 (2) 5 (3) 0 (4) 1

48
37. Calculate the de-Broglie wave length of the electron in the ground state of hydrogen atom , given that
its kinetic energy is 13.6 eV - (1eV = 1.602 x 10– 19J)
(1) 3.328 x 10–10 m (2) 2.338 x 10–10 m (3) 3.328 x 1010 m (4) 2.338 x 10m

38. In H-atom, if ‘x’ is the radius of the first Bohr orbit, de Broglie wavelength of an electron in 3 rd orbit is :
9x x
(1) 3  x (2) 6  x (3) (4)
2 2
39. The uncertainty in position of an electron & helium atom are same. If the uncertainty in momentum for
the electron is 32 × 105, then the uncertainty in momentum of helium atom will be :
(1) 32 × 105 (2) 16 × 105 (3) 8 × 105 (4) None

40. A golf ball has a mass of 40 g and a speed of 45 m/s. If the speed can be measured within accuracy of
2%, then uncertainty in position will be approximately :
(1) 10–33 m (2) 10–40 m (3) 10–20 m (4) 10–50 m

41. Which of the following transition neither shows absorption nor emission of energy in case of Hydrogen
atom:
(1) 3px 3 ࢮs (2) 3dxy ࢮ 3dyz (3) 3s 3 ࢮdxy (4) All the above

42. The number of electrons in the M-shell of the element with atomic number 24 are :
(1) 24 (2) 12 (3) 8 (4) 13

43. Magnetic moment of Xn+ (Z = 26) is 24 B.M. Hence number of unpaired electrons and value of n
respectively are :
(1) 4, 2 (2) 2, 4 (3) 3, 1 (4) 0, 2
44. For a shell of principal quantum number n = 4 , which is incorrect :
(1) 16 orbitals (2) 4 subshells
(3) maximum 32 electrons (4)  = 0, 1, 2, 3, 4

45. Which set has the same number of s-electrons

(1) C, Cu2+ , Zn (2) Cu2+ , Fe2+,Ni2+ (3) S2– , Ni2+,Zn (4) None of these

46. Which electronic configuration does not follow the pauli's exclusion principle
(1) 1s2, 2s2 2p4 (2) 1s2, 2s2 2p4 3s2 (3) 1s2, 2p4 (4) 1s2, 2s2 2p6 3s3

47. For a d-electron, the orbital angular momentum is

(1) 6 (2) 2 (3) (4) 2

48. Which of the above statement (s) is/are false.

I Orbital angular momentum of the electron having n = 5 and having value of the azimuthal
h
quantum number as lowest for this principle quantum number is .

II If n = 3,  = 0, m = 0, for the last valence shell electron, then the possible atomic number may
be 12 or 13.
7
III Total spin of electrons for the atom 25Mn is ± .
2
IV Spin magnetic moment of inert gas is 0
(1) I, II and III (2) II and III only (3) I and IV only (4) None of these
49. Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum
numbers,  = 1 and 2 are, respectively :

(1) 16 and 5 (2) 12 and 5 (3) 16 and 4 (4) 12 and 4

49
50. Match List-I with List-II and select the correct answer using the codes given below the lists ( and m are

respectively the azimuthal and magnetic quantum no.)

List-I List-II
(1) Number of value of  for an energy level (1) 0, 1, 2, ............. (n - 1)

(2) Value of  for a particular type of orbital (2) + to – through zero

(3) Number of values of m for  = 2 (3) 5

(4) Value of 'm' for a particular type of orbital (4) n

Code :
A B C D A B C D
(1) 4 1 2 3 (2) 4 1 3 2
(3) 1 4 2 3 (4) 1 4 3 2

51. The orbital angular momentum corresponding to n = 4 and m = –3 is :

h 6h 3h
(1) 0 (2) (3) (4)
2 2 

52. A compound of Vanadium has a spin magentic moment 1.73 BM. Work out the electronic configuration
of the Vanadium ion in the compound :
(1) 1s22s22p63s23p64s1 (2) 1s22s22p63s23p63d2
(3) 1s22s22p63s23p63d1 (4) nothing can be said with certainty

53. A neutral atom of an element has 2K, 8L, 9M and 2N electrons. Which of the following is/are incorrectly
matched :
(1) Total number of s electrons - 8 (2) Total number of p electrons - 12
(3) Total number of d electrons -1 (4) Number of unpaired electrons in element - 3

54. For an electron, with n = 3 has only one radial node. The orbital angular momentum of the electron will
be
h h  h 
(1) 0 (2) 6 (3) 2 (4) 3  
2 2  2 

55. The maximum probability of finding electron in the dxy orbital is :

(1) Along the x-axis (2) Along the y-axis
(3) At an angle of 45 from the x and y axis
0
(4) At an angle of 900 from the x and y axis.

50
 PART - I : NEET / AIPMT QUESTION (PREVIOUS YEARS )
* Marked Questions are having more than one correct option.
1. The energy of a photon is 3.03 × 10–19 J, then wavelength of this photon is:
(Given, h = 6.63 × 10–34 Js, c = 3.00 × 108 ms–1) [AIPMT 2000]
(1) 6.56 nm (2) 65.6 nm (3) 656 nm (4) 0.656 nm

2. Maximum number of orbitals in an atom which can have the quantum numbers n = 3, l = 2, m = + 2 are:
[AIPMT 2001]
(1) 1 (2) 2 (3) 3 (4) 4

3. Energy of first excited state in hydrogen atom is – 3.4 eV then, kinetic energy of electron in same orbit
of hydrogen atom is : [AIPMT 2002]
(1) + 3.4 eV (2) + 6.8 eV (3) – 13.6 eV (4) + 13.6 eV

4. 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 of a quantum of light with frequency of 8  1015 s–1 ?
[AIPMT 2003]
(1) 2  10 –25
(2) 5  10 –18
(3) 4  10 1
(4) 3  10 7

5. The frequency of the radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom
will be (Given ionization energy of H = 2.18  10–18 J atom–1 and h = 6.625  10–34 Js) [AIPMT 2004]
(1) 1.54  1015 s–1 (2) 1.03  1015 Js–1 (3) 3.08  1015 s–1 (4) 2.0  1015 s–1

6. Among the following transition metal ions, the one set where all the metal ions have 3d 2 electronic
configuration is [At Nos. Ti = 22, V = 23, Cr = 24, Mn = 25] [AIPMT 2004]
(1) Ti3+, V2+, Cr3+, Mn4+ (2) Ti4+, V4+, Cr+6, Mn7+ (3) Ti4+, V3+, Cr2+, Mn3+ (4) Ti2+, V3+, Cr4+, Mn5+

7. The energy of the second Bohr orbit of the hydrogen atom is –328 kJ mol–1 ; hence the energy of fourth
Bohr orbit would be [AIPMT 2005]
(1) – 1312 kJ mol–1 (2) – 82 kJ mol–1 (3) – 41 kJ mol–1 (4) – 164 kJ mol–1

