JEE NEET Physics Question Bank - Atom and Nucleus

18 Atom and Nucleus

D – Scattering experiment and Rutherford's atomic model :
– Radioactive source 83 Bi 214 emitts, D  particles of energy 5.5 MeV, incident on a thin gold foil of

thickness 2.1 u 107 m . Scattered D  particles observed on circular scintillation screen of
ZnS. [zinc sulphide]

– 105 D  particles scattered at  15q , 0.1% Scattered at ~150q , about only 1 out of 10
4

D  particle was scattered at 180q .
– Path of scattered D  particle can be determined with the help of Coloumb’s law and Newton’s

second law. Repulsive force acting between D  particle and gold nucleus is F
1 (2 e) (79 e)
4S 0
.
r2
The magnitude and direction of the force on D  particle continuously changes as it approaches the
nucleus.
– The perpendicular distance of the initial velocity vector of the D  particle from the centre of the
nucleus is known as the impact parameter (b).
– For head on collision b 0.
– The minimum distance of the D  particle from the centre of the nucleus (for b 0 ) is known as
the distance of closest approach.
– Radius of nucleus is about 10–15 m.

If 't' is thickness of foil and 'N' are No. of scattered D  particles then, = constant,
N
l
t

.
N1 N2
t1 t2

According to Rutherford, orbit of an electron is not circular but spiral and motion of an electron

ends inside the nucleus. In this case the atom cannot remain stable. Thus Rutherford’s atomic

model failed to explain the stability of atom.

(1) An D  Particle of 10 MeV is moving for a head on collision. What will be the distance of

closest approach from the nucleus of atomic number Z = 60 ?

(A) 1.44 u 1014 m (B) 2.88 u 1014 m (C) 0.53 u 1014 m (D) 1.728 u 1014 m

(2) An D  particle with some energy is moving for head on collision with nucleus of Z = 85, if the

distance of closest approach is 1.85 u 1014 m , find the energy of D  particle.

(A) 23.13 MeV (B) 13.2 MeV (C) 10 MeV (D) 20 MeV

466

(3) Distance of closest approach of an D  particle with energy 27 MeV is 1.10 × 10–14 m. Find
atomic no of an atom ?

(A) 100 (B) 103 (C) 105 (D) 90
(4) If thickness of foil in D  scattering experiment increases from 2 × 10–7 m to 2.5 × 10–6 m,
Find the increased number of scattered D  particles ?

(A) about 12 times (B) 100 times (C) remains constant (D) 10 times
(5) If number of scattered D  particles increased by 40%, what is the percentage change in
thickness of foil ?

(A) 40 % (B) 80 % (C) 10 % (D) 20 %

(6) If thickness of foil is t1 , number of scattered D  particles are 8500 and thickness of foil is t2 ,
No. of scattered D  particles are 27,500 then,

(A) t1 3.2 t2 (B) t2 3.2 t1 (C) t2 1.6 t1 (D) t2 t1

Ans. : 1 (D), 2 (B), 3 (B), 4 (A), 5 (A), 6 (B)

Bohr’s atomic model and energy levels of H-atom
Hypothesis-1 : Electron can revolve only in those orbits in which its orbital angular momentum is

an integral multiple of 2S . These orbits are known as stationary or stable orbits. In such orbit
h

electron does not radiate energy. 2S = and h Plank’s constant = 6.625 u 10 34 Js .
h

Hypothesis-2 : When electron transit from higher energy (Ei ) orbit to lower energy (E k ) orbit. It
radiates photon of frequency f. Similarly when electron absorbs a photon of frequency f, it makes
transition from lower energy state (E k ) to higher energy state (Ei ) . Ei  Ek hf .

nh
l mvr n=
2S

Ei  E k
hc
hf
O
l Radius of an orbit is,

n 2 h2 0
r where, n principle quantum no.
S m Z e2

n2
r v 0 permittivity of vaccum
Z
m = mass of an electron, Z = atomic number

467

For H – atom r v n2 .

l Kinetic energy for electron,

1 1 Ze2
K mv 2
2 8S 0 r

1 (Ze) (e)
Potential energy, 4S 0
U
r

1 Ze2
4S 0 r

Total energy (E) = K  U

1 Ze 2
E
8S 0 r

K
U
E
2

me 4 Z2
l by substituting value of r, E En , by substituting value of m, e, 0 , h
8 02 h2 n2

13.6 Z2  Z2
En eV , E n D
n2 n2

13.6 1
For H  atom Z 1 Ÿ E n 2
eV En v  .
n n2

Energy levels of H – atom :

E(eV)
0 n f
–1.51
n=3 n 3 (Second excited state)
–3.4
n=2 n 2 (First excited state)

–13.6 n 1 (Ground State)

468

l Transition of an electron from orbit with quantum number ni and energy Ei to orbit with
quantum number nk and energy E k , E i  E k hf

§ 1 1 ·
? f ¨  ¸
me4
8 02 h3 ¨© n 2k n i2 ¸¹

§ 1 1 ·
? ¨  ¸
1 me4
O 8 02 C h3 ¨© n k2 n i2 ¸
¹

§ 1 1 ·
? R ¨  ¸
1
O ¨ n n i2 ¸
© k ¹
2

FÝë_ me4
R = Rydberg constant = 10973700 m 1.
8 02 c h3

l Rydberg constant for an atom,

Rf
R atom
m
1
M atom

Rf 10973700 m1 , m mass of an electron, M mass of nucleus of atom.

l Frequency of electron in a orbit with principle quantum number ‘n’ for hydrogen atom,

me4 Sme4
f , angular frequency Z
4 02 h3 n3 2 02 h3 n3

Zv
2 Rc 1
f
3
n n3

(Periodic time) T v n3
1 2R
O n3

f D n 3 , O v n3

l If electron transit form n = 3 to n = 2 and n = 2 to n = 1

n=3
hf31 hf32  hf 21
f31 f32
? f31 f32  f 21
n=2
? 
1 1 1
f 21 O31 O32 O 21
n=1

469

Success of the Bohr model :

l Stability and energy of Hydrogen-like atoms can be calculated.

 2 3
l Atomic spectra of Hydrogenic atoms e.g. He , Li , Be can be explain.

l It is useful for the confirmation of some principles, for the invention of “heavy hydrogen” or
“deuterium”.

Limitations :

l The orbits of an electron need not to be circular.

l There is an odd combination of classical and quantum mechanics.

l Unable to explain the relative intensities of the spectral lines.

l Fine structure of spectral lines can not be explained.

l Unable to explain the arrangement of electrons in atoms.

(7) The ratio of energies in Fourth and third excited state for Hydrogen atom is .

(A) 4 : 5 (B) 16 : 25 (C) 25 : 16 (D) 1 : 1
(8) Momentum of photon of red light with frequency 400 × 1012 Hz is .

(c = 3 × 108 ms–1)

(A) Zero (B) 8.8 × 10–28 kgms–1
(C) 11.65 × 10–6 MeV C–1 (D) insufficient information
(9) Calculate the radius of an electron in most external orbit, when Phosphorus atom is added in
Silicon (whose Di-electric constant = 12).

(A) 380.9 pm (B) 390.8 pm (C) 930.8 pm (D) 830.9 pm
(10) Linear speed of an electron in Hydrogen atom for ground state (first orbit) is .

(A) 2c (B) 11
c
(C) 137
c
(D) 274
c

(11) When electron transit from n = 5 to n = 1 in Hydrogen atom. Find the speed of emitted photon.

(A) 104 ms1 (B) 2 u 102 ms 1 (C) 4 ms1 (D) 8 u 102 ms 1

(12) If R, v, T and E are radius of orbit, speed of electron, periodic time of revolution and total energy of
an electron respectively. Which option is not directly proportional to quantum number ‘n’.

(A) vR (B) RE (C) Ev (D) RT

470

(13) Total energy of an electron in first excited state for Hydrogen atom is –3.4 eV. What is kinetic
energy of this electron ?
(A) 0 (B) 3.4 eV (C) –3.4 eV (D) 6.8 eV
(14) Angular momentum of an electron in Hydrogen atom for ground state is L1 and for fourth
excited state is L4 then L4  L1 = .

