1

Assignment -1 : Basic Properties of Nuclei

GATE Previous Year's Questions
1. The mean momentum of a nucleon in a nucleus with mass number A varies as
[GATE 2000]
(a) A (b) A2 (c) A2/3 (d) A1/3

2. The masses of hydrogen atom, neutron and 238 are given by 1.0078, 1.0087 and
92 U

238.0508 a.m.u., respectively. The binding energy of 238 is therefore approximately

92 U
equal to : (Taking i a. m.u. = 931.64 MeV) [GATE 2000 & 03]
(a) 120 MeV (b) 1500 MeV (c) 1600 MeV (d) 1800 MeV
3. The volume ofa nucleus in an atom is proportional to the : [GATE 2004]
(a) mass number (b) proton number (c) neutron number (d) electron number
56
4. As one moves along the line of stability from Fe to 235 U nucleus, the nuclear binding
energy per particle decreases from about 8.8 MeV to 7.6 MeV. This trend is mainly due
to the: [GATE 2004]
(a) short range nature of the nuclear forces
(b) long range nature of the coulomb forces
(c) tensor nature of the nuclear forces
(d) spin dependence oft he nuclear forces
5. The order of margnitude of the binding energy per nucleon in a nucleus is: [GATE 2006]

(a) 105 MeV (b) 103 MeV (c) 1.0MeV (d) 10MeV
6. The experimentally measured spin g-factors of proton and neutron indicate that :
(a) both proton and neutron are elementary point particles [GATE 2006]
(b) both proton and neutron are not elementary point particles
(c) while proton is an elementary point particle, neutron isnot
(d) while neutron is an elementary point particle, proton is not
4 3
7. An 16 is spherical and has a charge radius R and a volume V SR .
8 O nucleus 3
According to empirical observations of the charge radii the volume of the 128 nucleus
54 Xe
assumed to be spherical is : [GATE 2008]
(a) 8V (b) 2V (c) 6.75V (d) 1.89V
15
8. A particle is confined within a spherical region of radius one femtometer 10 m . Its
momentum can be expressed to be about : [GATE 2010]

keV keV MeV MeV

(a) 20 (b) 200 (c) 200 (d) 2
c c c c

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9. The mean kinetic energy of a nucleon in a nucleus of atomic weight A varies as An, where
n is ____(upto two decimal places) [GATE 2015]

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

8.(c) 9.(–0.66 to –0.68)
CSIR-UGC-NET Previous Year's Questions

1. The radius of a 64
29 Cu
nucleus is measured to be 4.8 u 1013 cm . [NET June 2011]

(A) The radius of a 27 nucleus can be estimated to be

12 Mg

(a) 2.86 u 1013 cm (b) 5.2 u1013 cm (c) 3.6 u1013 cm (d) 8.6 u 1013 cm
(B) The root-mean-square (rms) energy of a nucleon in a nucleus of atomic number A in
its ground state varies as :
(a) A4/3 (b) A1/3 (c) A1/3 (d) A2/3

2. The intrinsic electrric dipole moment of a nucleus A [NET Dec. 2013]

Z X

(a) increases with Z, but independent of A (b)decreases with Z but independent of A

(c) is always zero (d) increases with Z and A

1.(A)(c) (B)(c) 2.(c)

TIFR Previous Year's Questions
16
1. A proton is accelerated to a high energy E and shot at a nucleus of Oxygen 8 O . In
order to penetrate the Coulomb barrier and reach the surface of the Oxygen nucleus, E
must beat least : [TIFR 2012]
(a) 3.6 MeV (b) 1.8 MeV (c) 45 keV (d) 180 eV

1.(a)
JEST Previous Year's Questions

1
1. The stable nucleus that has the radius of 189
3 Os nucleus is [JEST 2016]

(a) 7 (b) 16 (c) 4 (d) 14

Li O He N

1.(a)

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Other Examinations Previous Year's Questions

1. If M e , M p and M H are the rest masses of electron, proton and hydrogen atom in the
ground state (with energy –13.6 eV), respectively, which of the following is exactly true?
(c is the speed of light in free space)

13.6eV
(a) M H M p  Me (b) M H M p  Me 
c2

13.6eV
(c) M H M p  Me 
c2

13.6eV
(d) M H M p  M e  K , where K z r or zero
c2

2. A neutron of mass mn 10 27 kg is moving inside a nucleus. Assume the nucleus to be a

cubical box of size 1014 m with impenetrable walls. Take h 1034 Js and 1 MeV

| 1013 J . An estimate of the energy in MeV of the neutron is:

1 1
(a) 80 MeV (b) MeV (c) 8 MeV (d) MeV
8 80
3. The following histogram represents the binding energy per particle (B.E./A) in MeV as a
function of the mass number A of a nucleus.

A nucleus with mass number A = 180 fissions into two nuclei of equal masses. In the
process :
(a) 180 MeV of energy is released (b) 180 MeV of energy is absorbed
(c) 360 MeV of energy is released (d) 360 MeV of energy is absorbed

4. A nucleus has a size of 1015 m . Consider an electron bound within a nuclens. The estimated
energy of this electron is of the order of :

(a) 1 MeV (b) 102 MeV (c) 104 MeV (d) 106 MeV

5. Two spherical nuclei have mass numbers 216 and 64 with their radii R1 and R2 respectively..
R1
The ratio R is :
2
(a) 1.0 (b) 1.5 (c) 2.0 (d) 2.5

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6. According to the uncertainty principle, the kinetic energy of an electron confined to a

spherical region of volume 1033 m3 is of the order of :

(a) 10 10 J (b) 10 12 J (c) 10 14 J (d) 10 16 J

7. The radius of the nucleus of the Ra atom, which carries an electric charge +88e, is
7.0 u 1015 m . What shouid roughly be the speed of a proton, if it has to reach as close as
1.0 u1014 m from the centre of the nucleus? [The radius of the cloud of orbital electrons
of the Ra atom is approximately 5.0 u1011 m ]

(a) 6.7 u 109 m s (b) 3.1 u 108 m s (c) 1.4 u 105 m s (d) 4.9 u 107 m s

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

Assignment-2 : Nuclear Models
GATE Previous Year's Questions

1. The angular momentum and parity of 17 nucleus, according to the nuclear shell model
8 O
(including spin-orbit coupling), is: [GATE 1998]

1 3 5
(a) 0 (b) (c) (d)
2 2 2
2. The asymmetry terms in the Weizsacker semi-empirical mass formula is because of:
(a) non-spherical shape of the nucleus [GATE 2002]
(b) non-zero spin of nucleus
(c) unequal number of protons and neutrons inside the nucleus
(d) odd number of protons inside the nucleus

3. The nuclear spins of 6 C14 and 12 Mg

25 nuclei are respectively,, [GATE 2002]

(a) zero and half integer (b) half integer and zero
(c) an integet and half integer (d) both half integer

4. The spin and parity of 4 Be 9 nucleus as predicted by the sheel model are respectively..

(a) 3/2 and odd (b) 1/2 and odd (c) 3/2 and even (d) 1/2 and even
Common Data for Q.5 and Q.6. :

The nucleus 44
Ca can be described by the single particle shell model.
5. The single particle states occupied by the last proton and the last proton and the last
nuetron, respectively are given by [GATE 2004]
(a) d 5/2 and f 7/2 (b) d 3/2 and f 5/2 (c) d 5/2 and f 5/2 (d) d3/2 and f 7/2

6. The ground state angular momentum and parity of 41

Ca are : [GATE 2004]

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§7· 3 5 5
(a) ¨ ¸ (b) (c) (d)
©2¹ 2 2 2
7. Which o f the following expressions for total binding energy B of a nucleus is correct
a1 , a2 , a3 , a4 ! 0 ? [GATE 2005]

2
2/3 Z Z 1 A  2Z
(a) B a1 A  a2 A  a3  a4 G
A1/3 A

2
2/3 Z Z 1 A  2Z
(b) B a1 A  a2 A  a3  a4 G
A1/3 A

2
1/3 Z Z 1 A  2Z
(c) B a1 A  a2 A  a3  a4 G
A1/3 A

2
Z Z 1 A  2Z
(d) B a1 A  a2 A1/3  a3  a4 G
A1/3 A

8. Ccording to shell model, the ground state of 15 nucleus is : [GATE 2005]

8 O

3 1 3 1
(a) (b) (c) (d)
2 2 2 2

9. According to the shell model the ground state spin of the 17

O nucleus is : [GATE 2007]

3 5 3 5
(a) (b) (c) (d)
2 2 2 2
10. The following gives a list of pairs containing :
(i) a nucleus (ii) one of its properties
Find the pair which is in appropriate.

(a) (i) 10 Ne 20 nucleus, (ii) stable nucleus;

(b) (i) a spheroidal nucleus, (ii) an electric quadrupole moment;

1
(c) (i) 8 O16 nucleus, (ii) nucleus spin J
2

(d) (i) U 238 nucleus, (ii) binding energy =1785 MeV, (approx)

11. The four possible configurations of neutrons in the ground state of 4 Be 9 nucleus according
to the shell model, and the associated nuclear spin are listed below. Choose the correct
one. [GATE 2008]

1 2 1 3 3 1 2 1 2 1 1 3
(a) s1/2 p3/2 ;J (b) s1/2 p1/2 p3/2 ; J
2 2

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2 2 1 2 2 1 1
1 1 1 1 1
(c) s1/2 p3/2 ;J (d) s1/2 p3/2 p1/2 ; J
2 2

12. The mass difference between the pair of mirror nuclei 6 C11 and 5B
11 is given to be

'MeV / c 2 . According to the semi empirical mass formula the mass difference between
the pair of mirror nuclei 17 and 20 will approximately be (rest mass of proton
9 F 2 O

mp 938.27 MeV c 2 and rest mass of neutron mu 939.57 MeV c 2 ). [GATE 2008]

(a) 1.39' MeV c 2 (b) 1.39'  0.5 MeV c 2

(c) 1.86' MeV c 2 (d) 1.6'  0.87 MeV c 2

13. Consider the following expression for the mass of a nucleus with Z protons and A nucleons.

1 ª
M A, Z 2 ¬
f A  yZ  zZ 2 º
c ¼

Here, f A is a function of A. [GATE 2009]

y  4a A

z ac A1/3  4a A A 1

a A and ac are constants of suitable dimensions. For a fixed A, the expression of Z for
the most stable nucleus is :

A/2 A/2
Z Z
ª § ac · 2/3 º ª § ac · 2/3 º
(a) «1  ¨ ¸A » (b) «1  ¨ ¸A »
¬ © aA ¹ ¼ ¬ © 4a A ¹ ¼

A/ 2
Z A
ª § ac · 2/3 º Z
(c) «1  ¨ ¸A » (d) 1  A2/3
¬ © 4a A ¹ ¼

14. Consider a nucleus with N neutrons and Z protons. If m p , mn and B.E. represents the
mass of the proton, the mass of the neutron and the binding energy of the nucleus
respectively and c is the velocity of light in free space, the mass of the nucleus is given by:
[GATE 2009]

(a) Nmn  Zm p (b) Nm p  Zmn

B.E. B.E.
(c) Nmn  Zm p  2 (d) Nm p  Zmn 
c c2

15. In the nuclear shell model the spin parity of 15 is given by : [GATE 2010]
7 N

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1 1 3 3
(a) (b) (c) (d)
2 2 2 2

16. The first three energy level of 238

Th90 are shown below : [GATE 2010]

The expected spin-parity and energy of the next level are given by :
   
(a) 6 ;400keV (b) 6 ;300keV (c) 2 ;400keV (d) 4 ;300keV

17. The semi-empirical mass formula for the binding energy of nucleus contains a surface
correction term. This term depends on the mass number A of the nucleus as : [GATE
2011]
(a) A1/3 (b) A1/3 (c) A2/3 (d) A
18. According to the single particle nuclear shell model, the spin parity of the ground state of
17 is : [GATE 2011]
8 O