8. The Uncertainity involved in the measurement of velocity of electron within a distance of 0.1 Å is :
(Given : The mass of electron is 9.11  10–31 kg, planck constant is 6.626  10–34 J s) [AIPMT 2006]
(1) 5.79  108 m s–1 (2) 5.79  105 m s–1 (3) 5.79  106 m s–1 (4) 5.79  107 m s–1

9. The orientation of an atomic orbital is governed by : [AIPMT 2007]

(1) azimuthal eqantum number (2) spin quantum number
(3) magnetic quantum number (4) principal quantum number

10. Consider the following sets of quantum numbers :

n l m s
(i) 3 0 0 +1/2
(ii) 2 2 1 +1/2
(iii) 4 3 –2 –1/2
(iv) 1 0 –1 –1/2
(v) 3 2 3 +1/2
Which of the following sets of quantum numbers is not possible ? [AIPMT 2007]
(1) (i) and (iii) (2) (ii), (iii) and (iv) (3) (i), (ii), (iii) and (iv) (4) (ii), (iv) and (v)

51
11. The uncertainity in measurement position of the electron is associated with an uncertainity in
momentum, which is equal to 1  10–18 g cm s–1. The Uncertainity in electron velocity is, (mass of an
electron is 9  10–28 g) [AIPMT 2008]
(1) 1  10 cm s
9 –1
(2) 1  10 cm s
6 –1
(3) 1  10 cm s
5 –1
(4) 1  10 cm s–1
11

12. If the uncertainity in position and momentum are equal, then uncertainity in velocity is : [AIPMT 2008]
1 h h 1 h h
(1) (2) (3) (4)
2m  2 m  

13. Maximum number of electrons in a subshell of an atom in determined by the following : [AIPMT 2009]
(1) 4 l + 2 (2) 2 l + 1 (3) 4 l – 2 (4) 2 n2

14. A photon of energy 4.4 × 10–19 J collides with A2 molecules. If bond energy of A2 is 4.0 × 10–19 J. Then
kinetic energy of per atom of A will be : [AIPMT 2009]
(1) 2.0 × 10–20 J (2) 2.2 × 10–19 J (3) 2.0 × 10–19 J (4) 4.0 × 10–20 J

15. Which of the following is not permissible set of quantum numbers of electrons in an atom?
[AIPMT 2009]
(1) n = 4, l = 0, m = 0, s = – 1/2 (2) n = 5, l = 3, m = 0, s = + 1/2
(3) n = 3, l = 3, m = 0, s = – 1/2 (4) n = 3, l = 2, m = – 2, s = – 1/2

16. A 0.66 kg ball is moving with a speed of 100 m/s. The associated with ball wavelength will be : (h = 6.6
× 10–34 Js) [AIPMT 2010]
–32 –34 –35 –32
(1) 6.6 × 10 m (2) 6.6 × 10 m (3) 1.0 × 10 m (4) 1.0 × 10 m

17. The total number of atomic orbitals in fourth energy level of an atom are : [AIPMT 2011]
(1) 8 (2) 16 (3) 32 (4) 4

18. The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. The relation between their
wavelengths (i.e. 1 and 2) will be : [AIPMT 2011]
1
(1) 1 = 2 (2) 1 = 22 (3) 1 = 42 (4) 1 = 
2 2
19. If n = 6, the correct sequence for filling of electrons will be : [AIPMT 2011]
(1) ns → (n – 2)f → (n – 1)d → np (2) ns → (n – 1)d → (n – 2)f → np
(3) ns → (n – 2)f → np → (n – 1)d (4) ns → np(n – 1)d → (n – 2)f
20. According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to
the least energetic photon ? [AIPMT 2011]
(1) n = 6 to n = 1 (2) n = 5 to n = 4 (3) n = 6 to n = 5 (4) n = 5 to n = 3

21. Maximum number of electrons in a subshell having n = 4 and  = 3 are : [AIPMT 2012]
(1) 14 (2) 16 (3) 10 (4) 12
22. The correct set of four quantum numbers for the valence elecron of rubidium atom (Z=37) is :
[AIPMT 2012]
(1) 5, 1, + 1/2 (2) 6, 0, 0 + 1/2 (3) 5, 0, 0 + 1/2 (4) 5, 1, 0 + 1/2

23. The orbital angular momentum of a p-electron is given as : [AIPMT 2012]

h h 3h h
(1) (2) 3 (3) (4) 6.
2 2 2 2

52
24. The value of Planck's constant is 6.63 10–34 Js. The speed of light is 3  1017 nm s–1. Which value is
closest to the wavelength in nanometer of a quantum of light with frequency of 6  1015 s–1 ?
[NEET 2013]
(1) 25 (2) 50 (3) 75 (4) 10

25. What is the maximum numbers of electrons that can be associated with the following set of quantum
numbers ? n = 3,  = 1 and m = –1. [NEET 2013]
(1) 6 (2) 4 (3) 2 (4) 10

 Z2 
26. Based on equation E = –2.178  10–18J  2  , certain conclusions are written. Which of them is not
n 
correct ? [NEET 2013]
(1) Larger the value of n, the larger is the orbit radius.
(2) Equation can be used to calculate the change in energy when the electron changes orbit.
(3) For n = 1, the electron has a more negative energy than it does for n = 6 which mean that the
electron is more loosely bound in the smallest allowed orbit.
(4) The negative sign in equation simply means that the energy or electron bound to the nucleus is
lower than it would be if the electrons were at the infinite distance from the nucleus.

27. What is the maximum number of orbitals that can be identified with the following quantum number
n = 3,  = 1, m = 0 [AIPMT 2014]
(1) 1 (2) 2 (3) 3 (4) 4

28. Calculate the energy in corresponding to light of wavelength 45 nm : (Planck's constant h = 6.63 × 10 –34
Js : speed of light c = 3 × 108 ms–1) [AIPMT 2014]
–15 –18
(1) 6.67 × 10 15
(2) 6.67 × 10 11
(3) 4.42 × 10 (4) 4.42 × 10

29. Be2+ is isoelectronic with which of the following ions? [AIPMT 2014]
(1) H + (2) Li + (3) Na + (4) Mg 2+

30. Magnetic moment 2.84 B.M. is given by [AIPMT 2015]

(At. nos, Ni =28, Ti= 22, Cr =24, Co = 27 )
(1) Ti3+ (2) Cr2+ (3) Co2+ (4) Ni2+

31. The number of d-electrons in Fe2+ (Z=26) is not equal to the number of electrons in which one of the
following? [AIPMT 2015]
(1) p - electrons in Cl (Z=17) (2) d-electrons in Fe (Z=26)
(3) p-electrons in Ne (Z=10) (4) s-electrons in Mg (Z=12)