(A) 5L1 (B) 3L1 (C) 2L1 (D) 4L1
(15) Energy of photon in Hydrogen atom is 12.1 eV, its angular momentum is .

(A) 1.05 u 10 34 Js (B) 2.11 u 10 34 Js (C) 3.16 u 10 34 Js (D) 4.22 u 10 34 Js
(16) Kinetic energy of an electron in n = 1 for H-atom is 13.6 eV, Total energy of an electron for
n = 2 for He 2 is .

(A) 13.6 eV (B) 3.4 eV (C) 13.6 eV (D) 3.4 eV
(17) Find the ratio of orbital periodic time for an electron in n = 1 and n = 2.
(A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 1 : 8
(18) Ratio of orbital area for an electron in first excited state and ground state in Hydrogen atom
is .
(A) 2 : 1 (B) 4 : 1 (C) 8 : 1 (D) 16 : 1
(19) Energy of an electron in ground state for H – atom is (R = Rydberg’s constant) .

(A)  Rch (B) Rh1c (C) Rhc (D) hRc

(20) Radius of first orbit in Hydrogen atom is 0.528 A , radius of second orbit is .
D

(A) 4.752 A (B) 2.112 A (C) 0.071 A (D) 0.142 A
D D D D

(21) Find the speed of photon during transition n = 5 to n = 1 in Hydrogen atom.

(A) 4.718 ms 1 (B) 7.418 ms 1 (C) 4.178 ms1 (D) 7.148 ms 1
(22) If principle quantum number n > 4 is not possible, then number of possible elements are .
(A) 4 (B) 32 (C) 60 (D) 64
(23) For energy levels A, B and C, E A  EB  EC . O1 , O 2 and O3 are wave lengths for A, B, and
C. Which option is true for a transistion in Figure.
C
O1 O1 O 2
O3 (A) O3 O1  O 2 (B) O 3 O1  O 2
B

O2 (C) O32 O12  O 22 (D) O1  O 2  O3 0
A
471

(24) Electron transit from fourth orbit to first excited state in Hydrogen atom, Find the frequency of

radiation. R 107 m 1

(A) 16 u 105 Hz (B) 16 u 1015 Hz (C) 16 u 1015 Hz (D) u 1015 Hz
3 3 9 3
4

(25) Total energy of an electron in excited state of H – atom is –3.4 eV. calculate the De-broglie
wavelength.

(A) 6.6 u 1010 m (B) 6.6 u 1011 m (C) 6.6 u 109 m (D) 6.6 u 1012 m

(26) Orbital angular momentum quantum number l = 7, What is orbital angular momentum ?

(A) 72 Sh (B) 42
2S
h
(C) 7 h
2S
(D) 56 h
2S

Ans. : 7 (C), 8 (B), 9 (A), 10 (C), 11 (C), 12 (B), 13 (B), 14 (D), 15 (B), 16 (C),
17 (D), 18 (D), 19 (C), 20 (B), 21 (C), 22 (C), 23 (B), 24 (C), 25 (A), 26 (D)

Hydrogen spectrum :
When Hydrogen electrically discharged at low pressure, atom excited and emits radiation of certain
wavelengths. The group of these radiations is called Hydrogen spectrum.
l Different series

§ ·
Lyman series : O
1 R¨1  1 ¸
©12
n ¹
2

where n 2, 3, 4 ...

n 2 o HD or LD line

n 3 o HE or LE line

n 4 o HJ or LJ line

l In Lyman series for maximum wavelength n = 2 and minimum wavelength n f.
l It is seen in ultraviolet region.

§ ·
Balmar series : O
1 R¨ 1  1 ¸
©2 n ¹
l 2 2

where n 3, 4, 5, 6 ...

n 3 o HD for maximum wavelength o n 3

n 4 o HE for minimum wavelength o n f

l It is seen in visible region.

472

§ ·
Paschen series O
1 R¨ 1  1 ¸
©3 n ¹
l 2 2

Where n 4, 5, 6, 7 ...

l n 4 o HD , n 5 o H E , ....

l for maximum O o n 4, for minimum O o n f
l In near infrared region.

§ ·
Brackett series O
1 R¨ 1  1 ¸
©4 n ¹
l 2 2

where n 5, 6, 7, 8 ...

l n 5 o HD , n 6 o H E , ....

l for O max o n 5, for O min o n f
l In infrared region.

§ ·
Pfund series : O
1 R¨ 1  1 ¸
©5 n ¹
l 2 2

where n 6, 7, 8 ...

l n 6 o HD , n 7 o HE , ....

l for O max o n 6, for O min o n f In far infrared region

n n 1
l When electron transits in lower orbit from nth orbit no. of emitted spectrat lines = .
2

§ 1 1 ·
R Z2 ¨ n 2  n 2 ¸ .
Emitted wavelength for any atonic spectra, O ¨ k i ¸
1
© ¹
l

4  6 
(27) Hydrogen 1H
1
, deuterium 1H
2
, ionized helium 2 He and ionized lithium 3 Li are

given. Their wavelengths are O1, O 2 , O 3 , O 4 respectively compare their wavelengths for a
transistion of an electron from n 2 to n 1.

(A) O1 2O 2 3O3 4O 4 (B) 4O1 2O 2 2O3 O 4

(C) O1 2O 2 2O3 O 4 (D) O1 O 2 4O3 9O 4

(28) Calculate the number of emitted spectral lines, when hydrogen atom having principle quantum
number = 4 moves from ground state to specific excited state.

(A) 3 (B) 5 (C) 6 (D) 2

473

(29) Find the ratio of minimum and maximum wavelength for H-spectra.
(A) 81.86 (B) 86.81 (C) 0.012 (D) 0.12
(30) Find the ratio of maximum and minimum wavelength for Brackett series.
(A) 3.6 (B) 0.36 (C) 78.2 (D) 2.78
(31) Maximum wave number in infrared series is m–1.
(A) 12.18 × 105 (B) 12.18 × 1010 (C) 18.12 × 105 (D) 8204
(32) Find the ratio of O of D  line, for Balmer and Lyman series.
(A) 27 : 5 (B) 5 : 27 (C) 1 : 4 (D) 20 : 27
(33) Wavelength of first line for Balmar series .

(A) 6563 A (B) 6365 A (C) 6563 m (D) 6563 cm
D D

(34) f1 and f 2 are frequency of last and first line of Lyman series. f3 is frequency of last line of
Balmar series then,

(A) f1  f2 (B) f 2  f1 (C) f3 f1  f 2 (D) f1  f 2
1
f3 f3 f3
2

(35) Minimum wavelength of Lyman series is 912 A , then maximum wavelength is A.
D D

(A) 1216 A (B) 1824 A (C) 2434 A (D) 3648 A
D D D D

Ans. : 27 (D), 28 (C), 29 (C), 30 (D), 31 (A), 32 (A), 33 (A), 34 (A), 35 (A)

Excitation and Ionization energy and potential
l Electron revolving in a stationary orbit absorbs specific energy and jumps to an orbit of higher
energy, absorbed energy is called excitation energy and corresponding potential is called
excitatoion potential.
0 n= f

–1.51
n=3
eV
12.09 eV
–3.39 n=2
eV
10.21 eV
–13.6
n=1
eV
energy (eV) = potential (V)
n 1o n 2 required energy = 3.39  ( 13.6) 10.21 eV
n 1o n 3 required energy = 1.51  (13.6) 12.09 eV

474

l The minimum energy required to remove an electron from an atom (to send is it n f ) is called
ionization energy and corresponding potantial is called ionization potential. Ionization energy
(eV) = Ionization patential (V) .
O n= f

1.51eV
–1.51 eV n=3

3.39eV
–3.39 eV n=2

13.6 eV
–13.6 eV n=1
l n 1o n f required energy = 0  ( 13.6) 13.6 eV

l n 2o n f 0  (3.39) 3.39 eV
Emission and absorption spectra
l During the transition of an electron from higher energy (Ei ) orbit to lower energy (Ek ) orbit,
the emitted radiation is called emission spectrum Ei  Ek hf .
l Intensity of such a spectral lines increases as atomic density increases and it decreases as
temperature increases.
l Wavelength of spectraline depends on atomic number.
l Radiation of continuous wavelength is incident on atomic gas to send electron from lower
energy orbit to higher energy orbit. In incident radiation certain wavelength absorbs, these
appear as dark lines in the spectrum, such a spectrum is known as “absorption spectrum”.
l For example, radiation emitted by the lower layer of photosphere in the Sun, certain
wavelength are absorbed hence dark lines are observed. These lines are called fraun
hoffer lines.
l X-ray : Discovered by Rontgen. wavelength between 0.001 to 1 nm.
l X-ray emitted when there is a collision between electron and anode of Cu, Tungston and M o .
Electrons accelerated with 20  40 kV .
l Relative intensity

KD

X-ray
continuous spectrum
KE

Omin

O (1012 m)
30 40 50 60 70 80 90

475

l All the wavelengths (frequencies) emitted in X – ray radiation causes continuous spectrum.
l During the head on collision of electron with anode, radiation of minimum wavelength and
maximum frequency emitted.