1 3 3 5
(a) (b) (c) (d)
2 2 2 2

19. Total binding energies of O15 , O16 and O17 are 111.96 MeV, 127.62 MeV and 131.76
1 1
MeV, respectively. The energy gap between p 1 and d 3 neutron shells for the nuclei
2 2
whose mass number is close to 16 is:
(a) 4.1 MeV (b) 11.5 MeV (c) 15.7 MeV (d) 19.8 MeV

2 q
20. The electromagnetic form factor F q
2
of a nucleus is given by F q exp  .
2Q 2

4S f
q ³0
2
where Q is a constant. Given that F q rdr .U r sin qr [GATE 2013]

³d
2
rU r 1

where U r is the charge density, the root mean square radius of the nucleus is given by

(a) 1 / Q (b) 2 /Q (c) 3/Q (d) 6 /Q

21. A proton is confined to a cubic box, whose sides have length 1012 m . What is the minimum
kinetic energy of the proton? The mass of proton is 1.67 u 1027 kg and Planck’s constant
is 6.63 u 1034 Js . [GATE 2013]

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(a) 1.1 u 1012 J (b) 3.3 u1017 J (c) 9.9 u1017 J (d) 6.6 u 1017 J
Statement for Linked Answer Q. 22 and Q. 23 :
In the Schmidt model of nuclear magnetic moments, we have,

G eh G G
P gi l  g s S
2 Mc
where the symbols have their usual meaning
22. For the case J l  1 / 2 , where J is the total angular momentum, the expectation value
G G
of S ˜ J in the nuclear ground state is equal to, [GATE 2011]

(a) J  1 / 2 (b) J  1 / 2 (c) J / 2 (d)  J / 2

23. For the O17 nucleus ( A 17, Z 8 ) the effective magnetic moment is given by :

eh
Peff gJ
2 Mc
where g is equal to, ( g s 5.59 for proton and = 3.83 for neutron) [GATE 2013]
(a) 1.12 (c) –0.77 (c) –1.28 (d) 1.28

1 G G
24. In the nuclear shell model, the pontential is modeled as V r mZ2 r 2  OL ˜ S , O ! 0 .
2
The parity and isospin assignments fot the ground state of 13
C is : [GATE 2015]

1 1 1 1 3 1 3 1
(a) , (b) , (c) , (d) ,
2 2 2 2 2 2 2 2
25. According to the nuclear shell model the respective ground state spin parity values of
15 and 17 nuclei are [GATE 2016]
8 O 8 O

1 1 1 5 3 5 3 1
(a) , (b) , (c) , (d) ,
2 2 2 2 2 2 2 2

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

8.(d) 9.(b) 10.(c) 11.(a) 12.(b) 13.(c) 14.(c)
15.(a) 16.(a) 17.(c) 18.(d) 19.(b) 20.(c) 21.(c)
22.(b) 23.(b) 24.(a) 25.(b)

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CSIR-UGC-NET Previous Year's Questions

1. The difference in the Coulomb energy between the mirror nuclei 49 and 49 is 6.0
24 Cr 25 Mn
MeV. Assuming that the nuclei have a spherically symmetric charge distribution and that
49
e2 is approximately 1.0 MeV-fm, the radius of the 25 Mn
nucleus is : [NET Dec. 2011]

(a) 4.9 u 1013 m (b) 4.9 u 1015 m (c) 5.1 u 1013 m (d) 5.1 u 1015 m

2. According to the shell model the spin and parity of the two nuclei 125 and 39 are,
51 Sb 38 Sr
respectiyely. [NET Dec. 2011]
       
§5· §5· §5· §7· §7· §5· §7· §7·
(a) ¨ ¸ and ¨ ¸ (b) ¨ ¸ and ¨ ¸ (c) ¨ ¸ and ¨ ¸ (d) ¨ ¸ and ¨ ¸
©2¹ ©2¹ ©2¹ ©2¹ ©2¹ ©2¹ ©2¹ ©2¹

3. The single particle enertgy difference between the p-orbitals (i.e. p3 2 and p 12 ) of the

nucleus 114 is 3 MeV. The energy difference between the states in its 1f orbital is :
50 Sn

[NET Dec. 2012]

(a) –7 MeV (b) 7 MeV (c) 5 MeV (d) –5 MeV
4. The binding energy of a light nucleus (Z, A) in MeV is given by the approximate formula
2
3 N Z
B A, Z | 16 A  20 A2/3  Z 2 A1/3  30 [NET Dec. 2013]
4 A
where N A  Z is the neutron number. The value of Z of the most stable isobar for a
given A is :
1 1
A§ A /3 · A A§ A /3 · A§ A4/3 ·
¨
(a) ¨ 1  ¸ (b) ¨
(c) ¨ 1  ¸ ¨
(d) 2 ¨ 1  ¸
2© 160 ¹¸ 2 2© 120 ¹¸ © 64 ¹¸
5. According to the shell mode the total angular momentum (in units of h) and the parity of
the ground state of the 37 Li nucleus is : [NET Dec. 2013]

3 3
(a) with negative parity (b) with positive parity
2 2

1 3
(c) with positive parity (d) with positive parity
2 2
6. A permanently deformed even-even nucleus with J P 2 has rotational energy 93 keV..
The energy of the next excited state is : [NET June 2014]
(a) 372 keV (b) 310 keV (c) 273 keV (d) 186 keV
7. If the binding energy B of a nucleus (mass number A and charge Z) is given by :
2
2Z  A ac Z 2
B aV A  aS A2/3  asym  where aV 16 MeV , aS 16MeV , asym
A A1/3
24MeV and ac 0.75MeV , then the Z for the most stable isobar for a nucleus with A =
216 is : [NET Dec. 2014]

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(a) 68 (b) 72 (c) 84 (d) 92

8. Let us approximate the nuclear potential in the shell model by a 3-dimensional isotropic
harmonic oscillator. Since the lowest two energy levels have angular momenta l 0 and
l 1 respectively, which of the following two nuclei have magic numbers of protons and
neutrons ? [NET June 2015]

4 16 2 8 4 8 4 12
(a) He and O (b) D and Be (c) He and Be (d) He and Be
2 8 1 4 2 4 2 6
9. Of the nuclei of mass number A 125 , the binding energy calculated from the liquid drop
model (given that the coefficients for the Coulomb and the asymmetry energy are
aC 0.7 MeV and asym 22.5MeV respectively) is a maximum for : [NET Dec. 2015]

(a) 125 (b) 125 (c) 125 (d) 125

54 Xe 53 I 52 Te 54 Sb

2 j 1 2
10. The electric quadrupole moment of an odd proton nucleus is 2 j  1 r , where j is the

total angular momentum. Given that R0 1.2 fm , what is the value, in beam, of the
quadrupole moment of the 27
AI nucleus in the shell model? [NET Dec. 2015]
(a) 0.043 (b) 0.023 (c) 0.915 (d) 0
11. Let ES denote the contribution of the surface energy per nucleon in the liquid drop model.
27 64
The ratio ES 15 Al : ES 30 Zn is : [NET June 2016]

(a) 2 : 3 (b) 4 : 3 (c) 5 : 3 (d) 3 : 2

12. 27
According to the shell model the nuclear magnetic moment of the 13 Al nucleus is (Given
that for a proton gl 1, g s 5.586 and for a nuetron gl 0, g s 3.826 ). [NET

(a) 1.913P N (b) 14.414P N (c) 4.793P N (d) 0 June 2016]

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

8.(a) 9.(c) 10.(a) 11.(b) 12.(c)
TIFR Previous Year's Questions
1. In the semi-empirical mass formula, the volume (V), surface (S), coulomb (C), and pairing
(P) contributions to the binding energy of a nucleus A vary with mass number A as :
Z X

(a) V v A, S v A2/3 , C v A1/3 , P v A3/4 [TIFR 2015]

(b) V v A, S v A1/3 , C v A1/3 , P v A3/4

(c) V v A, S v A2/3 , C v A1/3 , P v A3/4

(d) V v A2 , S v A2/3 , C v A1/3 , P v A3/4

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2. The Weizacker semi-empirical mass formula for an odd nucleus with Z protons and A
nucleons may be written as M Z , A D1 A  D2 A2/3  D3 A  D 4 Z 2 where the D , are
constants independent of Z, A. For a given A, if Z A is the number of photons of the most
stable isobar, the total energy released when an unstable nuclide undergoes a single E 
decay to Z A , A is : [FIER 2016]

(a) D3 (b) D 4 (c) D 4  D3 (d) D1  D 2

1.(a) 2.(b)
Other Examinations Previous Year's Questions
1. A nucleus may be modelled as a drop of liquid consisting of the nucleons (protons and
neutrons). In this model, the dominant contribution to the nuclear binding energy is from
the volume, which is proportional to A, the total number of nucleons. Then the two important
subdominant contributions from the surface tension and the coulomb repulsion of the
proton S are, proportional to

(a) A2/3 and Z / A1/3 respectively (b) A2/3 and Z 2 / A1/3 respectively

(c) A1/3 and Z 2 / A2/3 respectively (d) A1/2 and Z 2 / A1/3 respectively

1.(b)
GATE Previous Year's Question
1. A nucleus having mass number 240 decays by a emission to the grond state of its daughter
nucleus. The Q value of the process is 5.26 MeV. The energy (in MeV) of the a particle
is: [GATE 2005]
(a) 5.26 (b) 5.17 (c) 5.13 (d) 5.09
2. The plot of log At is versus time t, where At is activity, as shown in the figure, corresponds
to decay : [GATE 2005]

(a) from only one Idnd of radioactive nuclei having same half-life
(b) from only neutron activated nuclei
(c) from a mixture of radioactive nuclei having different half-lives
(d) which is unphysical

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3. The evidence for the non-conservation of parity in E  decay has been obtained from the
observation that the E  intensity.. [GATE 2005]
(a) anti-parallel to the nuclear spin directions is same as that a along the nuclear spin in
direction
(b) anti-parallel to the nuclear spin direction is not the, same as that along the nuclear spin
directions
(c) shows a continuous distribution is a function of momontum
(d) is independent of the nuclear spin direction

4. Which one of the following disintegration series of the heavy elements will give 209
Bi as
a stable nucleus? [GATE 2006]
(a) Thorium series (b) Neptunium series (c) Uranium series (d) Actinu series
5. Consider Fermi theory of E  decay :
(I) The number of final state of electrons corresponding to momenta between p and
p  dp is

(a) independent of p (b) proportion to pdp

(c) proportional to p 2 dp (d) proportional to p 3dp

(II) The number of emitted electrons with momentum (p) and energy (E) in the allowed
approximation is proportional to ( E0 is the total energy given up by the necleus)
2
(a) E0  E (b) p E0  E (c) p 2 E0  E (d) p E0  E

6. Half-life of a radioisotope is 4 u 108 years. If these are 103 radioactive nuclei in a sample
today, the number of such nuclei in the sample 4 u 109 years ago were. [GATE 2007]

(a) 128 u 103 (b) 256 u 103 (c) 512 u 103 (d) 1024 u 103
7. Fission fragments are generally radioactive as : [GATE 2007]
(a) they have excess in neutrons
(b) they have excess of protons
(c) they are products of radioactive nuclides
(d) their total kinetic energy is of the order of 200 MeV
Statement for Linked Answer Q. 8 and Q. 9 :
Consider the E  decay of a free neutron at rest in the laboratory
8. Which of the following configurations of the decay products correspond to the largest
energy of the anti-neutrino vG ? (rest mass of electron, me 0.51MeV / c 2 , rest mass of

proton m p 938.27 MeV / c 2 and rest mass of neutron mn 939.57 MeV / c 2 )

(a) In the laboratory, proton is produced at rest [GATE 2008]