32. The angular momentum of electron in 'd' orbital is equal to : [AIPMT 2015]
(1) 2 (2) 2 3 (3) 0 (4) 6

33. Two electrons occupying the same orbital are distinguished by : [NEET-1 2016]
(1) Spin quantum number (2) Principal quantum number
(3) Magnetic quantum number (4) Azimuthal quantum number

34. How many electrons can fit in the orbital for which n = 3 and  = 1 ? [NEET-2 2016]
(1) 14 (2) 2 (3) 6 (4) 10

35. Which of the following pairs of d-orbitals will have electron density along the axis ? [NEET-2 2016]
(1) dxy , dx 2 – y 2 (2) dz2 , dxz (3) dxz , dyz (4) dz2 , dx 2 – y 2

36. Which one is the wrong statement ? [NEET- 2017]

53
 h
(1) de-Broglie's wavelength is given by  = , where m = mass of the particle,  = group
m
velocity of the particle.
h
(2) The uncertainty principle is E × t 
4
(3) Half filled and fully filled orbitals have greater stability due to greater exchange energy,
greater symmetry and more balanced arrangement.
(4) The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms.

37. Which one is a wrong statement? [NEET- 2018]

(1) Total orbital angular momentum of electron is 's' orbital is equal to zero.
(2) The value of m for dZ2 is zero.
1s2 2s2 2px1 2py 2pz1
1
(3) The electronic configuration of N atom is-
(4) An orbital is designated by three quantum numbers while an electron in an atomis designated by
four quantum numbers.

38. Which of the following series of transitions in the spectrum of hydrogen atom falls in visible region ?
[NEET-1- 2019]
(1) Brackett series (2) Lyman series (3) Balmer series (4) Paschen series

39. 4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is :
[NEET-1- 2019]
(1) 5f > 6p > 4d > 5p (2) 5f > 6p > 5p > 4d (3) 6p > 5f > 5p > 4d (4) 6p > 5f > 4d > 5p
40. Orbital having 3 angular nodes and 3 total nodes is : [NEET-2- 2019]
(1) 5 p (2) 3 d (3) 4 f (4) 6 d
41. In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is:
[Given that Bohr radius, a0 = 52.9 pm] [NEET-2- 2019]
(1) 211.6 pm (2) 211.6  pm (3) 52.9  pm (4) 105.8 pm
42. The number of angular nodes and radial nodes in3s orbital are [NEET-2020(COVID-19)]
(1) 0 and 2, respectively (2) 1 and 0, respectively
(3) 3 and 0, respectively (4) 0 and 1, respectively
2+
43. The calculated spin only magnetic moment of Cr ion is : [NEET-2020]
1) 2.84 BM 2) 3.87 BM 3) 4.90 BM 4) 5.92 BM
44. A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1,368 kHz(kilohertz).
The wavelength of the electromagnetic radiation emitted by the transmitter is : [speed of light,
c = 3.0 108 ms −1 ] [NEET-2021]
1) 219.2 m 2) 2192 m 3) 21.92 cm 4) 219.3 m

PART - II : AIIMS QUESTION (PREVIOUS YEARS )

1. The isoelectronic pair is : [AIIMS 2005]
(1) Cl2O, ICl2– (2) ICl2–, ClO2 (3) IF2+, I3– (4) ClO2–, ClF2+

2.  – particles can be detected using : [AIIMS 2005]

(1) thin aluminium sheet (2) barium suplhate
(3) zinc sulphide screen (4) gold foil

3. The most probable radius (in pm) for finding the electron in He+ is : [AIIMS 2006]
(1) 0.0 (2) 52.9 (3) 26.5 (4) 105.8

4. The de-broglie wavelength associated with a ball of mass 1 kg having kinetic energy 0.5 J is.

54
 [AIIMS 2006]
(1) 6.626  10–34 m (2) 13.20  10–34m (3) 10.38  10–21 m (4) 6.626  10–34 Å

5. The uncertainties in the velocities of two particles, A and B are 0.05 and 0.02 ms –1, respectively. The
 x 
mass of B is five times of that of the mass of A. What is the ratio of uncertainties  A  [AIIMS 2008]
 xB 
(1) 2 (2) 0.25 (3) 4 (4) 1

6. The de–Broglie wavelength of helium atom at room temperature is : [AIIMS 2009]

(1) 6.6 × 10–34 m (2) 4.39 × 10–10 m (3) 7.34 × 10–11 m (4) 2.335 × 10–20 m

7. n and  for some electrons are given. Which of the following is expected to have least energy ?
[AIIMS 2009]
(1) n = 3,  = 2 (2) n = 3,  = 0 (3) n = 2,  = 1 (4) n = 4,  = 0

8. If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected
out with a velocity of 1.5 × 107 ms–1, what is the energy with which it is bound to the nucleus ?
[AIIMS 2010]
(1) 1.2 × 102 eV (2) 2.15 × 103 eV (3) 7.6 × 103 eV (4) 8.12 × 102 eV
9. Smallest wavelength occurs for : [AIIMS 2011]
(1) Lyman series (2) Balmer series (3) Paschen series (4) Brackett series
10. Which of the following is wrong for Bohr model ? [AIIMS 2011]
(1) It establishes stability of atom.
(2) It is inconsistent with Heisenberg uncertainty principle.
(3) It explains the concept of spectral lines for hydrogen like species.
(4) Electrons behave as particle and wave.

11. Ratio of energy of photon of wavelength 3000Å and 6000Å is : [AIIMS 2012]
(1) 3 : 1 (2) 2 : 1 (3) 1 : 2 (4) 1 : 3

n,h
12. Assertion : Angular momentum of an electron in any orbit is given by angular momentum = ,
2
where n is the principal quantum number.
[AIIMS 2012]
Reason : The principal quantum number, n, can have any integral value.
(1) If both assertion and reason are true and reason is the correct explanation of assertion.
(2) If both assertion and reason are true but reason is not the correct explanation of assertion.
(3) If Assertion is true but reason is false.
(4) If both assertion and reason are false.