O min where V = potential required to accelerate the electron.
hc
eV

l Characteristic X – ray spectrum :
KE

KD

n=1(K)

n=2(L)
n=3(M)

l Incident electrons penetrate deep into the atoms of the anode and knock out the electron
from the atom, from the inner shells, which creates vacancies. The electron from outer shells
experience transition to these vacancies and fill them. The radiation of definite frequencies are
emitted during such transition.

l The radiation is called KD , if it is emitted when electron of K – shell ( n 1 ) is thrown out

and the vacancy is filled by the electron from L – shell ( n 2 ). Similarly KE during the

transition from n 3 to n 1 , LD during the transition from n 3 to n 2 . The X – ray

spectrum formed by such lines is called the chracteristic spectrum.

l Due to screening of the charge of the nucleus Z o Z  1 .

m e4 (Z  1) 2
Energy En 
802 n2 h2

13.6 (Z  1) 2
 eV
n2

l For KD radiation : E L  E K hf K D

§1 1 ·
? 13.6 (Z  1) 2 ¨  u 1.6 u 1019 J
2¸
©1 2 ¹
hf kD
2

13.6 u 1.6 u 1019 § 3 ·
? ¨ ¸ (Z  1)
2
6.62 u 10 © ¹
fk
34 4

476

13.6 u 1.6 u 10 19 u 3
? (Z  1)
6.62 u 1034 u 4
f

? f C (Z  1)

? f CZ  C equation of straigth line.

where C
1
4.965 u 107 Hz 2

f o Z graph is straight lines and its slope is C.

Moseley’s expt. work :
l Moseley suggested that the elements should be arranged with respect to their atomic numbers,
which establish the relation between chemical properties and position of elements in the
periodic table.
l Missing positions in periodic table filled up with appropriate elements.
l Rare earth (Lanthanide) elements, elements coming after Uranium, are arranged properly.

l Information regarding to charge of nucleus can be obtained with the help of K radiation.

Other useful information :

O d 4A radiation, whose penetration is high, called hard X – ray.
D
l

O ! 4A radiction, whose penetration is low, called soft X – ray.
D
l

l Only 1 % energy of incident electron converts in energy of X – ray 99 % energy, waste in
form of Heat. So, arrangement of cooling is required in coolidge tube.
l If f is a fraction of kinetic energy of an electron, converts in X – ray then wavelength of

emitted X – ray is, O
hc
f eV0 .

K
radiation : EL  EK
(L o K)
l hfK

K
l radiation : EM  EK hf K
(M o K)

L
l radiation : E M  E L hf L
(M o L)

(36) Excitation energy in third orbit for Hydrogen atom is eV.
(A) 1.51 eV (B) 3.4 eV (C) 0.66 eV (D) 0.85 eV

477

(37) Energy required to send an electron from second orbit to third orbit is 47.22 eV for a given atom,
then its atomic number is .

(A) 1 (B) 3 (C) 4 (D) 5
(38) Excitation potential of Helium atom in second orbit is .

(A) 7.55 V (B) 21.7 V (C) 13.2 V (D) 10.21 V
(39) For an atom excitation potential in first is ‘V’ volt, its ionization potential is volt.

(A) 14 (B) 34 (C) 43 (D) 54

(40) Energy required to remove an electron from first excited state in Li  is eV.

(A) 122.4 (B) 30.6 (C) 13.6 (D) 3.4
(41) If accelerating potential in X – ray tube increases, .

(A) Intensity of X – ray increases. (B) minimum wavelength of X – ray increases.
(C) minimum wavelength of X – ray decreases. (D) Intensity of X – ray decreases.

(42) Wavelength of X – ray photon is 3.3 A , its corresponding energy is .
D

(A) 7.5 keV (B) 3.8 keV (C) 5.5 MeV (D) 3.7 MeV

(43) Minimum wavelength in X – ray tube of potential 4 MV is A.
D

(A) 1 (B) 0.0062 (C) 0.0031 (D) 10–5
(44) Changing of position of molecules and proper place of molecules can be explained with the help
of .

(A) Moseley’s law (B) Mendeleev’s law (C) Crompton effect (D) Hund’s law

(45) Wavelength K X-Ray is 0.76 A , then find the atomic number of element of anode.
D

(A) 20 (B) 60 (C) 41 (D) 80

(46) In two different emission of K radiation atomic number of target nucleus are 65 and 81. Find
the ratio of their wavelengths.

(A) 14 (B) 16 (C) (D) 16
1 2 25
5

(47) If accelerating potential in X – ray tube increases, speed of X – ray .

(A) Increases (B) Decreases (C) Does not change (D) nothing can be said

478

(48) Operating voltage in X – ray tube is 66 kV, in continuous spectrum of X – ray .
(A) 0.01 nm and 0.02 nm wavelengths remains present.
(B) Both the above wavelengths are absent.
(C) 0.01 nm wavelength would be present and 0.02 nm wavelength would be absent.
(D) 0.01 nm wavelength would be absent and 0.02 nm wavelength would be present.
(49) Maximum radiated frequency in X-ray tube is ‘f’ and operating voltage is ‘V’ volt. If operating

voltage becomes 2 , find maximum radiated frequency.
V

(A) (B) f (C) 2f (D) 4f
f
2
(50) Wavelength of K radiation from Z = 41 is O , wavelength of K radiation from Z = 21
is .
O
(A) 4 O (B) (C) 3.08 O (D) 0.26 O
4
(51) Electron beam of 80 keV energy is incident on Tungston in X – ray tube. If energy of an electron
in K – shell of Tungston is –72.5 keV then .

(A) Presence of continuous spectrum with minimum wavelength 0.155 A is observed.
D

(B) Continuous spectrum of all the wavelengths is observed.
(C) Characteristic spectrum of X – ray of tungston is observed.

(D) Continuous spectrum of minimum wavelength 0.155 A and characteristic spectrum of
D

X – ray is observed.
(52) Gragh of Intensity of X – ray o wavelength in Coolidge tube is as shown in Figure. O c = minimum
wavelength and O K = wavelength of, K . If Accelerating potential increases then .

I

Oc Ok O

(A) Oc Increases (B) O K decreases
(C) OK – Oc Increases (D) Oc , O K decreases but O K  Oc remain constant.
(53) Ionization energy of H–atom is 13.6 eV. H–atom in the ground state is excited with the help of
radiation of 12.1 eV energy. Number of radiated spectral lines are .
(A) One (B) Two (C) Three (D) Four

479

(54) Wavelength of K radiation in H-atom during the emission of X-ray is 0.32 A , then wavelength
D

of K radiation is .

(A) 0.21 A (B) 0.27 A (C) 0.34 A (D) 0.40 A
D D D D

(55) If Z is atomic number of element, Frequency of characteristic spectrum of X – ray is directly
proportional to .