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(b) In the laboratory, momenta of proton, electron and the anti-neutrino all have the same
magnitude
(c) In the laboratory, proton and electron fly-o ff with (nearly) equal and opposite momenta
(d) In the laboratory, electron is produced at rest
9. Using the result of above problem, answer the following. Which of the following represents
approximately the maximum allowed energy of the anti-neutrino vG ? [GATE 2008]
(a) 1.3 MeV (b) 0.8 MeV (c) 0.5 MeV (d) 2.0 MeV
10. The disintegration energy is defined to be the difference in the rest energy between the
initial and final states. Consider the following process: [GATE 2009]
240 236
94 Pu o92 U  24 He
The emitted D particle has a kinetic energy 5.17 MeV. The value of the disintegration
energy is :
(a) 5.26 MeV (b) 5.17 MeV (c) 5.08 MeV (d) 2.59 MeV
G
11. In the E  decay of neutron, n o p  e  v , the anti-neutino v1e escapes detection.
Itsexistence is inferred from the measurement of :
[GATE 2011]
(a) energy distribution of electrons (b) angular distribution of electrons
(c) helicity distribution of electrons
(d) forward backward asymmetry of electrons

12. In the E  decay process, the transition 2 o 3 , is : [GATE 2013]

(a) allowed both by Fermi and Gamow-Teller selection rule
(b) allowed by Fermi and but not by Gamow Teller selection rule
(c) not allowed by Fermi but allowed by Gamow teller selection rule
(d) not allowed both by Fermi and Gamow Teller selection rule
13. A nucleus X undergoes a first forbidden E  decay to a nucleus Y. If the angular momentum
T
(I) and parity (P), denoted by I P as for X when of the following is a possible I P
2
value for Y? [GATE 2014]

1 1 3 3
(a) (b) (c) (d)
2 2 2 2

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

8.(c) 9.(a) 10.(a) 11.(a) 12.(c) 13.(c)

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CSIR-UGC-NET Previous Year's Questions

1
1. The ground state of 207
82 Pb
nucleus has spin-parity J p , while the first excited state
2

5
has J p . The electromagnetic radiation emitted when the nucleus makes a transition
2
from the first excited state to the around state are : [NET June 2012]
(a) E2 and E3 (b) M2 and E3 (c) E2 and M3 (d) M2 and M3
2. A radioactive element X decays to Y, which in turn decays to a stable element Z. The
decay constant from X to Y is O1 and that from Y to Z is O 2 . If to begin with, there are
only N 0 atoms of X, at short times ( t  1 / O1 as well as 1 / O 2 ) the number of atoms of
Z will be : [NET June 2016]

1 O1O 2
(a) O1O 2 N0t 2 (b) 2 O  O N 0t (c) O1  O 2 2
N 0t 2 (d) O1  O 2 N 0 t
2 1 2

1.(c) 2.(a)
TIFR Previous Year's Questions
1. A detector is used to count the numbr of J -rays emitted by a radioactive source. If the
number of counts recorded in exactly 20 seconds is 10000, the error in the counting rate
per second is : [TIFR 2010]
(a) r5.0 (b) r22.4 (c) r44.7 (d) r220.0

2. A lead container contains 1 gm of a 60 radioactive source. It is known that a 60

27 Co 27 Co
nucleus emits a E -particle of energy 316 KeV followed by two J -emissions of energy
1173 and 1333 KeV respectively. Which of the following experimental methods would be
the best way to determine the life time of the 60 source? [TIFR 2010]
27 Co

(a) Measure the change in temperature of the source

(b) Measure the weight of the source now and again after one year
(c) Measure the recoil momentum ofthe nucleus during E -emission
(d) Measure the number of J - photons emitted by this source
3. An excited atomic electron undergoes a spontaneous transition 3d3/2 o 2 p1/2 . The
interaction responsible for this transition must be of the type : [TIFR 2011]
(a) electric dipole (E1) OR magnetic dipole (M2)
(b) eletric dipole (E1) OR magnetic dipole (M1)
(c) eletric dipole (E2) OR magnetic dipole (M2)
(d) eletric quadrupole (E2) OR magnetic dipole (M1)

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4. A cloud Chamber of width 0.01m is filled with pure nitrogen gas N 2 at normal
temperature and pressure. A beam of D  particles, when incident normally no the chamber
4
make tracks which are visible under strong illumination. Whenever as D  particle 2 He

has a nuclear collision with a 14

7 N
nucleus, the track shows a distinct bend. The radius of
a nucleus move at non relativistic speeds and the total number of incident D particles is
107 , the number of such distinct bends is approximately..
(a) 100 (b) 200 (c) 300 (d) 400
(e) 500 (f) 600

5. Let E N be the energy released when one mole of pure 235

U undergoes controlled
fission, and EC be the energy released when one mole o f pure carbon undergoes complete
combustion. The ratio EN / EC will have the order of magnitude [TIFR 2013]

(a) 104 (b) 108 (c) 109 (d) 106

6. A standard radioactive source is known to decay by emission of J -rays. The source is
provided to a student in a thick sealed capsule of unbreakable plastic and she is asked to
find out the half-life. Which of the following would be the most useful advice to the
student? [TIFR 2014]
(a) The half-life cannot be measured because the initial concentration of the source is not
given
(b) Mount the source in front of a gamma ray detector and count the number of photons
detected in one hour
(c) Measure the mass of the source at different times with an accurate balance having a
least count of 1 mg. Plot these values on a curve and fit it with an exponential decay law
(d) Mount the source in front of a gamma ray detector and count the number of photons
detected in a specific time interval. Repeat this experiment at different times and note
how the count changes
197
7. An alpha particle of energy E is shot towards a gold nucleus 29 Au . At distances much

larger than the nuclear size RN , the dominant force is the Coulomb repulsion, but at
distances comparable to the nuclear size the dominant force is the strong nuclear attraction.
These combine to form a potential barrier of height VC . If E  VC the probability that the
alpha particle will fuse with the gold nucleus can be written (in terms of a dimensionless
positive constant k) as [TIFR 2014]

kE § E · § kVC ·
(a) zero (b) 2 2 (c) k ¨1  V ¸ (d) exp ¨ ¸
k E  VC2 © C ¹ © E ¹

8. Which of the following radio active decay chains is it possible to observe? [TIFR 2015]
206 202 202 210 210 206
(a) 82 Pb o80 Hg o79 Au (b) 83 Bi o84 Po o82 Pb

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214 210 207 206 202 202

(c) 88 Ra o86 Rn o82 Pb (d) 82 Pb o80 Hg o79 Au

9. In an experiment, 192 nuclei were bombarded with neutrons leading to formation of

79 Au
198 , which is un-stable. The half-life of 198 was measured to be 2.25 days and it
79 Au 79 Au
was found later that this measured half-life was an underestimate by 10%. The
corresponding percentage error in the estimated population of 198 after 9 days is :
79 Au

(a) 10% (d) 15% (c) 2.5% (d) 1% [TIER 2015]

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

8.(b) 9.(b)
JEST Previous Year's Questions

138
1. U decays with a half-life of 4.51u 109 years, the decay series eventually ending at
206
Pb , which is stable. A rock smaple analysis shows that the ratio o f the numbers
ofatoms of 206 238 206
Pb to U is 0.0058. Assuming that all the Pb has been produced by
the decay of 238
U and that all other haif lives in the chain are negligible, the age of rock
sample is : [JEST 2013]

(a) 38 u 106 years (b) 48 u 106 years (c) 38 u 107 years (d) 48 u 107 years

2. In the mixture of isotopes normally found on the earth at the present time, 238 has on a
92 U

boundance of 99.3% and 235 has an abound ance of 0.7%. The measured lifetimes of
92 U

these isotopes are 6.52 u 109 years and 1.02 u 109 years respectively. Assuming that
they were equally aboundant when the earth was formed, the estimated age of the earth,
in years is : [JEST 2014]

(a) 6.0 u 109 (b) 1.0 u 109 (c) 6.0 u108 (d) 1.0 u 108
3. The half-life of a radioactive nuclear source is 9 days. The fraction of nuclei which are
left undercayed after 3 days is:

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

1.(a) 2.(a) 3.(d)

Other Exmainations Previous Year's Questions
1. The activity of a radioactive sample is decreased to 75% of the initial value after 30 days.
The half-life (in days) of the sample is approximately [You may use ln 3 = 1.1, In 4 = 1.4]

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(a) 38 (b) 5 (c) 59 (d) 69

2. 60 is a radioactive nucleus of half-1ife 2An2 u 108 s . The activity of 10 g 60 in

27 Co 27 Co
disintegration per second is :

1 1
(a) u 1010 (b) 5 u 1010 (c) u 1014 (d) 5 u 1014
5 5
3. The variation of binding energy per nucleus with respect to the mass of nuclei is shown in
the figure

Consider the following reactions:

(1) 238
92 U
206
o32 Pb  10 p  22 n (2) 238
92 U
206
o82 Pb  842 He  6e 

Which one of the following statements is true for the given decay modes of 238 ?
92 U

(a) both 1 and 2 are allowed (b) both 1 and 2 are forbidden
(c) 1 is forbidden and 2 is allowed (d) 1 is allowed and 2 is forbidden

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4. A particular radioisotope has a half-life of 5 days. In 15 days the probability of decay in

percentage will be ________

5. The radioactivity of a sample ot Co55 decreases by 4% every hour. (The decayproduct is

not radioactive). The half-life of Co55 is approximately :
(a) 1 hour (b) 1 day (c) 1 month (d) 1 year
6. In order to determine the.age. of ancient wooden tools, radiocarbon dating method is
used. This is done by measuring the fraction of radioactive iotopee 14 C of carbon compared
to the normal (non-radioactive) isotope 12
C in a sample. An old sample is found to
contain 1/10 times the fraction of 14
C as compared to a fresh piece of wood. Given that
the half-life of 14
C is 5570 years, the approximate age of the old sample is :
(a) 557 years (b) 12800 years (c) 18500 years (d) 55700 years

7. A radioactive nucleus X decays to Y with a mean lifetime W 2 W1 / 2 . If N 0 nuclei of X

(but no nuclei of Y) are present at t 0 , how many nuclei of Y will be there when the
number of X nuclei becomes N 0 / 2 ?

(a) N 0 / 2e (b) N 0 / 4 (c) N 0 / 2 (d) N 0 / e

8. Radon has a of 3.8 days. If we start with 10.24 gm of radon, the amount of it which will
be left after 38 days is :

(a) 10 4 gm (b) 10 2 gm (c) 10 6 gm (d) 10 3 gm

9. If the age of the universe is 1010 years, the humans have existed for 106 years. If we take
the age of the universe to be a day, how many seconds have the humans existed?
(a) 2.40 sec (b) 86.40 sec (c) 8.64 sec (d) 24 sec

1.(d) 2.(d) 3.(c) 4.(87 to 88) 5.(b) 6.(c) 7.(b)

8.(d) 9.(c)
GATE Previous Year's Questions
1. Typical energies released in a nuclear fission and a nuclear fusion reactions are respectively.
(a) 50 MeV and 1000 MeV (b) 200 MeV and 1000 MeV
(c) 1000 MeV and 50 MeV (d) 200 MeV and 10 MeV [GATE 2002]
235
2. An atomic bomb consisting of U explodes and releases an energy of 1014 J . It is
known that each 235
U which undergoes fission releases 3 neutrons and 200 MeV of
energy. Further, only 20% of the 235
U atoms in the bomb undergoes fission:
(i) the total number of neutrons released is about [GATE 2003]

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(a) 4.7 u 1024 (b) 9.7 u 1024 (c) 1.9 u 1025 (d) 3.7 u 1025

(ii) The mass of 235

U in the bomb is about :
(a) 1.5 kg (b) 3.0 kg (c) 6.1 kg (d) 12 kg

3. A thermal neutron having soeed v impinges on a U 235 nucleus. The reaction moss section
is proportionl to : [GATE 2004]

(a) v 1 (b) v (c) v1/2 (d) v1/2

4. The threshold temperature above which the thermonuclear reaction [GATE 2005]
3 3
2 He  2 He o 42 He  211 H  12.86MeV

can occur is (use e 2 / 4 SH 0 1.44 u 109 K e 2 / 4 SH 0 1.44 u 1015 MeV in)