13. A compound of metal ion MX+ ( Z= 24) has a spin only magnetic moment of 15 Bohr Magenetons. The
number of unpaired electrons in the compound are [AIIMS 2013]
(1) 2 (2) 4 (3) 5 (4) 3

14. Which of the following combinations of quantum numbers is allowed ? [AIIMS 2013]
n l m ms
(1) 3 2 1 0
1
(2) 2 0 0 −
2
1
(3) 3 -3 -2 +
2

55
 1
(4) 1 0 1 +
2

15. The electrons, identified by quantum numbers n and l (i) n = 4,  = 1 (ii) n = 4,  = 0 (iii) n = 3, = 2

(iv) n = 3, = 1 can be placed in order of increasing energy, from the lowest to highest, as
[AIIMS 2014]
(1) (iv) < (ii) < (iii) < (i) (2) (ii) < (iv) < (v) < (iii) (3) (i) < (iii) < (ii) < (iv) (4) (iii) < (i) < (iv) < (ii)

16. In hydrogen atomic spectrum, a series limit is found at 12186.3 cm–1. Then it belong to [AIIMS 2014]
(1) Lyman series (2) Balmer series (3) Paschen series (4) Brackett series

1 1
17. Assertion : Spin quantum number can have two values, + and – [AIIMS 2014]
2 2
Reason : + and – signs signify the positive and negative wave functions.
(1) If both assertion and reason are true and reason is the correct explanation of assertion.
(2) If both assertion and reason are true but reason is not the correct explanation of assertion.
(3) If Assertion is true but reason is false.
(4) If both assertion and reason are false.

−RH
18. The degeneracy of hydrogen that has energy equal to is : [AIIMS 2015]
9
(1) 6 (2) 8 (3) 5 (4) 9

19. The electrons identified by qunatum numbers n and  : [AIIMS 2016]

(a) n = 4,  = 1 (b) n = 4,  = 0 (c) n = 3,  = 2 (d) n = 3,  = 1

can be placed in the order of increasing energy as :
(1) 3 > 4 < 2 < 1 (2) 4 < 2 < 3 < 1 (3) 2 < 4 < 1 3 (4) 1 < 3 2 < 4

20. Which transition in the hydrogen atomic spectrum wil have the same wavelength as the balmer
transition (i.e., n = 4 to n = 2) of He+ spectrum ? [AIIMS 2017]
(1) n = 4 to n = 3 (2) n = 3 to n = 2 (3) n = 4 to n = 2 (4) n = 2 to n = 1

21. Wave length of particular transition for H atom is 400 nm. What can be wavelength of He + for same
transition : [AIIMS 2018]
(1) 400 nm (2) 100 nm (3) 1600 nm (4) 200 nm

22. A gas metal in bivalent state have approximately 23e– what is spin magnetic moment in elemental state
(1) 2.87 (2) 5.5 (3) 5.9 (4) 4.9 [AIIMS 2018]

23. What is maximum wavelength of line of Balmer series of Hydrogen spectrum (R = 1.09 x 10 7 m–1) :
(1) 400 nm (2) 654 nm (3) 486 nm (4) 434 nm [AIIMS 2018]

24. In second orbit of H atom what is velocity of e– [AIIMS 2018]

(1) 2.18 × 106m/sec (2) 3.27 × 106m/sec (3) 10.9 × 105m/sec (4) 21.8 × 106m/sec

25. When on metal sheet fall 1 light will eject electron with V1 velocity and with 2 light eject electron of v2
velocity, what is v 22 – v12 value [AIIMS 2018]

2hc  1 1 hc  1 1 2hc  1 1  m  1 1
(1)  –  (2)  –  (3)  –  (4)  – 
m   2 1  m   2 1  m  1  2  2hc   2 1 

56
 PART - III : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
1. Which of the following ions has the maximum magnetic moment? [AIEEE 2002, 3/225]
(1) Mn+2 (2) Fe+2 (3) Ti+2 (4) Cr+2.

2. Energy of H-atom in the ground state is –13.6 eV, hence energy in the second excited state is :
[AIEEE 2002, 3/225]
(1) – 6.8 eV (2) – 3.4 eV (3) – 1.51 eV (4) – 4.53 eV
3. Uncertainity in position of a particle of 25 g in space is 10–15 m. Hence, Uncertainity in velocity (in m.sec–
1
) is: (plank’s constant, h = 6.6  10–34 Js) [AIEEE 2002, 3/225]
(1) 2.1  10 –18
(2) 2.1  10–34
(3) 0.5  10–34
(4) 5.0  10–24

4. The de-Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 m/s is
approximately (planck’s constant, h = 6.63  10–34 J-s) [AIEEE 2003, 3/225]
(1) 10–33 m (2) 10–31 m (3) 10–16 m (4) 10–25 m

5. In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one
of the following inner-orbit jumps of the electron for Bohr orbits in an atom of hydrogen ?
[AIEEE 2003, 3/225]
(1) 3 2 ࢮ (2) 5 2 ࢮ (3) 4 1ࢮ (4) 2 5 ࢮ

6. The numbers of d-electrons retained in Fe2+ (atomic number Fe = 26) ion is [AIEEE 2003, 3/225]
(1) 3 (2) 4 (3) 5 (4) 6

h
7. The orbital angular momentum for an electron revolving in an orbit is given by ( + 1) . This
2
momentum for an s-electron will be given by [AIEEE 2003, 3/225]
1 h h h
(1) + . (2) Zero (3) (4) 2.
2 2 2 2

8. The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to
stationary state 1, would be (Rydberg constant = 1.097  107 m–1) [AIEEE 2004, 3/225]
(1) 91 nm (2) 192 nm (3) 406 (4) 9.1  10–6 nm

9. Which of the following set a of quantum numbers is correct for an electron in 4f orbital?
[AIEEE 2004, 3/225]
(1) n = 4, l =3, m = +4, s = +1/2 (2) n = 4, l = 4, m = –4, s = –1/2
(3) n = 4, l = 3, m = +1, s = +1/2 (4) n = 3, l=2, m =–2, s = +1/2

10. Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum
numbers,  = 1 and 2 are, respectively [AIEEE 2004, 3/225]
(1) 12 and 4 (2) 12 and 5 (3) 16 and 4 (4) 16 and 5

11. In a multi-electron atom, which of the following orbitals described by the three quantum numbers will
have the same energy in the absence of magnetic and electric field ? [AIEEE 2005, 3/225]
(i) n = 1, l = 0, m = 0 (ii) n =2, l = 0, m = 0 (iii) n = 2, l = 1, m = 1 (iv) n = 3, l = 2, m =1
(v) n = 3, l = 2, m = 0
(1) (iv) and (v) (2) (iii) and (iv) (3) (ii) and (iii) (4) (i) and (ii)

12. Which of the following statements in relation to the hydrogen atom is correct ? [AIEEE 2005, 4½/225]

57
 (1) 3s, 3p and 3d orbitals all have the same energy
(2) 3s and 3p orbitals are of lower energy than 3d orbital
(3) 3p orbital is lower in energy than 3d orbital
(4) 3s orbital is lower in energy than 3p orbital

13. Uncertainity in the position of an electron (mass = 9.1  10–31 Kg) moving with a velocity 300 m.sec–1,
Accurate upto 0.001%, will be : (h = 6.63  10–34 J-s) [AIEEE 2006, 3/165]
(1) 19.2  10–2 m (2) 5.76  10–2 m (3) 1.92  10–2 m (4) 3.84  10–2 m
14. According to Bohr’s theory, the angular momentum to an electron in 5th orbit is : [AIEEE 2006, 3/165]
h h h h
(1) 25 (2) 1.0 (3) 10 (4) 2.5
   