(A) Z 2 (B) (Z  1) 2 (C) Z (D) Z1

Ans. : 36 (C), 37 (D), 38 (A), 39 (C), 40 (B), 41 (C), 42 (B), 43 (C), 44 (A), 45 (C),
46 (D), 47 (C), 48 (D), 49 (A), 50 (A), 51 (D), 52 (C), 53 (C), 54 (B), 55 (B)

Atomic mass and the constitution of Nucleus :
l The entire mass and entire positive charge are concentrated at the central region, known as
nucleus, includes protons and neutrons.
l Proton and neutron are is also called nucleon.
l Atomic number (Z) = number of protons (P)
l Atomic mass number (A) = [P + n]
l neutron number (N) = A – Z

Symbol for element Z X A or ZX
A
l

l The 12th (twelfth) part of the mass of unexcited 6 C12 atom is called 1 amu.

1u 1 amu 1.66 u 1027 kg

l mass spectrometer is used to measured mass of an atoms accurately.
l Isotops having same chemical propertics but different mass. same number of protons but
different number of neutrons.
l For example in case of Cl, the proportion of 34.98 u is 75.4 % and 36.98 u is 24.6 %. Hence
the mass of Cl atom is obtained from its weighted average.

75.4 u 34.98  24.6 u 36.98
mass of Cl Atom =
100

= 35.47 u

l The nuclei for which the neutron number (N = A – Z) is same are called isotones.
l For nuclei Z, A, N are same but radio active properties are different called isomers.
l For nuclei atomic mass number (A) are same, called isobars.

480

l e.g. of isotops : 92 U 235 , 92 U
238

l e.g. of isotops : 36 Kr86 , 37 Rb
87

l e.g. of isobars : 82 Pb 214 , 83Bi
214

l
35 Br
80 having a pair of isomers.

Naclear forces, nuclear radius and stability of nucleus

l P–P, P–n, n–n strong nuclear force.

l It is also called quark-quark force because P and n are made up of quarks.

l Types of quarks : up, down, charm, strange, top, bottom.

l Nuclear forces depends on orientation of spin.

Average radius of nucleus is given by R R0 A 3 where R0
1
l 1.1 fm

l Density of nucleus is about 2.3 × 1017 kgm–3, 1014 times than density of water.
l As atomic number increases, coloumbian force increases multiplyingly, to balance it number of
neutron increases so that strong nuclear force icreased.

For e.g. 6 C12 P 6, n 6 but in 92 U 235 P 92 , n 143

It is essential condition for the stability of nucleus.

l Mass - energy : 1 u 1.66 u 1027 kg ; 1 eV 1.6 u 10 19 J

1u 931.48 MeV o mass-energy equivalency

E m c2

(1 u ) (3 u 108 ms 1 )2

(1.66 u 1027 ) (9 u 1016 ) 931.48 u 106
1.6 u 1019 931.48 MeV

Binding energy of nucleus

Mass of the nucleus is always less than the total mass of its constituents in the free state. This
decrease in the mass is called mass defect,

M  Zmp  Nmn

'm Zmp  Nmn  M

481

l The energy equivalent to mass defect is called binding energy of nucleus. Eb = (' m) c2 .

l When it is divided with number of nucleons, we get average binding energy per nucleon-

§ Eb ·
¨ A ¸ , this energy is actually measurement of stability of the nucleus.
© ¹
E bn

l Ebn is an average energy per nucleon to release all the constituent particles from the nucleus.

l Average binding energy per nucleon (Ebn ) is measurement of stability of the nucleus.

l In case of deutron, 1 H 2 its mass is 2.0141 u and sum of the mass of proton and neutron in
free state is 2.0165 u. ' m 2.0165  2.0141 0.0024 u , the energy equivalent to this mass
defect is 0.0024 u 931.48 2.24 MeV. Which is called binding energy.

l Hence, binding energy of 1 H 2 is 2.24 MeV.

l Thus to liberate proton and neutron from 1 H 2 , 2.24 MeV energy has to be supplied to it from
outside. Conversely if one proton and one neutron coalease 2.24 MeV energy is emitted out.
MeV
For 1 H 2 E bn
2.24
l 2
1.12 .
necleon
Ebn

10 Fe

8

6

4

2

A
0 50 100 150 200 250

§ MeV ·
Ebn is maximum ¨ | 8.8 ¸ for the nucleus of (Fe) > A 56@ .
© nucleon ¹
l

l Ebn is small for A < 30 and A > 170.

l For 30  A  170 Ebn is almost constant. These nuclei are the most stable. It is due to the
saturation property of the nuclear forces.

l When heavy nucleus ( A ! 170 ) gets divided in two lighter nuclei, energy released, this
process is called nuclear fission.

l Two lighter nuclei (A  10) are fused to form a heavier nucleus and energy is produced, this
process is called nuclear fusion.

482

l The energy to be supplied to separate a nucleon from a nucleus is called separation energy.

l Mass defect per nucleon is called packing fraction (f).

'm
Ÿ E bn
MeV
f f c 2 or f u 931.48 .
A necleon

(56) For the same separation of 1 fm in the nucleus, the force acting between two proton is F1,
between two neutron is F2 and between proton and neutron is F3 then,

(A) F1 < F2 < F3 (B) F2 < F1 < F3 (C) F1 < F2 = F3 (D) F1 = F2 < F3

(57) 1040 deuterons are present in a star initially. Energy released according to reaction given below :

1H
2
 1H 2 o 1H 3  p and 1 H 2  1H 3 o 2 He 4  n

If emitted power is 1016 W. The time to destroy quantity of deuteron is .

(A) 106 s (B) 108 s (C) 1012 s (D) 1016 s

m 1 H2 2.014 u, m( P) 1.007 u , m (n) 1.008 u, m (2 He4 ) 4.001 u .

(58) Reaction of nuclear fission, 92 U 236 o X117  Y117  n  n . For X and Y E bn 8.5 MeV

and E bn 7.6 MeV for U 236 then produced energy MeV.

(A) 2000 MeV (B) 200 MeV (C) 20 MeV (D) 2 MeV
(59) Packing fraction (f) = .

(A) MA A (B) A A
M
(C) MM A (D) MA A

(60) If u = 1 amu and A = atomic mass number If mass of an atom is Au then A = .

(A) 1 (B) 12 (C) 16 (D) between 1 to 110
(61) Mass in the proton is completely converted in energy, then energy is MeV.

(A) 9310 (B) 931 (C) 10078 (D) 100
(62) A and B are Isotops, B and C are Isobars. dA , dB and dC are the densities of their nuclei
then .

(A) d A ! d B ! dC (B) d A  d B  dC (C) d A dB dC (D) d A d B  dC

(63) The energy required to separate a nucleon from a nucleus is called .

(A) Binding energy per nucleon (B) Binding energy
(C) Reaction energy (D) Separation energy

483

(64) Calculate the released energy Q in nuclear fusion. 1 H2  1H 2 o 2 He4  Q

m 1 H2 2.0141 u , m 2 He4 4.0024 u .

(A) 12 MeV (B) 6 MeV (C) 24 MeV (D) 48 MeV

(65) If velocity of light becomes 23 , the energy releases in nuclear fission decreases in multiple

of .

(A) 23 (B) 94 (C) 59 (D) 5
9

(66) Mass of two isotops of Boron 5 B10 and 5 B11 are 10.01294 u and 11.00931 u respectively.
Mass of an atom of Boron is 10.811 u. Find the proportion of these two isotops.
(A) 30 %, 70 % (B) 40.12 %, 59.88 % (C) 72.05 %, 27.95 % (D) 19.90 %, 80.10 %
(67) Mass defect in nuclear fission is 0.03 %. Energy released in fission of 1 kg mass is .

(A) 2.7 u 1013 J (B) 27 u 1014 J (C) 0.27 u 10 13 J (D) none of these

(68) In a nuclear reaction given below, Z X A .

7N
14
 2 He4 o Z X A  1H1

(A) 7 N16 (B) 7 N17 (C) 8 O16 (D) 8 O17

(69) For 8 O16 , Ebn = MeV.u , m p 1.007825 amu , mN 1.008665 amu . M 8 O
16

= 15.9949 amu .,
(A) 7.973 (B) 79.73 (C) 0.79 (D) none of these

(70) Find the number of electron, proton and neutron in 12 g, 6 C12 .