(a) 1.28 u 1010 K (b) 1.28 u 105 K (c) 1.28 u 103 K (d) 1.28 u 107 K

5. By capturing an electron 54 transforms into 54 releasing [GATE 2006]

25 Mn29 24 Cr30

(a) a neutrino (b) an anti-neutrino (c) and D particle (d) a positron

6. Fission fragments are generally radioactive as : [GATE 2007]
(a) they have excess of neutrons (b) they have excess of protons
(c) they are products of radioactive nuclide
(d) their total kinetic energy is of the order of 200 MeV
7. A neutron scatters elastically from a heavy nucleus, the initial and final states of the
neutron have the : [GATE 2007]
(a) same energy (b) same energy and linear momentum
(c) same energy and angular momentum (d) same linear and angular momenta
Linked Answer Q.8 and Q.9 :

A 16PA beam of alpha particles, having cross-sectional area 104 m2 , is incident on a

Rhodium target of thickness 1Pm . This produces neutrons through the reaction,

D 100 Rh o101 Pd  3n
8. The number of D particles hitting the target per second is : [GATE 2007]

(a) 0.5 u 1014 (b) 1.0 u 1014 (c) 2.0 u 1020 (d) 4.0 u 1020

9. The neutrons are observed at the rate of 1.806 u 108 s 1 . If the density of rhodium is
approximated as 10 4 kg m 3 the cross-section for the reaction (in barns) is:[GATE 2007]

(a) 0.1 (b) 0.2 (c) 0.4 (d) 0.8

10. The energy released in the fission of 1 kg Uranium (approximately [in Joule]:[GATE 07]

(a) 1014 (b) 1017 (c) 1016 (d) 1010

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11. The atomic masses of 152 152 1

63 Eu ,62 Sm,1 H and neutron are 151.921749, 15019756, 1.007825
and 1.008665 in atomic mass units (a.m.u.), respectively. Using the above information,
152
 n o152 3 a.m.u. (upto three
the Q-value of the reaction 63 Eu 62 Sm  p is_____ u10
decimal places) [GATE 2015]
53 
12. Consider the reaction 25 Mn  e o54
24 Cr  X . The particle X is [GATE 2016]

(a) J (b) ve (c) n (d) S0

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

8.(a) 9.(a) 10.(a) 11.(2.833) 12.(b)
TIFR Previous Year's Questions
235
1. The binding energy per nucleon for U is 7.6 MeV. The 235U nucleus undergoes fission
to produce two fragments, both having-Eirding energy per nucleon 8.5 MeV. The energy
released, in Joules, from the complete fission of 1 kg of 235
U is therefore, [TIFR 2010]
(a) 8000 (b) 1035 (c) 450 (d) 20000

(e) 8.7 u 1013 (f) 5.0 u108

2. A fast-moving 14
N nucleus collides with an a particle at rest in the laboratory frame,
giving rise to the reaction
14
N  D o17 O  p [TIFR 2011]

14
Given the masses 14.00307 a.m.u. and 16.99913 a.m.u. for N and 17 O nuclei
respectively, and 4.00260 a.m.u. and 1.00783 a.m.u. for a and p respectively, the minimum
kinetic energy in the laboratory frame of the 14
N nucleus must be :
(a) 4.20 MeV (b) 1.20 MeV (c) 5.41 MeV (d) 1.55 MeV
239
3. In a nuclear reactor, Plutonium 94 Pu is used as fuel, releasing energy by its fission

146 91
into isotopes of Barium 54 Ba and Strontium 38 Sr through the reaction :

239 1
94 Pu  0 n o146 91 1
56 Ba  38 Sr  3 u0 n [TIER 2012]

The binding energy (B.F.) per nucleon of each of these nuclides is given in the table
below :

239 146 91
Nuclide 94 Pu 54 Ba 38 Sr
B.E.per nucleon MeV 7.6 8.2 8.6

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Using this information, one can estimate the number of such fission reactions per second
in a 100 MW reactor as :

(a) 3.9 u 1018 (b) 7.8 u 1018 (c) 5.2 u 1019 (d) 5.2 u 1018

(e) 8.9 u 1017

4. The process of electron capture p  e o n  ve [TIFR 2013]

takes place at the quark level through the Feynman diagram.

5. A gold foil, having N 0 number of 197 2

Au nuclides per cm , is irradiated by a beam of
thermal neutrons with a flux of F neutrons-cm–2-s–1. As a result, the nuclide 198
Au , with
a half-life W of several years, is produced by the reaction : [TIER 2013]
197
Au  n o198 Au  J

which has a cross section of a Vcm2 . Assuming that the gold foil has 100% abundancy of
197
Au nuclides, the maximum number of 198 Au nuclides that.can accumubte at any time
in the foil is proportional to :

W 1 VF
(a) VWFN 0 (b) N 0 (c) N 0 (d) N 0
VF VWF W
6. In a beta decay experiment, an elecromagnet M and a detector D are used to measure

the energy of electrons E , as shown in the figure. [TIFR 2014]

 
The detector D is capable of detecting either electons E or positrons E . Now the

E  source is replaced with a E source, and we would like to measure the energy of the

positrons E using the same setup. Which of the following is correct?

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(a) This can be done quite easily, if the polarity of current in the coils of the electromagnet
is reversed
(b) This is can done trivially, without changing anything, since the detector D can detect
either E  or E

(c) There is no way to do this with the given set up, since E will have to be converted

into E  , which is obviously not possible

(d) This cannot be done since the magnet does not have a symmetric shape
7. It is well-known that the energy of the Sun arises from the fusion of hydrogen nuclei
(protons) inside the core of the Sun. The takes place through several mechanisms each
resulting in emission of energy. [TIFR 2014]
Which of the following reactions is NOT possible during the proton fusion inside the Sun?

(a) 11 H 11 H o 22 He (b) 12 H 11 H o32 He

(c) 11 H 11 H o12 H  e   ve (d) 11 H 11 H 11 H 11 H o 42 He  2e 

226
8. Consider a process in which atoms of Actinium 226 89 Ac get converted to atoms of

226
Radium-226 88 Ra and the yield of energy is 0.64 MeV per atom. This occurs through
[TIFR 2016]

(a) Both p o n  e   ve and p  e  o n  ve

(b) Both p o n  e   ve and n o p  e  ve

(c) Only p o n  e   ve (d) Only p  e  o n  ve

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

8.(d)
JEST Previous Year's Questions
1. If hydrogen atom is bombarded by energetic electrons, it will emit. [JEST 2014]
(a) K a X  rays (b) E  rays (c) Neutrons (d) None of the above

1.(d)
GATE Previous Year's Questions

1. Identify the reaction which has the same transition probability as S  p o S  p

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(a) S  n o S  n (b) S  p o S0  n [GATE 2000]

(c) S  n o S  n (d) S0  p o S  n

2. Which of the following functions describes the nature of interaction potential V(r) between
two quarks inside a nucleon? (r is the distance between the quarks and a and b are
positive constants of suitable dimensions). [GATE 2001]

a a* a a
(a) V r  br (b) V r  br (c) V r  br (d) V r   br
r r r r
3. Which of the following reactions violates lepton number conservation?

(a) e   e  o ve  vGe (b) e   p o ve  n

(c) e   n o p  ve (d) P  o e   ve  vGe

4. The cross-sections of the reactions : [GATE 2001]

p  S o 6   K  and p   S o 6   K 
at a given energy are the same due to :
(a) baryon number conservation (b) time reversal invariance
(c) charge conjugation (d) parity conservation
5. The Baryon number of proton, the lepton number of proton, the baryon number of electron
and the lepton nunber of the electron are respectively. [GATE 2002]
(a) zero, zero, one and zero (b) one, one, zero and one
(c) one, zero, zero and one (d) zero, one, one and zero

6. The nucleus of the atom 9 Be4 consists of : [GATE 2003]

(a) 13 up quarks and 13 down quarks (b) 13 up quarks and 14 down quarks
(c) 14 up quarks and 13 down quarks (d) 14 up quarks and 14 down quarks
7. Which one of the following nuclear reactions is possible? [GATE 2003]

(a) 14
N 2 o13 C6  E   ve (b) 13
N 7 o13 C6  E   ve

(c) 13
N 7 o13 C6  E (d) 13
N 7 o13 C7  E   ve
8. Suppose that a neutron at rest in free space decays into a proton and an electron. This
process would violate [GATE 2003]
(a) conservation of charge (b) conservation of energy
(c) conservation of linear momentum (d) conservation of angular momentum
9. A stationary particle in free space is oberved to spontaneously decay into two photons.
This imples that : [GATE 2003]
(a) the particle carries electric charge
(b) the spin of the particle must be greater than or equal to 2

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(c) the particle is a boson

(d) the mass of the particle must be greater than or equal to the mass of the hydrogen
atom
10. Choose the particle with zero Baryon number from the list given below : [GATE 2004]
(a) pion (b) neutron (c) proton (d) ' 
11. Which one of the following reactions is allowed? [GATE 2004]

(a) p o n  e  (b) p o e   ve (c) p o S  J (d) p  n o S   S 0

12. Which of the following decay is forbidden? [GATE 2005]

(a) P  o e   vP  ve (b) P  o P   vP

(c) P  o e   ve (d) P  o e   e  e 

13. The interaction potential between two quarks, separated by a distance r inside a nucleon,
can be described by (a, b and E are positive cosntants) [GATE 2006]

a a a
(a) aeEr (b)  br (c)   br (d)
r r r
14. Which one of the following nuclear processes is forbidden? [GATE 2006]

(a) v  p o n  e  (b) S  o e   ve  S0

(c) S  p  o n  K   K  () P  o e   ve  vP

15. Weak nuclear forces act on : [GATE 2006]

(a) both hadrons and leptons (b) hadrons only
(c) all particles (d) all charged particles
16. The strangeness quantum number is conserved in : [GATE 2007]
(a) strong, weak and electromagnetic interactions
(b) weak and electromagnetic interactions
(c) strong and weak interactons
(d) strong and electromagnetic interactions

17. A relativistic particle travels a length of 3 u103 m in air before decaying. The decay
process of the particle is dominated by. [GATE 2007]
(a) strong interactions (b) electromagnetic interactions
(c) weak interactions (d) gravitational interactions
18. The strange baryon 6 has the quark structure : [GATE 2007]
(a) uds (b) uud (c) uus (d) us
19. According According to the quark model,the K meson is commonsed which of the following
quarks. [GATE 2008]

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(a) uud (b) uc (c) us (d) su

20. Choose the correct statement from the following : [GATE 2008]

(a) The reaction K  K  o pp  can proceed irrespective of the kinetic energies of K 

and K 

(b) The reaction K  K  o 2 J ICK- pp is forbidden by the Baryon number conservation

(c) The reaction K  K  o 2 J is forbidden by strangeness conservation

(d) The decay K 0 o S S err- proceeds via weak interaction

0
21. A neutral pi meson S has a rest-mass of approximately 140 MeV/c2 and a lifetime of

W sec. A S0 produced in the laboratory is.found to decay after 1.25 W sec into two
photons. Which of the following sets represents a possible set of energies of the two
photons as seen in the laboratory? [GATE 2008]
(a) 70 MeV and 70 MeV (b) 35 MeV and 100 MeV
(c) 75MeV and 100 MeV (d) 25 MeV and 150 MeV
22. In the quark model which one of the following represents a proton? [GATE 2009]
(a) udd (b) uud (c) ub (d) cc

1 1 1 1
23. Let n and p denote the isospin state with I , I3 and I , I3  of a
2 2 2 2
nuclear respectively. Which one of the following two nuclear state has I 0, I 3 0?