15. The ‘spin-only’ magnetic moment [in units of Bohr magneton ()] of Ni2+ in aqueous solution would be
(Atomic number : Ni = 28) [AIEEE 2006, 3/165]
(1) 2.84 (2) 4.90 (3) 0 (4) 1.73

16. Which of the following nuclear reactions will generate an isotope ? [AIEEE 2007, 3/120]
(1) Neutron particle emission (2) Positron emission
(3) -particle emission (4) -particle emission

17. The ionisation enthalpy of hydrogen atom is 1.312  106 J mol–1. The energy required to excite the
electron in the atom from n1 = 1 to n2 = 2 is [AIEEE 2008, 3/105]
(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
18. Which of the following set of quantum numbers represents the highest energy of an atom ?
[AIEEE 2008, 3/105]
1 1
(1) n = 3, l = 0, m = 0, s = + (2) n = 3, l = 1, m =1, s = +
2 2
1 1
(3) n = 3, l = 2, m = 1, s = + (4) n = 4, l = 0, m = 0, s = +
2 2

19. The energy required to break one mole of Cl – Cl bonds in Cl2 is 242 kJ mol–1 . The longest wavelength
of light capable of breaking a single Cl – Cl bond is [AIEEE 2010, 4/144]
(c = 3 × 108 m s–1 and NA = 6.02 × 1023 mol–1)
(1) 594 nm (2) 640 nm (3) 700 nm (4) 494 nm
20. Ionisation energy of He+ is 19.6 × 10–18 J atom–1 . The energy of the first stationary state (n = 1) of Li2+ is
[AIEEE 2010, 4/144]
(1) 4.41 × 10–16 J atom–1 (2) – 4.41 × 10–17 J atom–1
(3) – 2.2 × 10–15 J atom–1 (4) 8.82 × 10–17 J atom–1

21. A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm,
the other is at : [AIEEE 2011, 4/120]
(1) 1035 nm (2) 325 nm (3) 743 nm (4) 518 nm

22. The frequency of light emitted for the transition n = 4 to n = 2 of He + isequalto the transition in H atom
corresponding to which of the following? [AIEEE 2011, 4/120]
(1) n = 2 to n = 1 (2) n = 3 to n = 2 (3) n = 4 to n = 3 (4) n = 3 to n = 1

23. The electrons identified by quantum numbers n and  : [AIEEE 2012, 4/120]
(a) n = 4,  = 1 (b) n = 4,  = 0 (c) n = 3,  = 2 (d) n = 3,  = 1
can be placed in order of increasing energy as :
(1) (c) < (d) < (b) < (a) (2) (d) < (b) < (c) < (a) (3) (b) < (d) < (a) < (c) (4) (a) < (c) < (b) < (d)

58
  Z2 
24. Energy of an electron is given by E = –2.178  10–18J.  2  Wavelength of light required to excite an
n 
electron in an hydrogen atom fr/om level n = 1 to n = 2 will be : (h = 6.62  10–34 Js and c = 3.0  108
ms–1) [JEE(Main)2013]
(1) 1.214  10 m
–7
(2) 2.816  10 m
–7
(3) 6.500  10 m –7
(4) 8.500  10–7 m

25. The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is :
[JEE(Main)2014, 4/120]
1 1 1 1
(1) 5, 0, 0, + (2) 5, 1, 0, + (3) 5,1, 1, + (4) 5, 0, 1, +
2 2 2 2

26. Which of the following is the energy of a possible excited state of hydrogen ? [JEE(Main) 2015, 4/120]
(1) +13.6 eV (2) –6.8 eV (3) –3.4 eV (4) +6.8 eV
27. A stream of electrons from a heated filament was passed between two charged plates kept at a
potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value
of h/ (where  is wavelength associated with electron wave) is given by : [JEE(Main) 2016, 4/120]
(1) 2meV (2) meV (3) 2meV (4) meV

28. The radius of the second Bohr orbit for hydrogen atom is : [JEE(Main) 2017, 4/120]
(Planck's Const. h = 6.6262 × 10–34 Js; mass of electron = 9.1091 × 10 –31 kg; charge of electron e =
1.60210 × 10–19 C; permittivity of vacuum  = 8.854185 × 10–12 kg–1m–3A2)
(1) 4.76 Å (2) 0.529 Å (3) 2.12 Å (4) 1.65 Å
 1
28. For emission line of atomic hydrogen from ni = 8 to nf = n, the plot of wave number (  ) against  2  will
n 
be (The Rydberg constant, RH is in wave number unit) [JEE(Main) 2019, 4/120]
(1) Linear with intercept –RH (2) Linear with slope –RH
(3) Non linear (4) Linear with slope RH
29. Which of the following combination of statements is true regarding the interpretation of the atomic
orbitals? [JEE(Main) 2019, 4/120]
(a) An electron in an orbital of high angular momentum stays away from the nucleus than an
electron in the orbitals of lower angular momentum.
(b) For a given value of the principal quantum number, the size of the orbit is inversely proportional
to the azimuthal quantum number.
h
(c) According to wave mechanics, the ground state angular momentum is equal to .
2
(d) The plot of  Vs r for various azimuthal quantum numbers, show peak shifting towards higher
value.
(1) (b), (c) (2) (a), (c) (3) (a), (d) (4) (a), (b)
30. Which of the graphs shown below does not represent the relationship between incident light and the
electron ejected from metal surface ? [JEE(Main) 2019, 4/120]

K.E. of No. of Electrons

(1) Electron (2)

0 0 Frequency
Energy of Light
of Light

59
 K.E. of K.E. of
(3) Electron (4) Electron

0 0
Frequency of Light Intensity of Light

31. The ground state energy of hydrogen atom is –13.6 eV. The energy of second excited state of He+ ion

eV. Is: [JEE(Main) 2019, 4/120]

(1) –27.2 (2) –54.4 (3) –3.4 (4) –6.04

32. The de Broglie wavelength () associated with a photoelectron varies with the frequency () of the

incident radiation as, [0 is threshold frequency] : [JEE(Main) 2019, 4/120]

1 1 1 1
(1)   (2)   (3)   (4)  
(v –
3
v0 2) (v 0 – v 0 ) (v –
1
v0 2)
1
(v – v 0 ) 4

33. Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line

of H-atom is suitable for this purpose ? [JEE(Main) 2019, 4/120]

[RH = 1 × 105 cm–1, h = 6.6 × 10–34 Js, c = 3 × 108 ms–1 ]

(1) Paschen,  → 3 (2) Paschen, 5 → 3 (3) Balmer,  → 2 (4) Lyman,  → 1

34. If the de Broglie wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to 1.5 a0

(a0 is Bohr radius), then the value of n/z is: [JEE(Main) 2019, 4/120]