(A) 6 u 1023 (each) (B) 12 u 1023 (each)

(C) 18 u 1023 (each) (D) 36 u 1023 (each)

(71) Mass-defect for the nucleus with Z = 2 and A = 4 is 0.04 u. Calculate the binding per nucleon.
(A) 931 MeV (B) 93.1 MeV (C) 9.31 MeV (D) 0.04 MeV
(72) Energy produced due to destruction of proton is 3724 MeV in a nuclear reaction, then these number
of protons = .
(A) One (B) Two (C) Three (D) Four

484

(73) Atomic mass number of Helium and sulpher are 4 and 32 respectively. The radius of Sulphur
nucleus is times the radius of Helium nucleus.
(A) 2 (B) 4 (C) 8 (D) 12

(74) 1H
2
 1H 3 o 2 He 4  0 n1 . Binding energies of 1 H 2 , 1H 3 and 2 He 4 are a, b and
c (MeV) respectively. Energy releases in the reaction = MeV.

(A) a  b  c (B) c  a  b (C) c  a  b (D) a  b  c
(75) E bn for the nucleus of 3 Li 7 and 2 He 4 is 5.60 MeV and 7.06 MeV respectively. Energy
released = MeV.

3 Li
7
 p o 2 2 He4

(A) 19.6 (B) 17.3 (C) 8.6 (D) 2.4
(76) Energy releases in a given fusion = .

1H
2
 1H 2 o 1H 3  1H1

(A) 1 erg (B) 1 eV (C) 4 MeV (D) 4 keV

(77) Energy released in each fission of U 235 is 200 MeV. Output power in nuclear reactor is
1.6 MW. Find the rate of fission. (number of fission per second).
(A) 5 × 1016 (B) 1017 (C) 1.6 × 1013 (D) 1019

Ans. : 56 (C), 57 (C), 58 (B), 59 (D), 60 (B), 61 (B), 62 (C), 63 (D), 64 (C), 65 (C), 66 (D),
67 (A), 68 (D), 69 (A), 70 (D), 71 (C), 72 (D), 73 (A), 74 (C), 75 (B), 76 (C), 77 (A)

Natural Radio activity

l Becquerel found that radiations of certain specific properties are emitted naturally from
Uranium, this phenomenon is called narural radio activity. Those radiations were initially
known as Becquerel rays.

l Madam curie separated two new elements from the ore of Uranium called pitch blende. They
were named Polonium and Radium, their activities are several times that of Uranium.

l The emission of radioactive radiations is spontaneous, instantaneous and continuous. It is not
affected by temperature, pressure, electric field and magnetic field.

l One can not stop the emission of radioactive radiations or can not change the rate of
emission.

l Heavy element emits radioactive radiation to become stable from unstability.

l D  particles : D  particle is a nucleus of - 2 He 4 with 2-proton and 2 neutron. Charge is +2 e.

485

l Almost the nucleus for which Z > 83 emits D  particles.

ZX
A
o Z  2 Y A 4  2 He4 ( D )

e.g., 92 U
238
o 90Th 234  2 He4 ( D )

l The disintegrating nucleus is called the parent nucleus and newly formed nucleus is called
the daughter nucleus.
l In the emission of D  particle, atomic number decreases by 2 and atomic mass number
decreases by 4.

l E  particle : E  particles are electrons emitted from nucleus. Their velocity depends on the
nuclide emitting them. In other reaction of E  decay positrons are also emitted.

l e = E and e = E .

l When a neutron converts into proton and electron is produced in the nucleus which can not

live in a nucleus so emits as E

n o p  e  X (anti neutrino)

ZX
A
o Z 1Y A  e E  X

Atomic number increases by one and Atomic mass number does not change.

l When a proton converts in to a neutron positron, is emitted as E . Atomic number decreases by

one and atomic mass number does not change.

p o n  e   X (neutrino)

ZX
A
o Z 1Y A  e  E  X

l J  rays : They are electromagnetic waves.
l Photon of J  rays emitted during the transition of nucleus from higher energy state to lower

energy state, wavelength O .
hc
E

l When nucleus emits D and E particles it is in a excited state, according to need, by emitting
photon of J  ray comes to a stable state.

l e.g., Due to emission of E 27 Co60 converts into 28 Ni60 , during this emits J  rays of
1.17 MeV and 1.33 MeV step by step.

486

l All these raido active radiations affect the photographic plate, produce fluorescence.

D E J

Relative ionizing power : 10000 100 1
Relative penetration power : 1 100 10000
l Energy value in D  emission

§ A ·
Q KD ¨ ¸ Where, kD = kinetic energy of D , A = atomic mass number
© A4¹

Nuclear Reactions
l By bombarding suitable particles of suitable energy on a stable element, transformed into
another element is called artificial nuclear reaction. Energy gain or loss is denoted as
Q – value.
A  a o B  b  Q

7N
14
 D 2 He
4
o 8 O17  1H1  Q

A o Target nucleus o 7 N14

a o projectile particle o D 2 He
4

B o product nucleus o 8O17

b o product (emitted) particle o 1H
1

Q > m A  ma  m B  m b @ c2
Q ! 0 exoergic reaction, Q  0 endoergic reaction
l In nuclear reaction momentum, electric charge and energy each one is conserved.
l Q  value of reaction = energy equivalent to decrease in mass in the reaction = increase in
the kinetic energy.
Nuclear Fission :
Neutron is good projectile because it is charge less and does not have to face the coloumb repulsive forces.
l Disintegration of nucleus in which enormous energy is produced. This process was named
nuclear fission.

e.g., 92 U
235
 0 n1 o 92 U 236 o 51Sb133  41Nb99  4 0 n1  Q

92 U
235
 0 n1 o 92 U 236 o 56 Ba144  36 Kr89  3 0 n1  Q

92 U
235
 0 n1 o 92 U 236 o 54 Xe140  38Sr 94  2 0 n1  Q

487

l Fission fragments having Z values between 36 to 56 and A values between 90 to 95. They

converts into stable nuclei with emission of E and J .

l Energy of incident neutron is about 2 MeV.

l Energy produced is about 200 MeV.

Nuclear chain reaction and Nuclear reactor

l With the help of neutrons produced nuclear fission, more nuclei is accomplished. So, we get
more energy and more neutrons. A series of such processes is called nuclear chain reaction.
If such a process is properly controlled, then energy can be obtained continuously at steady
rate. Nuclear reactor is the illustration of this.

Difficulties encountered in nuclear reactor and their removal :

l To stop neutrons from escaping reflecting surfaces, moderators like water, heavy water,
Graphite and Beryllium are used.

l In a chain reaction enormous heat energy is produced and the temperature is likely to become
106 K. Hence coolants like water, liquid sodium are used.

l The ratio of number of neutrons produced at any stage to the number of neutrons incident is called
multiplication factor (K). It is a measure of the growth of number of neutrons.
If K = 1 the reactor is said to be critical. If K > 1 is said to be super critical in such a condition
explosion can take place, if K < 1 said to be subcritical, the process slows down and eventually stops.

l In order to control the value of K, rods of Boron and Cadmium are kept, called control rods.

Reactor : As a fuel 3 % 92 U 235 and remaining 92 U 238 .

l
92 U
235
 0 n1 o 93 Np 239  1e0  X

93 Np
239
o 94 Pu 239  1e0  X

Plutonium is intense radio active and fissionable by slow neutron.

l In Pressurised Water Reactor normal water is used as moderator and also as coolent.

l Water is pushed into the core of the reactor at temperacure 600 K and 150 atm pressure,
generated steam operates the turbine, which produces electric power.

Thermonuclear Fusion in Sun and other Stars :

l The sun emitting energy at the rate of 3.8 u 1026 Js 1 .

l When two proper light nuclei are fused at a very high temperature to form a heavy nucleus,
enormous energy is produced, such process is called thermonuclear fusion.
488

l Proton - Proton cycle :

1 1 2 0
1 H  1H o 1H  1e  X  0.42 MeV

1 e
0
 1e0 o 2 J  1.02 MeV

1H
2
 1H1 o 2 He3  J  5.49 MeV

2 He
3
 2 He3 o 2 He 4  1H1  1H1  12.86 MeV

l First three reaction ocur twice.
l Total produced energy = 2 u 0.42  2 u 1.02  2 u 5.49  12.86 26.7 MeV.