1 1
(a) nn  pp (b) nn  pp [GATE 2009]
2 2

1 1
(c) np  pn (d) np  pn
2 2

24. The basic process underlying the neutron E -decay is : (GATE 2009]

(a) d o u  e   ve (b) d o u  e (c) s o u  e   ve (d) u o d  e  ve

25. Match the reactions on the left with the associated interactions on the right: [GATE 2010]

(1) S o P   vP (i) strong

(2) S0 o J  J (ii) elctromagnetic

(3) S0  n o S  p (iii) weak

(a) 1-iii, 2-ii, 3-i (b) 1-i,2-ii, 3-iii (c) 1-ii, 2-i, 3-iii (d) 1-iii, 2-i, 3-ii

26. The quark content of 6 , K  S and p is indicated. [GATE 2010]

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6 uus ; K  su ; S du udd

In the process S   p o K  6 

Considering strong interactions only, which of the following is true?

(a) the process is allowed because, 'S 0
(b) the process is allowed because, 'I 3 0

(c) the process is not allowed because, 'S z 0, 'I3 z 0

(d) the process is not allowed because; 'S z 0 and 'I 3 z 0

27. A S0 meson at rest decays into two photons, which move long the x-axis. They are both
detected simultaneously after a time, t 10 s . In an inertial frame moving with a velocity
v 0.6c in the direction of one of the photons, the time interval between the two detections
is : [GATE 2010]
(a) 15 s (b) 0 s (c) 10 s (d) 20 s
28. The isospin and the strangeness of :  baryon are : [GATE 2011]
(a) 1, –3 (b) 0, –3 (c) 1, 3 (d) 0, 3
29. Which one of the following sets corresponds to fundamental particles ? [GATE 2012]
(a) proton, electron and neutron (b) proton, electron and photon
(c) electron, photon and neutrino (d) quark, electron and meson
30. Choose the CORRECT statement from the following : [GATE 2012]
(a) Neutron interacts through eletromagnetic interaction
(b) Electron does not interact through weak interaction
(c) Neutrino interacts through weak and electromagnetic interaction
(d) Quark interacts through strong interaction but not through weak-interaction

31. The decay process n o p   e   ve violates : [GATE 2013]

(a) baryon number (b) lepton number (c) isospin (d) strangeness
32. The isospin (I) and baryon number (B) of the upquark is : [GATE 2013]
(a) I 1, B 1 (b) I 1, B 1 / 3 (c) I 1 / 2, B 1 (d) I 1 / 2, B 1 / 3

33. Consider the decay of a pion into a muon and an anti-neutrino S o P   v e in the pion

rest frame. mS 139.6MeV / c 2 , mP 140 MeV / c 2 , mv 0 . The energy (in MeV), of

the emitted neutrino, to the nearest integer is ____ [GATE 2013]

34. Consider the process P   P  o S  S . The minimum kinetic energy of the muons
P in the centre of mass frame required to produce the pion S pairs at rest is __ MeV..

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(Given : m 105MeV / c 2 , mS 140 MeV / c 2 ) [GATE 2014]

35. Which one of the following three-quark states (qqqq), denoted by X, CANNOT be a
possible baryon? The corresponding electric charge is indicated in the superscript [GATE
2014]
(a) X  (b) X  (c) X  (d) X 
36. Which one of the following high energy p rocesses is allowed by conservation laws?

(a) p  p o A0  A0 (b) S   p o S0  S [GATE 2014]

(c) n o p  e   ve (d) P  o e   J

37. If the halt-life of an dement any particle moving with speed 0.9c in the laboratory frame
8 8
is 5 u108 s , then the proper half-life is _______ u10 s c 3 u 10 s

38. Which one ofthe following I.is a fermion? [GATE 2014]

(a) D particle (b) 4 Be 7 nucleus (c) hydrogen atom (d) deuteron

39. The decay P  o e   J is forbidden, because it violets. [GATE 2015]

(a) momentum and lepton number conservations

(b) baryon and lepton number conservations
(c) angular momentum conservation
(d) lepton number conservation

40. In the SU 3 quark model, the triplet of measons S  S
0
S has : [GATE 2016]

(a) Isospin = 0, Strangeness = 0 (b) Isopin = 4, Strangeness = 0

(c) Isospin = ½, Strangeness = +1 (d) Isospin = ½, Strangeness = –1

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

8.(d) 9.(c) 10.(a) 11.(d) 12.(d) 13.(c) 14.(b)
15.(c) 16.(d) 17.(c) 18.(c) 19.(c) 20.(d) 21.()
22.(b) 23.(c) 24.(a) 25.(a) 26.(c) 27.() 28.(b)
29.(c) 30.(a) 31.(c) 32.(d) 33.(30) 34.(35) 35.(d)
36.(b) 37.(2.2) 38.(b) 39.(d) 40.(b)
CSIR-UGC-NET Previous Year's Questions

1. A beam of pions S is incident on a proton target, giving rise to the process

S  p o n  S  S  [NET June 2011]

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(A) Assuming that the decay proceeds through strong interactions, the total isospin I and
its third component I 3 for the decay products, are

3 3 5 5 5 3 1 1
(a) I , I3 (b) I , I3 (c) I , I3 (d) I , I3 
2 2 2 2 2 2 2 2
(B) Using isosp in symmetry, the cross-section for the above process can be related to
that o f the process

(a) S n o p S  S (b) S p o n S S  (c) S n o pS S (d) S p o nS S

2. Consider the decay process W  o S   VW in the rest frame of the W . The masses of

W , S and vW are M W , M n and zero repectively.. [NET June 2012]

(A) The energy of S is:

M W2  M S2 c2 M W2  M S2 c 2
2 2 2
(a) (d) (c) M W  M S c (d) M WM S c2
2M W 2M W

(B) The velocity is S is:

M W2  M S2 c M W2  M S2 c M Sc M Wc
(a) (b) (c) M (d) M
M W2  M S2 M W2  M S2 W S

3. The dominant interactions underlying the following processes : [NET June 2012]

A. K   p o 6   S  B. P   P  o K   K  C. 6  o p  S0 are

(a) A: strong, B: electromagnetic and C weak (b) A: strong, B : weak and C : weak
(c) A: weak, B: electromagnetic and C: strong
(d) A: weak, B: electromagnetic and C: weak

v
4. If a Higgs boson of mass mH with a speed E decays into a pair photons, then the
c
invariant mass of the photon pair is : [NET June 2012]
[Note: The invariant mass of a system of two particles with four momenta p1 and p2 is
2
p1  p2 ]

(a) EmH (b) mH (c) mH 1  E2 (d) E mH 1  E2

5. Consider the following particles: the proton p, the neutron n, the neutral pion S0 and the
delta resonance '  . When ordered in terms of decreasing lifetime, the correct
arrangement is as follows : [NET Dec. 2012]

(a) S0 , n, p, '  (b) p , n, '  , S0 (c) p , n, S0 , '  (d) '  , n, S 0 , p

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6. The muon has mass 105 MeV/c2 and mean lifetime 2.2Ps in its rest frame. The mean
distance traversed by muon of energy 315 MeV/c2 before decaying is approximately.
[NET Dec. 2012]

(a) 3 u105 km (b) 2.2 cm (c) 6.6Pm (d) 1.98 km

7. A spin-1/2 particle A undergoes the decay A o B  C  D where it is known that B and

C are also spin-1/2 particles. The complete set of allowed values of the spin of the particle
D is : [NET June 2013]

1 3 5 1 1 3 5 7
(a) ,1 , 2, ,3,.... (b) 0, 1 (c) only (d) , , , ,....
2 2 2 2 2 2 2 2
8. Muons are produced through the annihilation of particle a and its antiparticle;:nattrely the
process a  aG o P   P  . A ninon has a rest mass of 105 MeV/c2 and its proper life time
is 2Ps . If the center of mass energy of the collision is 2.1 GeV in the laboratory frame
that coincides with the center of mass frame, then the fraction of muons that will decay
before they reach a detector placed 6km away from the interaction point is : [NET June

(a) e1 (b) 1  e1 (c) 1  e2 (d) e10 2013]

* U o S  S0
9. Consider the following ratios of the partial decay widths R1 and
* U o S   S 0

* '  o S   p
R2 . If the effects of electromagnetic and weak interactions are
* '  o S  n

neglected, then R1 and R2 are, respectively,, [NET Dec. 2013]

(a) 1 and 2 (b) 1 and 2 (c) 2 and 1 (d) 1 and 1

10. The recently-discovered Higgs boson at the LHC experiment has a decay mode into a
photon and a Z boson. If the rest masses of the Higgs and Z boson are 125 GeV/c 2 and
90 GeV/c2 respectively, and the decaying Higgs particle is at rest, the energy of the
photon will approximately be : [NET June 2014]

(a) 35 3 GeV (b) 35 GeV (c) 30 GeV (d) 15 GeV

11. In a classical model, a scalar (spin 0) meson consists of a quark and an antiquark bound
b
by a potential V r ar  , where a = 200 MeV fm–1 and b = 100 MeV fm. If the
r
masses of the quark and antiquark are negligible, the mass of the meson can be estimated
as approximately. [NET June 2014\
(a) 140 MeV/c2 (b) 283 MeV/c2 (c) 353 MeV/c2 (d) 425 MeV/c2
12. Consider the four process [NET Dec. 2014\

(i) p  o n  e   ve (ii) / 0 o p   e   ve

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(iii) S o e   ve (iv) S0 o J  J

Which of the above is/are forbidden for free particles?

(a) only (ii) (b) (ii) and (iv) (c) (i) and (iv) (d) (i) and (ii)
13. In deep inelastic scattering electrons are scattered ott protons to determine if a proton has
any internal structure. The energy of the electron for this must be at-least
[NET Dec. 2014]

(a) 1.25 u 109 eV (b) 1.25 u 1012 eV (c) 1.25 u 106 eV (d) 1.25 u 108 eV

2 2
14. The reaction 1 D 1 D o 24 He  S0 cannot proceed via strong interactIons because it
violaies conservations of : [NET June 2015]
(a) angular momentum (b) electric charge (c) baryon number (d) isospin
15. The charm quark is assigned a charm quantum number C 1 . How should the Gellmann-
Nishijima formula for electric charge be modified for four flavours of quarks?

1 1
(a) I3  B S C (b) I 3  B  S  C [NET June 2015]
2 2

1 1
(c) I 3  B  S C (d) I 3  BS C
2 2
16. Consider the following processes involving free particles. [NET Dec. 2015]

(i) n o p  e   ve (ii) p  n o S 

(iii) p  n o S  S0  S0 (iv) p  ve o n  e 

Which of the following statements is true?