(1) 0.75 (2) 0.40 (3) 1.0 (4) 1. 50

60
 EXERCISE - 1
SECTION (A)
1. (1) 2. (4) 3. (3) 4. (2) 5. (2) 6. (4) 7. (1)
8. (3) 9. (2) 10. (4) 11. (3) 12. (1) 13. (3) 14. (4)
15. (4)
SECTION (B)
1. (2) 2. (3) 3. (1) 4. (1) 5. (4) 6. (1) 7. (1)
8. (1) 9. (4) 10. (3) 11. (3) 12. (3) 13. (2) 14. (3)
15. (1)
SECTION (C)
1. (2) 2. (2) 3. (2) 4. (2) 5. (3) 6. (1) 7. (2)
8. (2) 9. (1) 10. (2) 11. (1) 12. (4) 14. (2) 14. (4)
15. (1) 16. (4) 17. (1) 18. (1) 19. (2) 20. (2) 21. (2)
22. (2) 23. (4) 24. (3) 25. (4) 26. (1) 27. (3)
SECTION (D)
1. (4) 2. (1) 3. (2) 4. (1) 5. (4) 6. (4) 7. (4)
8. (3) 9. (3) 10. (2) 11. (3) 12. (2) 13. (3) 14. (1)
SECTION (E)
1. (3) 2. (1) 3. (3) 4. (3) 5. (1) 6. (2) 7. (4)
8. (1) 9. (4) 10. (2) 11. (2) 12. (2) 13. (3) 14. (3)
15. (4)
SECTION (F)
1. (3) 2. (4) 3. (1) 4. (4) 5. (4) 6. (2) 7. (4)
8. (3) 9. (4) 10. (1) 11. (4) 12. (2) 13. (1) 14. (2)
15. (1) 16. (2) 17. (2) 18. (2) 19. (3) 20. (3) 21. (2)
22. (2) 23. (4) 24. (3) 25. (2) 26. (3) 27. (4)
SECTION (G)
1. (1) 2. (4) 3. (3) 4. (3) 5. (4) 6. (2) 7. (3)
8. (1) 9. (3) 10. (3) 11. (4) 12. (2)
EXERCISE - 2
1. (4) 2. (1) 3. (4) 4. (2) 5. (2) 6. (4) 7 (1)
8. (1) 9. (2) 10. (3) 11. (3) 12. (1) 13. (2) 14. (1)
15. (3) 16. (4) 17. (3) 18. (2) 19. (3) 20. (2) 21. (1)
22. (3) 23. (1) 24. (3) 25. (3) 26. (4) 27. (2) 28. (4)
29. (1) 30. (1) 31. (3) 32. (3) 33. (1) 34. (2) 35. (2)
36. (1) 37. (1) 38. (2) 39. (1) 40. (1) 41. (4) 42. (4)
43. (1) 44. (4) 45. (2) 46. (4) 47. (1) 48. (1) 49. (2)
50. (2) 51. (4) 52. (3) 53. (4) 54. (3) 55. (3)
EXERCISE - 3
PART-I
1. (3) 2. (1) 3. (1) 4. (3) 5. (3) 6. (4) 7. (2)
8. (3) 9. (3) 10. (4) 11. (1) 12. (1) 13. (1) 14. (1)
15. (3) 16. (3) 17. (2) 18. (2) 19. (1) 20. (3) 21. (1)
22. (3) 23. (1) 24. (2) 25. (3) 26. (3) 27. (1) 28. (4)
29. (2) 30. (4) 31. (1) 32. (4) 33. (1) 34. (2) 35. (4)
36. (4) 37. (3) 38. (3) 39. (2) 40. (3) 41. (2) 42. (1)
43. (3) 44. (4)
PART-II
1. (4) 2. (3) 3. (3) 4. (1) 5. (1) 6. (3) 7. (3)
8. (3) 9. (1) 10. (4) 11. (2) 12. (2) 13. (4) 14. (2)
15. (1) 16. (3) 17. (3) 18. (4) 19. (2) 20. (4) 21. (2)
22. (3) 23. (2) 24. (3) 25. (1)
PART-III
1. (1) 2. (3) 3. (1) 4. (1) 5. (2) 6. (4) 7. (2)
8. (1) 9. (3) 10. (2) 11. (1) 12. (1) 13. (3) 14. (4)
15. (1) 16. (1) 17. (4) 18. (3) 19. (4) 20. (2) 21. (3)
22. (1) 23. (2) 24. (1) 25. (1) 26. (3) 27. (3) 28. (3)
28. (4) 29. (2) 30. (3) 31. (4) 32. (3) 33. (1) 34. (1)

61
 Self Practice Paper (SPP)

1. A 5g orbital has
(1) Zero angular node and zero radial node (2) Zero radial node and two angular nodes
(3) 4 radial nodes and 4 angular nodes (4) Zero radial node and 4 angular nodes

2. The threshold wavelength (0) of sodium metal is 6500Å. If UV light of wavelength 360Å is used, what
will be kinetic energy of the photoelectron in ergs?
(1) 55.175 × 10–12 (2) 3.056 × 10–12 (3) 52.119 × 10–12 (4) 48.66 × 10–10

3. An electron beam can undergo diffraction by crystals. Through what potential should a beam of
electrons be accelerated so that its wavelength becomes equal to 1.54 Å?
(1) 54.3 volt (2) 63.3 volt (3) 66.2 volt (4) None of these

4. Radiation corresponding to the transition n = 4 to n = 2 in hydrogen atoms falls on a certain metal (work
function = 2.5 eV). The maximum kinetic energy of the photo-electrons will be :
(1) 0.55 eV (2) 2.55 eV (3) 4.45 eV (4) None of these

5. Calculate the number of photons emitted by a 100 W yellow lamp in 1.0 s. Take the wavelength of
yellow light as 560 nm and assume 100 percent efficiency.
(1) 6.8 × 1020 (2) 4 × 1012 (3) 4 × 1020 (4) 2.8 × 1020

6. In a photoelectric experiment, kinetic energy of photoelectrons was plotted against the frequency of
incident radiation (), as shown in figure. Which of the following statements is correct?

(1) The threshold frequency is 1.

(2) The slope of this line is equal to Plank’s constant.
(3) As the frequency of incident wavelength increases beyond threshold frequency, kinetic energy of
photoelectrons decreases.
(4) It is impossible to obtain such a graph.