(78) What is produced in a consecutive step in a given reaction ?

ZX
A
o Z 1Y A o Z 1T A4 o Z 1T A4

(A) D , E, J (B) E, D , J (C) J , D , E (D) D , J , E

(79) Find the number of D and E  particles in the conversion 92 X 235 o 88Y 219 .

(A) 4, 4 (B) 5, 5 (C) 6, 6 (D) 4, 8
(80) If PD , PE , PJ are penetrating power of D , E, J then .

(A) PD PE PJ (B) PD ! PE ! PJ (C) PD  PE  PJ (D) PD PE  PJ

(81) D and E particles are emitted from the ends A and B respectively in a wire,
then .
(A) Electric current flows from A to B
(B) Electric current flows from B to A
(C) Electric current is not produced.
(D) Electric current flows toward the mid-pt. from every side.
(82) During the radio active emission of element having atomic number 90 and atomic mass number

232, final product is 82 Pb 208 . Find the emitted number of D and E – particles.

(A) D 3, E 3 (B) D 6, E 4 (C) D 6, E 0 (D) D 1, E 6

(83) Nuclear reaction : 5 B10  2 He 4 o 7 N13  _________

(A) Proton (B) neutron (C) electron (D) D
(84) 6C
12 absorps neutron and emits E  particle. Find the final product.

(A) 7 N14 (B) 5 B13 (C) 7 N13 (D) 6 C13

489

(85) Arrange the ionizing power of D , E and J in descending order.

(A) J , D , E (B) J, E, D (C) D, E, J (D) E, J , D
(86) Energy produced in nuclear fission is 200 MeV. If output power of reactor is 5W then find the
rate of fission.

(A) 1.56 u 1010 s 1 (B) 1.56 u 1011 s 1

(C) 1.56 u 10 16 s 1 (D) 1.56 u 1017 s 1

(87) D  particle is emitted by nucleus Z X A . Final nucleus emits E . Then find the atomic number
and atomic mass number of final nucleus.
(A) Z – 3, A – 4 (B) Z – 1, A – 4 (C) Z – 2, A – 4 (D) Z, A – 2

(88) Three D  particles and one E  particle are emitted by 86 R n 236 , then for product nucleus X
.
(A) Z = 83, A = 224 (B) Z = 84, A = 218
(C) Z = 84, A = 220 (D) Z = 82, A = 223

(89) Calculate the release energy in reaction, 88 Ra 226 o 86Ra 222  2He 4 . Kinetic energy of

D  particle is 4.78 MeV . Parent element is stable.
(A) 8 MeV (B) 4.78 MeV (C) 4.87 MeV (D) none of these

(90) Energy emitted from a star is 2.7 u 1036 Js 1 . Find the decrement in its mass.

(A) 3 u 1018 kg s 1 (B) 3 u 1019 kg s 1 (C) 3 u 10 20 kg s 1 (D) 3 u 10 21 kg s 1

(91) Series of emitted particles from the nucleus (Z 92) is D, D, E , E , D, D, D, D,

E , E , D, E , E , D . Then Z of final nucleus = .

(A) 78 (B) 82 (C) 74 (D) 76
(92) 1H
2
 1H3 o 2 He4  n , potential energy due to repulsive force between two nucleus in

nuclear fusion is 7.7 u 10  14 J . How much temperature should be given to the gas so this
reaction be possible ?
(A) 105 K (B) 103 K (C) 109 K (D) 107 K
(93) What this reaction suggest ?

4 1H1 o 2 He4  2 1e0  26 MeV.

(A) E  decay (B) J  decay (C) Fusion (D) Fission

490

(94) D  particles are emitted from stable radio active element having atomic mass mumber 208.
Energy of emitted D  particles is E. Find the energy of disintegration.

(A) 51 E (B) 52 E (C) 52 E (D) E
52 51

Ans. : 78 (B), 79 (A), 80 (C), 81 (A), 82 (B), 83 (B), 84 (C), 85 (C), 86 (B), 87 (A),
88 (A), 89 (C), 90 (B), 91 (A), 92 (C), 93 (C), 94 (A)

Radio active constant and Activity
l In a specimen of radio active material, if the number of undisintegrated nuclei of an element
at time t is N and there after if 'N nuclei disintegrate in time interval 't , then

lim 'N dN
is called the rate of disintegration or the decay rate or activity (I) of
' t oo ' t dt
that element at time t. Activity means the number of nuclei decaying per unit time.

dN
l The decay rate is proportional to the number of undisintegrated nuclei at that time. vN
dt

?  ON (negative sign indicates that as time passes N decreases.)
dN
I
dt

O Radioactive constant = decay constant, unit = s–1) its value depends on the type of
disintegrating element. For different unstable isotops of the same element, the values of O are
different.
l O o large o I o more o shortlived elements.

O o small o I o less o longlived elements.

l O is independent of temperature pressure, electric field and magnetic field.

§ dN ·
l For the nucleus of a given element O ¨ O ¸ shows the probability of disintegration per
© Ndt ¹
unit time.
Units of Activity
l 1 disintegration oceur in one second, then activity of body is called 1 Becquerel.
1 Bq = 1 disintegration / sec.
l If 3.7 × 1010 disintegration per second take place, the activity of a substance called 1 curie (Ci).
1 Ci = 3.7 × 1010 disintegration / sec.
1 Ci = 3.7 × 1010 Bq.
Exponential law of Radio active Disintegration

 ON Ÿ  O dt
dN dN
dt N

491

ª t 0 o N N0 º
³  O ³ dt
N t
« t t o N N »¼
dN
N0
N
0 ¬

? > ln N @ NN0  O > t @0
t

? ln N  ln N0  Ot

§ N·
? ln ¨  Ot
© N 0 ¸¹

? eO t Ÿ N N 0 e O t
N
N0

Similarly, I I 0 e O t and M M 0 e  O t M 0 = Initial mass, M = mass after disintegration.

l As time passes, the number of nuclei and activity decreases exponentially. This curve is called
decay curve. For I o t and M o t similar graph is obtained.

N N0

t
Half life :
l The time interval in which the number of nuclei of radioactive element becomes half of its
§W ·
value at the beginning, is called half life of that element ¨ 1 ¸ .
© 2¹

W1
N0 e O t , t and N
N0
N 2
2

 O W1 O W 1 OW 1
1
Ÿ 2 Ÿ 2
N0 2 2 2
N0 e e e
2

? O W1 Ÿ
ln2
O O
ln 2 W
0.693
1
2 2

l Half life of different radio active elements are from 10–7 s to 1010 Yr.

492

0
§1·
O t 0 time N N0 Ÿ
N
¨ ¸
N0 ©2¹
1
N0 N §1·
t W1 time N Ÿ ¨ ¸
2 2 N0 ©2¹
2
N0 N §1·
t 2 W 1 time N Ÿ ¨ ¸
2 4 N0 ©2¹
3
N0 N §1·
t 3 W 1 time N Ÿ ¨ ¸
2 8 N0 ©2¹

§1·
n W1 time o
n
¨ ¸ .
N
©2¹
t
2 N0

Where n
t given time
W1 Half life
2

Time undecayed part Decayed part
t=0 100 % 0%

t = W 12 50 % 50 %

t = 2W 12 25 % 75 %

t = 3W 12 12.5 % 87.5 %

t = 4W 12 6.25 % 93.75 %

Mean life W
The time interval during which the number of nuclei of a radioactive element becomes eth part of its
original, is called Mean or average life W of that element. (e = 2.718)

N0 e O t , t W and N , N 0 e O W Ÿ e1 e O W
N0 N0
l N
e e

Ÿ OW 1 Ÿ W
1
O

other informations : – W1 0.693 W
2

– W W 1 u 1.44
2

493

l For the emission of D and E , total O t O D  OE


1 1 1
? 
1 1 1
W( t ) WD WE Similarly, W 1 (t ) W 1 (D ) W 1 (E)
2 2 2

1 th
(95) Half life of At 215 is 100 Ps , The time in which activity of its specimen becomes of its
16
original is .