(a) Process (i) obeys all conservation laws
(b) Process (ii) conserves baryon number, but violates energy-momentum conservation
(c) Process (iii) is not allowed by strong interactions, but is allowed by weak interactions
(d) Process (iv) conserves baryon number, but violates lepton number conservation

17. The decay constants f p of the heavy pseudoscalar mesons, in the heavy quark limit, are

a
f
related to their masses in m p by the relation p m p , where a is an empirical

parameter to be determined. The values m p 6400 r 160 MeV and f p 180 r 15MeV
correspond to uncorrelated measurement of a meson. The error on the estimate of a is.
[NET June 2016]
(a) I 75 (MeV)3/2 (b) 900 (MeV)3/2 (c) 1200 (MeV)3/2 (d) 2400 (MeV)3/2
18. In the large hadron collider (LHC) two equal energy proton beams traverse in opposite
directions along a circular path of length 27 km. If the total center of mass energy of a

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proton-proton pair is 14 TeV, which of the following is the best approxamation for the
proper time taken by a proton to traverse the entire path? [NET June 2016]
(a) 12 ns (b) 1.2Ps (c) 1.2 ns (d) 0.12Ps

1.A.(c) B.(b) 2.A.(b) B.(a) 3.(a) 4.(b) 5.(c)

6.(d) 7.(d) 8.(b) 9.(d) 10.(c) 11.(b) 12.(d)
13.(a) 14.(d) 15.(d) 16.(b) 17.(c) 18.(a)
TIFR Previous Year's Question

1. A spin  1 2 particle A decays to two other particles B and C. If B and C are of spin  1 2 and
spin-1 respectively, then a complete list of the possible values of dre-01-bital angular
momentum of the final state (i.e. B + C) is : [TIFR 2013]

1 3
(a) 0, 1 (b) , (c) 0, 1, 2 (d) 0, ±1
2 2
2. The interaction strength of the recently-discovered Riggs boson (mass approximately 125
GeV/c2) with any other elementary pacticle is proportional to the mass of that particle.
Which of the following decay processes will have the greatest probability? [TIFR 2014]
(a) Higgs boson decaying to a top quark a top anti-quark
(b) Higgs boson decaying to a bottom quark + a bottom anti-quark
(c) Higgs boson decaying to an electron and a positron
(d) Higgs boson decaying to a neutrino-antineutrino pair
3. Cosmic ray anions generated at the top of the Earth’s atmosphere decay according, to the
radioactive decay law : [TIFR 2014]

1 3
where , is the number of muons at time t, and T1/2 1.52Ps is the proper half-life of
2 2
the muon. Immediately after generating most of these muons shoot down towards the
Earth's surface. Some of these muons decay on the way, but their interaction with the
atmosphere is negligible.
An observer on the top of a mountain of height 2.0 km above mean sea level detects
muons with the speed 0.98c over a period of time and counts 1000 muons. The number of
muons of the same speed detected by an observer at mean sea level in the same period of
time would be :
(a) 232 (b) 539 (c) 839 (d) 983
4. A spin-2 nucleus absorbs a spin ½ electron and is then observed to decay to a stable
nucleus in two stages, recoiling against an emitted invisible particle in the first stage and
against an emitted spin-1 photons in the second stage. If the stable nucleus is spinless,
then the set of all possible spin values of the invisible particle is : [TIFR 2014]
(a) {1/2, 5/2} (b) {3/2, 7/9} (c) {3/2, 5/2} (d) {1/2, 3/2, 5/2,7/2}
5. Consider the following reaction involving elementary paiticle :

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( A ) S  p o K   6 

(B) K   p o K   U

Which of the following statements is true for strong interactions? [TIFR 2015]
(a) (A) and (B) are both forbidden (b) (B) is allowed but (A) is forbidden
(c) (A) is allowed but (B) is forbidden (d) (A) and (B) are both allowed
6. In a fixed target experiment, a proton of total energy 200 GeV is bombarded on a proton
at rest and produces a nucleus Z and its anti-nucleus Z
AN AN

p  p o ZA N  ZA N  p  p

The heaviest nucleus Z that can be created has atomic mass number A =
AN

(a) 15 (b) 9 (c) 5 (d) 4

7. Consider the hyperon decay 1 A o S  S0 followed 2 S0 o JJ . If the isospin cmponent

baryon number and strangeness quantum numbers are denoted by I Z , B and S
respectively, then which of the following statements is completely correct ?[TIFR 2016]
(a) In (1) I Z is not conserved, B is conserved, S is not conserved;

In (2) I Z is conserved, B is conserved, S is conserved

(b) In (1) I Z is conserved, B is not conserved, S is not conserved;

In (2) I Z is conserved, B is conserved, S is conserved

(c) In (1) I Z is not conserved, B is conserved, S is not conserved;

In (2) I Z is not conserved, B is conserved, S is conserved

(d) In (1) I Z is not conserved, B is conserved, S is conserved;

In (2) I Z is conserved, B is conserved, S is conserved

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

JEST Previous Year's Questions
1. A K-meson (with a rest mass of 494 MeV) at rest decays into a muon (with a rest mass
of 106 MeV) and a neutrino. The energy of the neutrino, which can be taken as massless,
approximately : [JEST 2013]
(a) 120 MeV (b) 236 McV (c) 300 MeV (d) 388 MeV

2. The reaction e   e o J is forbidden because. [JEST 2015]

(a) lepton number is not conserved (b) linear mementum is not conserved

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(c) angular momentum is not conserved (d) charm is not conserved

1.(b) 2.(c)
Other Examinations Previous Year's Questions
1. The ratio FC / FG of the electostatic Coulomb force FC to the gravitational force FG
between the proton and the electron in the first Bohr orbit (or radius rB) of a hydrogen
atom is closest to the following value.

(a) 2 u1039 (b) 2 u1049 (c) 2 u 10 42 rB (d) 2 u 1035 / rB

2. In the large Hadron Collider (LHC) in CERN, protons will be accelerated to an energy of
7 TeV (tetraelectron volt), i.e., 7 u 1012 eV . (As a matter of fact, two proton beams, each
with 7 TeV will collide so that the energy of collision in the centre of mass frame is 14
TeV). The speed v of a proton of 7 TeV energy is in the range :
(a) 0.999999000c < v < 0.999999999c (b) 0.999900c < v < 0.99999c
(c) 0.9900c < v < 0.9999c (d) 0.90c < v < 0.99c
3. An 57
Fe nucleus decays from an excited state to the ground state by emitting a gamma
ray. The energy of the emitted photon is 14.4 eV when the nucleus is held fixed. If the
nucleus is free to recoil, then the energy of the photon emitted will be :
(a) lower by app roximately 2 keV (b) lower by approximately 2 meV
(c) higher by approximately 2 eV (d) higher by approximately 20 eV
(e) lower by approximately 20 meV
4. Muons of kinetic energy E are produced in collision with a target in a laboratory. The
mass of a muon is 106 MeV/c2 and its half-life is 1.4 u 106 s in its rest frame the minimum
value of E if more than half the muons created at the target are to each a detectro 840 m
away?
(a) 106 MeV (b) 4212 MeV (c) 130 MeV (d) 189 MeV
(e) 162 MeV
5. A free neutron is unstable and decays into a proton, an electron and anti neutrino :
n o p  e   ve . The rest masses of these particles may be taken to be mn 939.6 MeV ,
mp 938.3MeV , me 0.51MeV and mvG 0 , so that the change in the total test mass in
the decay is 0.79 MeV. If in a particular decay, the neutron as well as proton (created due
to decay) are at rest, then the energy of the anti-neutrino is :
(a) 1.05 MeV (b) 0.39 MeV (c) 0.78 MeV (d) 0.55 MeV
(e) 0.40 MeV

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

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GATE Previous Year's Questions

1. An admissible potential between the proton and the neutron in a deutron is : [GATE 2000]
(a) Coulomb (b) Harmonic oscillator
(c) Finite square potential (d) Infinite square well
2. Nuclear forces are : [GATE 20002]
(a) spin dependent and have no non-central part
(d) spin dependent and have a non-central part
(c) spin independent and have no non-central part
(d) spin independent and have a non-central part
3. With reference to nuclear forces which of the following statements is NOT true? The
nuclear forces are [GATE 2005]
(a) short range (b) charge independent
(c) velocity dependent (d) spin independent

4. To explain the observed magnetic moment of neutron 0.0857P N , its ground state wave
function is taken to be an admixture of S and D states, the expectation values of the z-
component of the magnetic momenta in pure S and pure D-states are 0.8797P N and
0.3101P N respectively. The contribution of the D-state to the mixed ground state is
approximately : [GATE 2006]
(a) 40% (b) 4% (c) 0.4% (d) 0.04%

5. In the deuteritun + tritium d  t fusion more energy is released as compared to deuterium

+ deuterium d  d fusion because : [GATE 2007]

(a) tritium is radioactive

(b) more nucleus participate in fusion

(c) the coulomb parrier is lower for the d  t syslem then d  d system

(d) the reaction product 4 He is more tightly bound

6. A heavy nucleus is found to contain more neutrons than protons. This fact is related to
which one of the fclowing statement : [GATE 2008]
(a) the nuclear force between neutrons is stronger than that between protons
(b) the nuclear force between protons is of a shorter range than those between neutrons,
so that a smaller number of protons are held together by held together by the nuclear
force
(c) protons are unstable, to their number in a nucleus dimisishes
(d) it costs more energy to add a proton to a (heavy) nucleus than a neutron because of
the Coulomb repulsion between protons
7. Pick the wrong statement : [GATE 2009]

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(a) the nuclear force is independent of electric charge

1 §m ·
(b) the Yukawa potential is proportinoal to r exp ¨ r ¸ Where, r is the seperation
©h ¹
between two nucleons.

(c) The range of nuclear force is of the order of 1015  1014 m

(d) the nucleons interact among each other by the exchange of mesons.
8. The ground state wavefunction of deuteron is in a superposition of s and d states. Which
of the following is NOT true as a consequence? [GATE 2010]
(a) It has a non-zero quadruple moment
(b) The neutron-proton potential is non-central
(c) The orbital wavefunction is not spherically symmetric
(d) The Hamiltonian does not conserve the angular momentum
9. To detect trace amounts of a gaseous species in a mixture of gases, the preferred probing
tools is : [GATE 2010]
(a) ionization spectroscopy with X-rays (b) NKR spectroscopy
(c) ESR spectroscopy (d) laser spectroscopy
10. A neutron passing through a detector is detected because of : [GATE 2011]
(a) the ionization it produces (b) the scintillation light it produces
(c) the electron-hole pair it produces
(d) the secondary particles produced in a nuclear reaction in the detector medium
11. In case of a Geiger-Muller (GM) counter, which one of the following statements is
CORRECT? [GATE 2012]

(a) Multiplication factor of the detector is of the order of 1010

(b) Type of the particles detected can be identified
(c) Energy o f the particles detected can be distinguished
(d) Operating voltage of the detector is few tens of Volts
12. Deuteron has only one bound state with spin parity V, isospin 0 and electric quadrupole
moment 0 286 efm2. These data suggest that the nuclear forces are having: [GATE 2012]
(a) only spin and isospin dependence
(b) no spin dependence and no tensor components
(c) spin dependence but no tensor comportents
(d)spindeppcieuce.alongyyith tensor component
13. Consider the scattering of neutrons by protons at very low energy due to a nuclear potential
of range r2 . Given that : [GATE 2013]
J
cot kr0  G
O

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where G is the phase shift, k the wave number and state wave function, the phase shift is:

k k S S
(a) G   kr0 (b) G  kr0 (c) G  kr0 (d) G   kr0
J J 2 2
14. The value of the magnetic field required to maintain non relativistic protons of energy 1
MeV in a circular orbit of radius 100 mm is ____ Tesla. [GATE 2014]

(Given: m p 1.67 u 10 27 kg , e 1.6 u 10 19 C )

15. Which of the following statements is NOT correct? [GATE 2016]

(a) A deuteron can be disintegrated by irradiating it with gamma rays of energy 4 MeV
(b) A deuteron has no excited states
(c) A deuteron has no electric quadrupole moment

(d) The 1 S0 state of deuteron cannot be formed

16. An alpha particle is accelerated in a cyclotron. it leaves the cyclotron with a kinetic
energy of 16 MeV. The potential different between the D electrodes is 50 kilovolts. The
number of revolutions the alpha particle makes in its spiral path before it leaves the
cyclotron is : [GATE 2016]

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

8.(d) 9.(c) 10.(d) 11.(a) 12.(d) 13.(a) 14.(1.44)
15.(c) 16.(80)
CSIR-UGC-NET Previous Year's Questions
1. An atom of mass M can be excited to a state of mass M  ' by proton capture. The
frequency of a photon which can cause this transition is : [NET Dec. 2011]
(a) 'c2 / 2h (b) 'c 2 / h

(c) ' 2c 2 / 2Mh (d) ' '  2 M c 2 / 2Mh

1.(d)
TIFR Previous Year's Questions
1. An atom is capable of existing, in two states a ground state of mass M and an excited
state of mass M  ' . If the transtion from the ground state to the excited state proceeds
by the absorption of a photon, the pheton frequency in the laboratory frame (where the
atom is initially at rest) is : [TIFR 2010]

'c 2 'c 2 § ' · Mc 2 'c 2 § ' ·

(a) (b) ¨1  ¸ (c) (d) ¨1  ¸
h h © 2M ¹ h h © 2M ¹

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Mc 2 § ' · Mc 2 § ' ·
(e) ¨1  ¸ (f) ¨1  ¸
h © 2M ¹ h © 2M ¹
1 3
2. Given that the ionization energies of hydrogen H and Lithium Li are 13.6 eV and
5.39 eV, respectively. the effective nuclear charge experienced by the valence electron
of a 3 Li atom may be estimated in terms of the proton charge e as: [TIFR 2011]
(a) 0.63 e (b) 1.26 e (c) 1.59 e (d) 3.00 e