7. Which of the following process not lead to formation of isobars ?

(1) 1  particle and 2 particles are emitted (2) Positron emission
(3)  particle (–1e0) emission (4) K-electron capture

8. A sample of hydrogen (in the form of atoms), is made to absorb white light. 52% of the hydrogen atoms
got ionised. In order to calculate the ionisation energy of hydrogen from its absorption spectrum
(assuming the electrons that got ejected have KE = 0), it is possible by measuring the frequency of the
(1) line of shortest wavelength (2) line of longest wavelength
(3) line of greatest intensity (4) line of smallest intensity
9. In what region of the electromagnetic spectrum would you look for the spectral line resulting from the
electronic transition from the tenth to the fifth electronic level in the hydrogen atoms? (R H = 1.10 × 105
cm–1)

62
 (1) Microwave (2) Infrared (3) Visible (4) Ultraviolet

10. Consider Xenon (Z = 54). The maximum number of electrons in this atom that can have the values for
1
their quantum numbers as n = 4,  = 3 and s = in its ground state is :
2
(1) Zero (2) 7 (3) 9 (4) 14

11. The increasing order for the values of e/m (charge/mass) is :

(1) e, p, n,  (2) n, p, e,  (3) e, , e (4) n,  , p, e
x
12. An electron in an atom jumps in such a way that its kinetic energy changes from x to . The change in
4
potential energy will be :
3 3 3 3
(1) + x (2) − x (3) + x (4) − x
2 8 4 4

13. What atomic number of an element “X” would have to become so that the 4th orbit around X would fit
inside the 1st Bohr orbit of Hydrogen ?
(1) 3 (2) 4 (3) 16 (4) 25
1
14. Select the incorrect graph for velocity of e– in an orbit VS. Z, and n :
n

(1) (2) (3) (4)

15 Which of the following is discreted in Bohr’s theory ?

(1) Potential energy (2) Kinetic energy
(3) Velocity (4) Angular momentum

16 The mass of a proton is 1836 times more than the mass of an electron. It a sub-atomic particle of mass
(m’) 207 times the mass of electron is captured by the nucleus, then the first ionization potential of H :
(1) decreases (2) increases (3) remains same (4) may be decrease or increase

17. In any subshell, the maximum number of electrons having same value of spin quantum number is :
(1) ( + 1) (2)  + 2 (3) 2 + 1 (4) 4 + 2

18. Which quantum number defines the orientation of orbital in the space around the nucleus ?
(1) Principal quantum number (n) (2) Angular momentum quantum number
(3) Magnetic quantum number (m) (4) Spin quantum number (ms)

19. For similar orbitals having different values of n :

(1) the most probable distance increases with increase in n
(2) the most probable distance decreases with increase in n
(3) the most probable distance remains constant with increase in n
(4) none of these

20. Maximum number of total nodes is present in :

(1) 5s (2) 5p
(3) 5d (4) All have same number of nodes

63
21. The possible set of quantum no. for the unpaired electron of chlorine is :
n  m n  m
(1) 2 1 0 (2) 2 1 1
(3) 3 1 1 (4) 3 0 0

22. Which of the following has the maximum number of unpaired electrons ?
(1) Mn (2) Ti (3) V (4) Al

23. The angular velocity of an electron occupying the seocond Bohr orbit of He + ion is (in sec–):
(1) 2.067 × 1016 (2) 3.067 × 1016 (3) 1.067 × 1018 (4) 2.067 × 1017

24. An exited state of H-atom emits a photon of wavelength  and returns in the ground state, the principal
quantum number of excited state is given by :
R (R − 1)
(1) R(R − 1) (2) (3) R(R − 1) (4)
( R − 1) R
25. Light of wavelength  strikes a metal surface with intensity X and the metal emits Y electrons per
second of average energy Z. What willl happen to Y and Z if X is havled?
(1) Y will be halved (2) Y will double
(3) Y will be remain same (4) Z will be halved

26. Neutron scattering experiments have shown that the radius of the nucleus of an atom is directly
proporitonal to the cube root of the number of nucleons in the nucleus. From 73 Li to 189
76 Os , the radius is
(1) Halved (2) the same (3) Doubled (4) Tripled

27. The nucleus of an atom is located at x = y = z = 0. If the probability of finding an electron in d X2–y2
orbital in a tiny volume around x=a , y =0 , z = 0 is 1 × 10 –5, what is the probablility of finding the
electron in the same size volume around x = 0 ,y = a, z = 0 ?
(1) 1 × 10–5 (2) 1 ×10–5 × a (3) –1 × 10–5 × a (4) zero

28. The energy of a I, II and III energy levels of a certain atom are E, 4E and 2E respectively. A photon of
3
wavelength  is emitted during a transition from III to I. What will be the wavelength of emission for
transition II to I ?

(1) (2)  (3) 2 (4) 3
2
29. A compound of vanadium has a magnetic moment of 1.73 BM. What will be the electronic
configurations:
(1) 1s2 2s2 2p6 3s2 3p6 3d1 (2) 1s2 2s2 2p6 3s2 3p6 3d2
(3) 1s2 2s2 2p6 3s2 3p6 3d3 (4) 1s2 2s2 2p6 3s2 3p6 3d4

30. Calculate the minimum and maximum number of electrons which may have magnetic quantum number,
1
m = +1 and spin quantum number, s = – in chromium (Cr) :
2
(1) 0, 1 (2) 1, 2 (3) 4, 6 (4) 2, 3
31. Which of the following pairs are isobars ?
235 239 83 84 19 19 139 140
(1) 92 U, 94 Pu (2) 36 Kr, 36 Kr (3) 10 Ne, 9 F (4) 58 Ce, 58 Ce

32. The ratio of the velocity of the electron in the third and fifth shell for He+ would be :
(1) 5 : 3 (2) 1 : 2 (3) 3 : 5 (4) 3 : 4

64
33. The light radiations with discrete quantities of energy are called :
(1) Photoelectric effect (2) photons (3) Photo emission (4) Photo-absorbtion

34. S1: The designation of an orbital, n = 4 and l = 0 is 4s.

S2: All the orbitals in a subshell are first occupied singly with parallel spins before pairing starts. This is
Hund’s rule.
S3: The energies of various subshells in the same shell are in the order of s > p > d > f
(1) TTF (2) FFT (3) TTT (4) FFF

35. Bohr's theory is not applicable to :

(1) H (2) He+ (3) Li2+ (4) H+

36. Difference in 17Cl35 and 17Cl37 is of :

(1) atomic number (2) number of neutron (3) number of electron (4) number of proton

37. Which of the following is known as alpha particle?

(1) Electron (2) Charged helium atom
(3) Proton (4) Positron

38. Which of the following do not travel with the speed of light?
(1) de-Broglie waves (2) X-rays (3) Gamma rays (4) All of the above

39. The ratio of area covered by second orbital to the first orbital is :
(1) 1 : 1 (2) 1 : 16 (3) 8 : 1 (4) 16 : 1

40. The probability of finding of three unpaired electron in nitrogen atom is defined by :
(1) Aufbau’s principle (2) Uncertainty principle (3) Pauli’s principle (4) Hund’s rule