(A) 400 Ps (B) 6.3 P s (C) 40 P s (D) 300 P s

(96) Activity of Radio active element at time t1 is R1 and at time t2 is R2 . t2 ! t1 . If the mean

life W then .

R1  R2
(A) R1 t1 (B) constant
t2  t1
R2 t2

§t t · § t ·
(C) R2 R1 exp ¨ 1 2 ¸ (D) R2
© W ¹
R1 exp ¨ 1 ¸
© W t2 ¹

(97) Quantity of Radon is 16 g and its half life is 3.8 days. Find the disintegration of Radon in 19 days ?

(A) 5 g (B) 0.5 g (C) 15.5 g (D) none of these

3 th
(98) Half-life of radio active element is 30 days. Find the time to disintetrate its mass.
4

(A) 15 days (B) 45 days (C) 30 days (D) 60 days

1 th
(99) Find the half life of radio active element whose activity becomes
16
of original in 30 years.

(A) 90 years (B) 120 years (C) 15 years (D) 7.5 years
(100) Find the decay constant of an element in Que : 99.

(A) 0.0924 Yr 1 (B) 9.24 Yr 1 (C) 924 Yr 1 (D) 0.0688 Yr 1

(101) Initially number of atoms in radio active element A and B are equal. Their half lives are 1 hr and
2 hr respectively. Find the ratio of rate of disintegration after 2 hr.

(A) 4 : 2 (B) 1 : 1 (C) 1 : 2 (D) 8 : 1

(102) Find the number of disintegration per second in 2.3 g, 90 Th 230 whose half life = 2.4 u 1011 s.

(A) 6 u 1021 (B) 0.73 u 1010 (C) 1.73 u 1010 (D) 109
494

(103) Half life of radio active isotop is 10 min. At a given moment number of radio active nucleus are
108 . Number of nucleus after 5 min = .
8 108
(A) (B) 104 (C) 2 u 107 (D)
10
2 2
(104) Half life of Pa218 is 3 min and mass of specimen is 16 g. How much mass is remaining after 15 min.

(A) 3.2 g (B) 2.0 g (C) 1.6 g (D) 0.5 g
(105) Half life of radio active element is 5 min. % of element remains undecayed after
25 min.

(A) 25 % (B) 75 % (C) 6.25 % (D) 3.125 %
(106) Activity of recenty bought bottle of radio active Tritium is 3 %, “seven years old” labelled on the
bottle, then is was made before how many years ? (Half life = 12.5 Yr)

(A) before 220 years (B) before 420 years (C) before 63 years (D) before 70 years
(107) Half life of radio active element is 10 years, then its average life is .

(A) 14.4 yr (B) 20 yr (C) 15 yr (D) 28.8 yr
(108) Activity of radio active sample at time t1 is I1 and at time t2 is I2. If the half life of sample is
W1
, the number of undecayed nucleus in t2  t1 time is directly proportional to .
2 I1  I 2
(A) I1 t2  I 2 t1 (B) I1  I 2 (C) W1 (D) ( I 2  I1 ) W 1
2 2
(109) Half life of radio active element is 20 min. Time to decay from 20% to 80 % is .

(A) 20 min (B) 40 min (C) 25 min (D) 30 min
(110) Half lives of radio active element for D and E decay are 8 year and 24 year respectively. After
12 year, its activity is how much percentage of its original activity ?

(A) 50 (B) 12.5 (C) 25 (D) 6.25
(111) There are 4 × 1016 nucleus in a radio active sample. Half life of element is 10 days. Find the
number of decayed nucleus in 30 days.

(A) 0.5 × 1016 (B) 2 × 1016 (C) 3.5 × 1016 (D) 1 × 1016
(112) Mass of radio active sample at t = 0 is 10 g. After two mean life mass of sample is
(A) 1.36 g (B) 2.50 g (C) 3.70 g (D) 6.30 g
(113) Half life of Radium is 1600 year. Mass of sample is 100 g. The time for which its mass becomes
25 g is years.

(A) 6400 (B) 2400 (C) 3200 (D) 4800
495

(114) After decayed two radio active nucleus P and Q becomes stable element R. At t = 0, the number
of nucleus in P and Q are 4 N0 and N0 respectively. Half lives of P and Q are 1 min and 2 min
respectively. When stable element R formed, number of nucleus in P and Q are equal. Then
number of nucleus in stable element R is .

(A) 2 N0 (B) 3 N0 (C) 3 N2 0 (D) 9 N2 0

(115) Half life of radio active isotop X is 50 year, after decayed it converts into stable element Y. If the
proportion of X and Y is 1 : 15, approximate life of rock is .

(A) 100 year (B) 150 year (C) 200 year (D) 250 year

(116) Sample of Cu decayed upto 8 in 15 min and converts in Zn, its Half life is .
7

(A) 10 min (B) 15 min (C) 5 min (D) 7.5 min
(117) A radio active element X decayed into new element Y. If creation rate of element Y is R then the
graph of R o t would be.
R R
(A) (B)

t t

R R
(C) (D)

t t

(118) Undecayed part of radio active sample in half time of half life is .

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

Ans. : 95 (A), 96 (C), 97 (C), 98 (D), 99 (D), 100 (A), 101 (B), 102 (C), 103 (D), 104 (D),
105 (D), 106 (D), 107 (A), 108 (D), 109 (B), 110 (C), 111 (C), 112 (A),
113 (C), 114 (D), 115 (C), 116 (C), 117 (C), 118 (A)

496

Assertion - Reason type Question :
Instruction : Read assertion and reason carefully, select proper option from given below.
(a) Both assertion and reason are true and reason explains the assertion.
(b) Both assertion and reason are true but reason does not explain the assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false and reason is true.

(119) Assertion : Z X A undergoes two D  decay, two E  decays and two J  decays and the
daughter product is Z  2 X A8 .

Reason : In D  decay the mass number decreases by 4 and atomic number decreases by 2.
In E – decay the mass number remain unchanged, but atomic number increases by
1 only.
(A) (a) (B) (b) (C) (c) (D) (d)
(120) Assertion : All nuclei are not of same size.
Reason : Size depends on atomic mass.
(A) (a) (B) (b) (C) (c) (D) (d)
(121) Assertion : X - rays are used for studying the structure of crystals.
Reason : The difference between the atoms of crystal is of the order of wavelength of X - ray.
(A) (a) (B) (b) (C) (c) (D) (d)
§ ·
W1 ¨ W 1  W .¸
(122) Assertion : If half life and mean life of radio active element are and t ¨ 2 ¸
2 © ¹

1
Reason : Mean life =
decay constant

(A) (a) (B) (b) (C) (c) (D) (d)
(123) Assertion : Isobars are the nuclei having same mass number A but different atomic number Z.
Reason : Neutrons and Protons are present inside nucleus.
(A) (a) (B) (b) (C) (c) (D) (d)
(124) Assertion : The ionisation potential of Hydrogen is 13.6 eV, the ionised potential of double
ionised lithium is 122.4 eV.