1.(b) 2.(b)
JEST Previous Year's Questions
1. The binding energy of the k shell electron in a Uranium atom (Z = 92, A = 238) will be
modified due to (i) screening caused by other electrons and (ii) the finite extent of the
nucleus as follows : [JEST 2013]
(a) increases due to (i), remains unchanged due to (ii)
(b) decreases due to (i), decreases due to (ii)
(c) increases due to (i), increases eut to (ii)
(d) decreases due to (i), remains unchanged due to (ii)
2. A hydrogen atom in its ground state is collided with an electron of kinetic energy 13 .377
eV. The maximum factor by which the radius of the atom would increase is[JEST 2014]
(a) 7 (b) 8 (c) 49 (d) 64
3. If the Rydberg constant of an atom of finite. nuclear mass is DRf where Rf is the
Rydberg constant corre-sponding to an infinite nuclear mass, the ratio of the electronic to
nuclear mass of the atom is : [JEST 2016]
1 D D 1 1
(a) (b) (c) 1  D (d)
D D D

1.(b) 2.(d) 3.(a)

Other Examinations Previous Year's Questions
1. A proton moves in a circular orbit in a plane which is perpendicular to a uniform magnetic
field of strength B. When the kinetic energy of the proton changes from 0.25m p c 2 to

m p c 2 (where m p is the rest mass of the proton) the fregency of rotation changes from
Z1 to Z2 . The ratio Z1 / Z2 is :

1.(c)

An Institute for NET/JRF Physical Science,

IIT-JAM, JEST TIFR & M.Sc. Entrances
Email : quantainstitute@gmail.com, 9694473047, 9136887760
 Most Important Problems
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Nuclear and Particle Physics

Assignments for CSIR-NET-JRF, GATE & other entrances

Shanu Arora
All India Rank -25 , IIT-JAM
Msc Physics , IIT bombay
Qualified :- CSIR-NET, JEST, TIFR & GATE exam.
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Nuclear and Particle Physics assignments
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INDEX
NET PYQs

● NET DEC 2023

● MET JUNE 2023
● NET JUNE 2022
● NET JUNE 2021
● NET JUNE 2020
● NET DEC 2019
● NET JUNE 2019
● NET DEC 2018
● NET JUNE 2018

GATE PYQs

● GATE 2024
● GATE 2023
● GATE 2022
● GATE 2021
● GATE 2020
● GATE 2019
● GATE 2018
 Nuclear and Particle Physics assignments 3
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10. Net June 21

Shanu Arora IIT-B

11. Net June 21

Shanu Arora IIT-B

12. Net June 21

Shanu Arora IIT-B

13. Net June 20

Shanu Arora IIT-B

14. Net June 20

Shanu Arora IIT-B

 Nuclear and Particle Physics assignments 4
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15. Net June 20

Shanu Arora IIT-B

16. Net Dec 19

Shanu Arora IIT-B

17. Net Dec 19

Shanu Arora IIT-B

18. Net Dec 19

Shanu Arora IIT-B

 Nuclear and Particle Physics assignments 5
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19. Net Dec 19

Shanu Arora IIT-B

20. Net June 19

Shanu Arora IIT-B

21. Net June 19

Shanu Arora IIT-B

22. Net June 19

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 Nuclear and Particle Physics assignments 6
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23. Net Dec 18

Shanu Arora IIT-B

24. Net Dec 18

Shanu Arora IIT-B

25. Net June 18

Shanu Arora IIT-B

26. Net June 18

Shanu Arora IIT-B
 Nuclear and Particle Physics assignments 7
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27. Net June 18

Shanu Arora IIT-B

28. Net June 18

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29. Net June 18

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Nuclear and Particle Physics assignments 8
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Ques Anskey Ques Anskey Ques Anskey

1 4 11 (a) 21 (b)
2 4 12 (d) 22 (c)
3 4 13 (c) 23 (b)
4 (b) 14 (c) 24 (b)
5 (a) 15 (d) 25 (c)
6 209 16 (b) 26 (c)
7 215 17 (c) 27 (a)
8 217 18 (a) 28 (b)
9 (d) 19 (a) 29 (d)
10 (b) 20 (a)
Nuclear and Particle Physics assignments
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GATE PYQs 2024 - 2018

 Nuclear and Particle Physics assignments 1
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1. Gate 24

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3. Gate 24

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 Nuclear and Particle Physics assignments 2
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4. Gate 24

Shanu Arora IIT-B

5. Gate 24
Shanu Arora IIT-B

6. Gate 24

Shanu Arora IIT-B

7. Gate 23

Shanu Arora IIT-B

 Nuclear and Particle Physics assignments 3
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8. Gate 23

Shanu Arora IIT-B

9. Gate 23

Shanu Arora IIT-B

10. Gate 23
Shanu Arora IIT-B
 Nuclear and Particle Physics assignments 4
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11. Gate 23
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12. Gate 23
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13. Gate 23

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 Nuclear and Particle Physics assignments 5
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14. Gate 23

Shanu Arora IIT-B

15. Gate 23
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16. Gate 22

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17. Gate 22
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18. Gate 22

Shanu Arora IIT-B

 Nuclear and Particle Physics assignments 6
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19. Gate 21

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20. Gate 21

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21. Gate 21

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22. Gate 21
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23. Gate 20

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 Nuclear and Particle Physics assignments 7
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24. Gate 20
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25. Gate 20

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26. Gate 20

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27. Gate 20
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29. Gate 20
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 Nuclear and Particle Physics assignments 8
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30. Gate 19

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35. Gate 19
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 Nuclear and Particle Physics assignments 9
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36. Gate 19
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37. Gate 18

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38. Gate 18
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 Nuclear and Particle Physics assignments 10
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39. Gate 18

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40. Gate 18

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41. Gate 18

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42. Gate 18

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Nuclear and Particle Physics assignments 11
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Shanu Arora IIT-B

Ques AnsKey Ques AnsKey Ques AnsKey

1 (b) 21 (b),(c) 41 300
2 (c) 22 8 42 25.995
3 (a) 23 (d)
4 (b) 24 (c)
5 ©,(d) 25 (d)
6 2.7 26 1
7 (C) 27 2
8 (B) 28 (c)
9 (A),(C) 29 6999
10 (A),(B),(C) 30 (c)
11 (A) 31 (c)
12 (A) 32 (a)
13 (A),(C) 33 (a)
14 340 34 (b)
15 4.917 35 (a)
16 (B) 36 40
17 (c) 37 (c)
18 3 38 (b)
19 (b) 39 (c)
20 (d) 40 (d)
TIFR PYQs 2024 - 2018
 Nuclear and Particle Physics assignments 1
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1. TIFR2024
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2. TIFR2024

Shanu Arora IIT-B

 Nuclear and Particle Physics assignments 2
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3. TIFR2024
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4. TIFR2024

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5. TIFR2024
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 Nuclear and Particle Physics assignments 4
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6. TIFR2024

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 Nuclear and Particle Physics assignments 5
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7. TIFR2023
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 Nuclear and Particle Physics assignments 6
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8. TIFR2023

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 Nuclear and Particle Physics assignments 7
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9. TIFR2023

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 Nuclear and Particle Physics assignments 8
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10. TIFR2023

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 Nuclear and Particle Physics assignments 9
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11. TIFR2023

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 Nuclear and Particle Physics assignments 10
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12. TIFR2023

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 Nuclear and Particle Physics assignments 11
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13. TIFR2023

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 Nuclear and Particle Physics assignments 12
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14. TIFR2022

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 Nuclear and Particle Physics assignments 13
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15. TIFR2022

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16. TIFR2022

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 Nuclear and Particle Physics assignments 14
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17. TIFR2022

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18. TIFR2022

Shanu Arora IIT-B

 Nuclear and Particle Physics assignments 15
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19. TIFR2021

Shanu Arora IIT-B

20 TIFR2021
Shanu Arora IIT-B
 Nuclear and Particle Physics assignments 16
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21. TIFR2020

Shanu Arora IIT-B

 Nuclear and Particle Physics assignments 17
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22. TIFR2020
Shanu Arora IIT-B

23. TIFR2020
Shanu Arora IIT-B
 Nuclear and Particle Physics assignments 18
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24. TIFR2020

Shanu Arora IIT-B

25. TIFR2019

Shanu Arora IIT-B

 Nuclear and Particle Physics assignments 19
SHANU ARORA SIR
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26. TIFR2019

Shanu Arora IIT-B

Nuclear and Particle Physics assignments 20
SHANU ARORA SIR
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Shanu Arora IIT-B

Ques AnsKey Ques AnsKey Ques AnsKey

1 (a) 11 (a) 21 (a)
2 (a) 12 (a) 22 (a)
3 (a) 13 (a) 23 (b)
4 (a) 14 (a) 24 (c)
5 (a) 15 (a) 25 (d)
6 (a) 16 (a) 26 (a)
7 (a) 17 (a)
8 (a) 18 (a)
9 (a) 19 (a)
10 (a) 20 (a)
 Most Important Problems
(MIP by Shanu Arora Sir )

Quantum Mechanics

Assignments for IIT- JAM, JEST & other entrances

Shanu Arora
All India Rank -25 , IIT-JAM
Msc Physics , IIT bombay
Qualified :- CSIR-NET, JEST, TIFR & GATE exam.
Educator IIT-JAM & CSIR-NET Unacademy

📌 master link: https://linktr.ee/shanuarora

📌 telegram: https://t.me/shanuiit
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To join all Groups Youtube TOP class Playlists

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 Quantum Mechanics assignments 1
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Nuclear Physics

1.
Shanu Arora IIT-B

2.
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Shanu Arora IIT-B

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Shanu Arora IIT-B

Ques AnsKey Ques AnsKey Ques AnsKey Ques AnsKey Ques AnsKey
1 (d) 27 (b) 53 (d) 79 (c) 105 (c)
2 87.5 28 (d) 54 (b) 80 (c) 106 (b)
3 (e) 29 (b) 55 (b) 81 (a) 107 (b)
4 (d) 30 (d) 56 (c) 82 (b) 108 (c)
5 (d) 31 (d) 57 (b) 83 (a) 109 (b)
6 (d) 32 (b) 58 (a) 84 (c) 110 (d)
7 (a) 33 (d) 59 (b) 85 (c) 111 (a)
8 (b) 34 (b) 60 (b) 86 (b) 112 (a)
9 (b) 35 (d) 61 (b) 87 (d) 113 (b)
10 (b) 36 (b) 62 (d) 88 (d) 114 (c)
11 (b) 37 (d) 63 (b) 89 2 115 (d)
12 (a) 38 (a) 64 (c) 90 1.33 116 (a)
13 (a) 39 (c) 65 (b) 91 11.48 117 (c)
14 (b) 40 (b) 66 (d) 92 10.27 118 (c)
15 (c) 41 (a) 67 (b) 93 7.3 119 (b)
16 (b) 42 (c) 68 (b) 94 24000 120 (c)
17 (d) 43 (d) 69 (d) 95 1 121 (a)
18 (d) 44 (c) 70 (c) 96 95 122 (c)
19 (b) 45 (a) 71 (c) 97 8 & 6 123 (c)
20 (c) 46 (b) 72 (a) 98 23.6 124 (b,c)
21 (c) 47 (a) 73 (c) 99 (a,d) 125 (a,b,c)
22 (d) 48 (b,c,d) 74 (d) 100 (a,b,c) 126 (a,b,c,d)
23 (b) 49 (a) 75 (b) 101 (a,b,c) 127 (a,d)
24 (d) 50 (b) 76 (c) 102 (a,b,d) Shanu Arora IIT-B