41. Electron occupy the available sub level which has lower (n + l) value. This is called :
(1) Aufbau rule (2) Paulli exclusion principle
(3) Hund’s rule (4) Heisenberg uncertainty principle

42. For n = 3 the value of I and m respectively are :

(1) 3, 9 (2) 3, 6 (3) 9,3 (4) 6, 3

43. The minimum real charge on any particle which can exist is :
(1) 4.8 × 10–10 coulomb (2) 1.6 × 10–19 coulomb (3) 1.6 × 10–10 coulomb (4) zero

44. In d-subshell electron is :

(1) less than p (2) equal to p (3) more than s (4) more than f

45. The mass of neutron is of the order of :

(1) 10–27 kg (2) 10–26 kg (3) 10–24 kg (4) 10–23 kg

SPP Answers

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

8. (1) 9. (2) 10. (1) 11. (4) 12. (1) 13. (3) 14. (4)
15 (4) 16 (2) 17. (3) 18. (3) 19. (1) 20. (4) 21. (3)

65
22. (1) 23. (1) 24. (2) 25. (1) 26. (4) 27. (1) 28. (4)
29. (1) 30. (4) 31. (3) 32. (1) 33. (2) 34. (1) 35. (4)
36. (2) 37. (2) 38. (1) 39. (4) 40. (4) 41. (1) 42. (1)
43. (2) 44. (3) 45. (1)

SPP Solutions

1. Total number of nodes = n – 1 = 5 – 1 = 4

Angular node =  = 4.
Zero radial node and 4 angular nodes.

2. The threshold frequency (0) corresponding to the wavelength 6500 Å is c/0.

Therefore, the threshold energy = h0 = hc/0.
Substituting for h, c and 0 we get, threshold energy = 3.056 × 10–12 ergs.
The energy of the incident photons is given by E = hc/0, since incident wavelength  = 360 Å.
Therefore, incident energy = 55.175 × 10–12 ergs.
The kinetic energy of the photoelectrons will be the difference of incident energy and threshold energy,
 KE = h – h0 = (55.175 × 10–12) – (3.056 × 10–12) ergs. = 52.119 × 10–12 ergs

1 h
3. For an electron mu2 = eV and =
2 mu
1 h2
Thus, m = 2 2 = eV
2 m 
1 h2 1 (6.62  10−34 )2
or V= = = 63.3 volt.
2 m  2 e 2  9.108  10−31(1.54  10−10 )2  1.602  10−19

13.6 13.6
4. En = – 2
eV; E2 =
n 22
13.6
E4 = – eV/atom
42
E = E4 – E2 = 2.55 eV
Absorbed energy = work function of metal + K.E. 2.55 = 2.5 + K.E. ; K.E. = 0.05 eV

E Pt  P t
5. The number of photon is N = = =
h h(c /  ) hc
Substitution of the data gives
(5.60  10−7 m)  (100Js−1 )  (1.0s)
N= = 2.8 × 1020
(6.626  10−36 s)  (3  108 ms −1 )

7. X
AZ → X–2BZ–4 + 2He4
X–2
BZ–4 → X–1CZ–4 + –1e0
X–1
CZ–4 → XDZ–4 + –1e0

1
9. Wave numbers are the reciprocals of wavelengths and are given by the expression v = .


66
 1 1 1
_ = 1.1 × 105  – 
  n1 n2 

2
11. Charge/mass for n = 0, for  = ,
4
1 1
for p = , for e– =
1 1/1837

2x 3
12. Change in P.E. = – + (2x)  x
4 2

n2
13. r1 = 0.529 Å; r4(X) = r1 × ;
Z
0.529  (4)2
r4(X)  ; Z = 160
Z

16 The ionization potential of an atom of nucleus of proton and the new sub-atomic particle system is
determined by replacing the mass of electron by reduced mass m, while
mpm' 1836me  207me
m= =
mp + m' 1836me + 207me
m = 186 me
I.E.  m, So, I.E. increases

17. Maximum number of electrons with same spin is equal to maximum number of orbitals, i.e., (2 +1).

20. Total nodes = n – 1.

21. Cl17 : [Ne] 3s2 3p5.

Unpaired electron is in 3p orbital.  n = 3,  = 1, m = 1, 0, – 1.

2Ze2
23. Velocity of an electron in He+ ion in an orbit = .....(i)
nh
n2h2
Radius of He+ ion in an orbit = ....(ii)
42me2 Z
By equations (i) and (ii),
u 83 Z2me4
Angular velocity () = = ....(iii)
r n3h3
8  (22 / 7)3  (2)2  (9.108  10−28 )  (4.803  10 −10 )4
= = 2.067 × 1016 sec–1.
(2)3  (6.626  10−26 )3

1  1 1
24. = R  2 – 2  ; n1 = 1, n2 = ?;
  n1 n2 

1 1 1  R
 – 2   n2 =
2
=R
  1 n 2  R  –1

R
 n2 =
R – 1

67
25. Number of emitted electron  Intensity of incident light.

1/ 3
ROs  189 
26. =   =3
RLi  7 

27. It would be same in x and y axis for dX2–y2 .

4E hc E hc
28. For II to I transition, E = –E= ; =
3  → 3   – 

hc hc
For  to  transition, E = 2E – E = or E =
 
hc hc
 =  = 3
3     –  –

29. Number of unpaired elecron are given by Magnetic moment = [n(n + 2)] where n is number of

unpaired electrons or 1.73 = [n(n + 2)] or 1.73 × 1.73 = n2 + 2n

 n = 1.
Now vanadium atom must have one unpaired electron and thus its confiuration is :
23
V4+ : 1s2 2s2 2p6 3s2 3p6 3d1

30.

1
Out of 6 electrons in 2p and 3p must have on electron with m = + 1 and s = but in 3d-subshell an
2
1 1
orbital having m = + 1 may have spin quantum no. – or + . Therefore, minimum and maximum
2 2
possible values are 2 and 3 respectively.

31. Isobars have same mass number.

V0 Z V3 5
32. Vn =  =
n V5 3

35. Bohr's theory is applicable only to atoms having one electron. Therefore, it is not applicable to H + as it
has no electron. According to Bohr.
−13.6 Z 2
En = eV / atom
n2
(Z = atomic number, n = number of the shell)
0.529 n2
rn = Å
Z

39. rn  n2  Area covered (An)  n4 (  Area = r2)

A 2 24 16
 = =
A1 14 1

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40. Hund’s rule

41. Electron occupy the available sub level which has lower (n + l) value. This called Aufbau rule.

42. Total value of l=n

Total value of m = n2
n=3
Total value of l=3
Total value of m=9

43. The minimum real charge on any particle which can exist is 1.6 × 10-19 C.

44. In d– subshell electron is more than s

subshell d = 10 electron
subshell s = 2 electron

45. The mass of neutron is of the order of 10–27kg

69