13.6
Reason : Energy in nth state of Hydrogen atom is E n eV.
n2
(A) (a) (B) (b) (C) (c) (D) (d)
(125) Assertion : Radio active nuclei emit E particle.
Reason : Electrons exist inside the nucleus.
(A) (a) (B) (b) (C) (c) (D) (d)

497

(126) Assertion : If the half life of radio active substance is 40 days, then 25 % substance decays in
20 days.

n t
§1·
Where
n
Reason : N N0 ¨ ¸ W1 .
© 2¹ 2

(A) (a) (B) (b) (C) (c) (D) (d)
(127) Assertion : Balmar series lies in the visible region of electro magnetic spectrum.
§ 1 1 ·
R ¨ 2  2 ¸ Where, n 3, 4, 5.
1
Reason :
©2 O n ¹
(A) (a) (B) (b) (C) (c) (D) (d)
(128) Assertion : A certain radio active substance has a half life of 30 days. Its disintegretion
constant is 0.0231 day–1.
Reason : Decay constant varies inversly as half-life.
(A) (a) (B) (b) (C) (c) (D) (d)
(129) Assertion : Energy is released in nuclear fission.
Reason : Total binding energy of the fission fragments is larger than the total binding energy
of the parent nucleus.
(A) (a) (B) (b) (C) (c) (D) (d)
(130) Assertion : Heavy water is preferred over ordinary water as a moderator in reactors.
Reason : Heavy water, used for slowing down the neutrons, has lesser absorption probability
of neutrons than ordinary water.
(A) (a) (B) (b) (C) (c) (D) (d)
Ans. : 119 (B), 120 (B), 121 (A), 122 (B), 123 (B), 124 (B), 125 (C), 126 (D), 127 (A),
128 (B), 129 (A), 130 (A)
Comprehension Type Questions :
Passage :
A single electron orbits arounds a stationary nucleus of charge + Ze, Where Z is a constant and
e is the magnitude of the charge. It requires 47.2 eV to excite the electron from the second orbit
to the third orbit. (The ionization energy of Hydrogen atom = 13.6 eV. radius = 5.3 × 10–11 m,
c = 3 × 108 ms–1, h = 6.6 × 10–34 Js) Based on the above facts, answer the following questions :
(131) The value of Z is = .
(A) 1 (B) 2 (C) 3 (D) 5
(132) The energy required to excite the electron from the second excited state to the third excited state
is = eV.
(A) 47.2 eV (B) 14.53 eV (C) 16.53 eV (D) 18.53 eV
(133) The energy required to excite the electron from the first excited state to the second excited state
is = eV.
(A) 47.2 eV (B) 16.53 eV (C) 255 eV (D) none of these
(134) The minimum wavelength of the electromagnetic radiation required to deport the electron first
orbit to an upper orbit is .

(A) 48.5 A (B) 36.4 A (C) 45.8 A (D) 34.6 A
D D D D

498

(135) The kinetic energy, potential energy, total energy and angular momentum of the electron in the
first orbit have respective values given by.

(A) 340 eV,  340 eV,  680 eV, 1.05 u 1034 Js

(B) 340 eV,  680 eV,  340 eV, 1.05 u 1034 Js

(C) 680 eV,  340 eV,  680 eV, 2.05 u 1034 Js

(D) 680 eV,  1360 eV,  680 eV, 2.05 u 1034 Js
(136) The radius of the first Bohr’s orbit is = .

(A) 0.53 A (B) 0.106 A (C) 5.3 A (D) 1.06 A
D D D D

Passage : For identical Hydrogen gas has some atoms in the lowest energy level A and some in a upper
energy level B. The atoms of the gas make transition to a higher energy level by absorbing
photon of energy 2.7 eV. Subsequently, the atoms emit radiation of only six different photon
energies. Some of the emitted photons have an energy of 2.7 eV, some have more energy and
some less than 2.7 eV. Based on the above facts, answer the following questions :
(137) The principle quantum number of the initially excited level B is .
(A) 2 (B) 4 (C) 6 (D) 8
(138) The ionization energy for the gas atom is eV.
(A) 14.4 eV (B) 13.6 eV (C) 3.4 eV (D) 1.51 eV
(139) The emitted photons will have energy value E (in eV). The photons emitted have maximum and
minimum energy values E and e respectively.
(A) e = 1.35 eV and E = 13.5 eV (B) e = 0.7 eV and E = 1.35 eV
(C) e = 0.7 eV and E = 13.5 eV (D) none of these
Passage :
Nuclei of a radio active element A are being produced at a constant rate D . The element has a
decay constant O . At time t 0 , there are N 0 , nuclei of the element. At time t, number of nuclei

of A is N. Also for D 2 N0 O , the number of nuclei of A after one half life is and the
N1
2

limiting value of N as t o f is N f . Based on the above facts, answer the following questions :

(140) N as a function of t is given by,
D
(A) N N0  1  e O t (B) N N0 e O t
O

1 ª
D  D  O N 0 e O t º
D
N0  1  e O t
O ¬ ¼
(C) N (D) N
O

499

(141) The value of in terms of N 0 is = .
N1
2

(A) N 12 (B) N 12 (C) N 12 (D) N 12
N0 3 N0 3 N0 3 N0
2 2 4 8

(142) The value of N f in terms of N 0 is = .

(A) Nf 3 N0 (B) N f o 0 (C) Nf N0 (D) Nf 2 N0

Passage : In the Bohr model of the Hydrogen atom, the electron revolves in a circular orbit of radius
ro = 0.53 A around the nucleus. Based on the above facts, answer the following question.
D

(143) The velocity of electron is nearly ms–1.

(A) 2.2 u 106 (B) 2.6 u 106 (C) 2.8 u 106 (D) 2.9 u 106
(144) The velocity of electron is n times the velocity of light. The value of n is = .

(A) 81 (B) 101 (C) 137 (D) 337
1 1 1 1

(145) The electric potential energy in eV.
(A) –13.6 (B) –27.2 (C) –36 (D) –54
(146) The kinetic energy in eV is .
(A) 13.6 (B) –27.2 (C) –36 (B) –54
(147) The total energy in eV is .
(A) 13.6 (B) –13.6 (C) 27.2 (D) –27.2
Ans. : 131 (D), 132 (C), 133 (A), 134 (B), 135 (B), 136 (B), 137 (B), 138 (A),
139 (C), 140 (D), 141 (B), 142 (D), 143 (A), 144 (C), 145 (B), 146 (A), 147 (B)

Match the columns :

Match the column-1 with column-2 Ñ

column-1 column-2

(a) In this reaction mass of product is less (p) D  decay

than the mass of reactants.

(b) Energy per nucleon Increases. (q) E  decay

(c) Conservation of atomic mass number. (r) Nuclear Fission

(d) Conservation of charge. (s) Nuclear Fusion

500

(A) (a ) o ( p, q, r , s), (b) o ( p, q ), (c) o (r , s), (d ) o ( p, q, r , s )
(B) (a ) o (q ), (b ) o (r ), (c ) o ( p ), ( d ) o ( s )

(C) (a ) o ( s ), (b ) o (r ), (c ) o ( q ), (d ) o ( p )

(D) (a ) o ( p ), (b) o ( s ), (c) o (q ), (d ) o ( r )
(149)
column-1 column-2

(a) Nuclear - Fusion (p) Some matter converts into energy

(b) Nuclear - Fission (q) Generally, possible for the nucleus
with less atomic number

(c) E  decay (r) Generally, possible for the nucleus
with high atomic number

(d) Endo thermic reaction (s) Possible due to weak nuclear force

(A) (a ) o ( p, r ), (b) o (q, r ), (c) o ( p, q, s ), ( d ) o ( r , s )
(B) (a ) o (q , s ), (b ) o ( p , s ), (c ) o (q , r , s ), (d ) o ( p, q , r , s )

(C) (a ) o ( s ), (b ) o ( s, p ), (c ) o ( p, q , r , s ), (d ) o (r )

(D) (a ) o ( p, q ), (b) o ( p, r ), (c ) o ( p, s ), (d ) o ( p, q, r )
(150)
column-1 column-2
(a) Transition between two energy levels of an atom. (p) Characteristics of X-ray
(b) emission of electron from the matter. (q) Photo-electric effect
(c) Moseley’s law. (r) Hydrogen spectrum
(d) Conversion of energy of photon in to energy of an (s) E  decay
electron

(A) (a ) o ( p, r ), (b) o ( p, q, s ), (c) o ( p ), (d ) o ( q )
(B) (a ) o ( p ), (b ) o ( s ), (c ) o ( r ), (d ) o ( q, s )

(C) (a ) o ( p, q, r , s ), (b) o ( s ), (c ) o ( p ), ( d ) o ( s )

(D) (a ) o ( s, r ), (b) o (r ), (c) o ( s, p ), (d ) o ( q, r )
Ans. : 148 (A), 149 (D), 150 (A)

l

501