25 (d) 51 (a) 77 (a) 103 (c,d)

26 (a) 52 (a) 78 (a) 104 (b)
 fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

Assignment 1 (Nuclear Properties)

Q1. Find the distance of closest approach of an  - particle of kinetic energy 1.0 MeV to a gold

nucleus of charge number 79 . Given that charge of an electron is 1.6  10 19 coulomb.
Q2.  
Assume that the nuclear radius R1 is given by 1.2  10 15  A1/ 3 , where A is mass number

of a nucleus. If mass of proton and neutron are each considered equal to 1.67  10 27 Kg ,

how many times is nuclear matter denser than that of water of density 103 Kg / m 3 .
Q3. Calculate the distance of closed approach for protons of 1MeV energy when meeting
Rutherford scattering by nuclei of gold Z  79 .
Q4. Find the ratio of the sizes of Pb82208 and Mg1226 nuclei.
Q5. Show that the mean momentum of a nucleon in a nucleus with mass number A varies as
A1/ 3 .
Q6. If the nucleus of Al 27 is 3.0 Fermi find the approximate nuclear radius of Cu 64 .
Q7. In a mass spectrometer, a single charged +ve ion is accelerated through a potential
difference of 1000 volt. It then travels through a uniform magnetic field for which
B  1000 gauss and is deflected into a circular path 18.2 cm in radius. What is
(a) The speed of the ion
(b) The mass of the ion, in gms and mu
(c) The mass number of the ion
Q8. The radius of a 29 Cu 64 nucleus is measured to be 4.8  10 13 cm. Estimate the radius of a

12 Mg 27 nucleus.

Q9. The difference in the Coulomb energy between the mirror nuclei 24 Cr 49 and 25 Mn 49 is
6.0 MeV . Assuming that the nuclei have a spherically symmetric charge distribution and

that e 2 is approximately 1.0 MeV - fm , find the radius of the 25 Mn49 nucleus.
Q10. Chlorine-33 decays by positron emission with a maximum energy of 4.3 MeV . Calculate
the radius of the nucleus.
Q11. The quadrupole moment of 64 Gd 155 is 130 fm 2 . Show that the Gd -155 nucleus is almost
spherical.

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1 189
Q12. (a) A stable nucleus has the radius of Os nucleus. Find the stable nucleus.
3
(b) The radius of Ge nucleus is measured to be twice the radius of 94 Be . How many

nucleons are there in Ge nucleus?

208 26
(c) Find approximately the ratio of the sizes of 82 Pb and 12 Mg nuclei.

(d) The radius of a 64

29 X nucleus is measured to be 4.8 10 11 cm . Find the radius of a 27
12 Y
nucleus.
Q13. An alpha of energy 5 MeV is scattered through 1800 by a fixed uranium nucleus. Find the
distance of closest approach.
Q14. In deep inelastic scattering electrons are scattered off protons to determine if a proton has
any internal structure. Find the approximate energy of the electron.
Q15. A nucleus has a size of 10 15 m . Consider an electron bound within a nucleus. Estimate the
energy of this electron.
Q16. The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus
of radius R is given by

3 Z  Z  1 e
2

E
5 4 0 R
1 15 15
The measured mass of the neutron, 1 H , 7 N and 8 O are 1.008665u ,1.007825u ,
15 15
15.000109u and 15.003065u respectively. Given that the radii of both the 7 N and 8 O

nuclei are same. 1u  931.5MeV / c 2 ( c is the speed of light) and e 2 /  4 0  

15 15
1.44 MeV fm . Assuming that the different between the binding energies of 7 N and 8 O is

purely due to the electrostatic energy, find the radius of either of the nuclei. ( 1 fm  10 15 m )

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Solution
Assignment-1 Basics Properties of Nuclei
Solution.1:
1 1
(a) R  ROs  R0  A   R0 189   A  7  3 Li 7
1/ 3 1/ 3

3 3

R  Ro  A and given that RGe  2 RBe  Ro  AGe   2Ro  9   AGe  72 .

1/3 1/3 1/3
(b)
1/3
R A   208 
1/3

(c) Since R  R0  A   8

1/3 1/3
 Pb   Pb    2
RMg  AMg   26 
1/ 3
A 
1/ 3
R  27 
(d) Since R  R0  A  Y   Y   
1/ 3

R X  AX   64 

RY 3 3
   RY   4.8  1013  3.6  1013 cm  4  1013 cm
RX 4 4

Solution.2:
Energy is conserved.
Loss in kinetic energy = Gain in potential energy
 Ze  2e   5  2 Ze 2
1
4 0 rmin
1.6 10  J  r
13
min 
1
4 0 5 1.6 1013

 9 10   2  92  1.6 10 

2
9 19

or rmin  13
 rmin  5.3 1014 m  rmin  5.3  10 12 cm
5  1.6 10
The distance of closest approach is of the order of 1012 cm

Solution.3:
The internal structure of proton can only be determined if the wavelength of the incoming
electron is nearly equal to the size of the proton
i.e.   R  1.2 A1/ 3  fm   1.2 fm  1.2 1015 m

h h
According to de-Broglie relation,  
p 2mE

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 0 150
This can be also written as    
  E  eV 

150 150
 E  eV     1.04  1012  E  1.04 1012 eV
1.2  10 
2 5 2
  0 
    
 
Solution.4:
h 6.6 1034
p   6.6 1019 kgm / sec
 10 15

p2 44 1038
E  31
 2.4  107 Joule
2me 2  9.110

2.4 107
E eV  1.5 1012 eV  1.5 106 MeV  150 104 MeV
1.6 1019
Solution.5:
3 8  7 e2 3 8 7 3 76
E0       1.44 MeV and EN    1.44 MeV
5 R 4 0 5 R 5 R

3 1.44
So E0  EN    7  2  ………………(i)
5 R
Now mass defect of N atom  7 1.007825  8 1.008665  15.000109  0.1239864u
So, binding energy  0.1239864  931.5 MeV
And mass defect of O atom  8 1.007825  7 1.008665  15.003065  0.12019044 u
So binding energy  0.1239864  931.5 MeV

So B0  BN  0.0037960  931.5 MeV ………(ii)

From (i) and (ii) we get

R  3.42 fm

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Solution.6:

(a) B.E.   ZmH  Nmn  m  12 H    931.5 MeV

 B.E.  11.0078  11.0087  2.0141  931.5 MeV

 B.E.  0.0024  931.5 MeV  2.2356 MeV

B.E. 2.2356
(b)   1.1178 MeV / nucleons
A 2

Solution.7:

(a) B.E.   ZmH  Nmn  m  238

92U    931.5 MeV


 B.E.  92 1.0078  146 1.0087  238.0508  931.5 MeV

 B.E.  1.937  931.5 MeV  1804 MeV

B.E. 1804
(b)   7.6 MeV / nucleons
A 238

Solution.8:
(a) 42
20 Ca 41
20 Ca 0 n ;
1

Total mass of the 41

20 Ca and 10 n  41.970943 u .

Mass defect m  41.970943  41.958622  0.012321 u

So, B.E. of missing neutron= m  931.5  11.48 MeV

(b) 42
20 Ca 19
41
K 11 p ;

Total mass of the 41

19 K and 11 p  41.969101 u .

Mass defect m  41.969101  41.958622  0.010479 u

So, B.E. of missing proton= m  931.5  10.27 MeV .

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Solution.9:
When total binding energy of products is more than that of reactants, energy is released in
the process. Effectively the total binding energy is increased.
Total B.E. = B.E. per nucleon  number of nucleons
Y
8.5 X
8.0 W
Binding Energy 7.5
in MeV
Nucleon 5.0

0 30 60 90 120
Mass number of nuclei

Reaction Reactant Product

(a) Y  2Z 60  8.5  510 MeV 2  30  5  300 MeV
Total binding energy is decreased, from 510 MeV to 300 MeV .

(b) W  X  Z 120  7.5  900 MeV  90  8  30  5  870 MeV

Total binding energy is decreased, from 900 MeV to 870 MeV .
(c) W  2Y 120  7.5  900 MeV 2  60  8.5  1020 MeV
Total binding energy is increased, in reaction W  2Y from 900 MeV to 1020 MeV

(d) X  Y  Z 90  8  720 MeV  60  8.5  30  5  660 MeV

Total binding energy is decreased, from 720 MeV to 600 MeV .

Solution.10:
(a) If  BE final   BE initial  0 ; Energy will not be released. (No)

(b) If  BE final   BE initial  0 ; Energy will be released. (Yes)

(c) If  BE final   BE initial  0 ; Energy will not be released. (No)

(d) If  BE final   BE initial  0 ; Energy will be released. (Yes)

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Assignment-2
(Liquid Drop Model)
Q1. According to liquid drop model consider binding energy formula. Find the value of
atomic number  Z 0  for the most stable nuclei among members of an isobaric

family  X.
A
Z

Q2. The binding energy of a light nucleus Z, A in MeV is given by the approximate
formula

B  A, Z   16 A  20 A 2/3 3
 Z 2 A 1 / 3  30
N  Z  2

4 A
where N  A  Z is the neutron number. Find the approximate value of Z of the most
stable isobar for a given A  27 .
Q3. The masses of pair of mirror nuclei 15
7 N ,
15
8 O are 15.000108 u , 15.003070 u and

proton  p  and neutron  n  are 1.0073 u and 1.0087 u respectively. Find the value of

coulomb coefficient ac in the semi-empirical mass formula.

Q4. Consider the following expression for the mass of a nucleus with Z proton and A
nucleons: (where A is some odd number)
M  Z, A   A  Z  Z2

where ,  and  are some constant with suitable dimensions.

(a) If M  Z0 , A  is nuclear mass of most stable nuclei of the isobaric family, then find the

difference in mass M  Z, A   M  Z0 , A  .

(b) If M  Z0 , A  is nuclear mass of most stable isobar after emitting   -particle, then

find the energy released in this process.

(c) If M  Z0 , A  is nuclear mass of most stable isobar after emitting   -particle, then find

the energy released in this process.

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Q5. The mass of a nucleus with Z protons and A nucleons is

1
M  Z , A  2   
 f A  yZ  zZ 2 
c

Here f  A is function of A , y  4aa and z  ac A

1
3
 4aa A1

aa and ac are constants of suitable dimensions. For fixed A , find the expression of Z for

the most stable nucleus.

Q6. Find out the energy needed to extract a neutron from 20 Ca 40 using liquid drop

model. (Given av  14.1 MeV , as  13.0MeV , ac  0.595MeV , aa  19MeV , a p  33.5MeV )

Q7. The given empirical mass formula is

M  A, Z   0.99198 A  0.000841 Z  0.01968 A 

0.0007668 Z 2
2/3
 0.09966
 Z  A / 2  8 2

1/ 3
A A
in atomic mass unit. Find the mass difference in  - decay for nuclei 12 Mg .
 27

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Solution
Assignment-2 Nuclear Models

Z(Z  1) (A  2Z) 2 ap
Solution.1: Since B  a v A  a s A 2/3
 ac  a a (  , 0)
A1/3 A A3/4

 dB  ac a
    1/3  2Z0  1  a 2(A  2Z0 )(2)  0
 dZ Z Z0 A A

ac 4a  ac 4a  ac
1/3 
 2Z0  1  a  A  2Z0   0  2Z0  1/3  a    A1/3  4a a
A A A A 
 ac 
 4a a  A1/3  1/3
 Z0     Z  4a a  a c A
 ac 4a  2a c A 1/3  8a a A 1
0
2  1/3  a
A A 

 A  2Z 
2
3
Solution.2: B  A, Z   16 A  20 A 2/3
 Z 2 A1/ 3  30
4 A
dB 3
 0   ZA1/ 3  120
 A  2Z  Z  0
dZ 2 A

 120
 A  2Z    3 ZA1/ 3  240Z  3 ZA1/ 3  120
A 2 A 2
240 Z  3 1   1  A
 1  2 / 3 
 120  Z 1  2 / 3 

A  480 A   160 A  2
1 1
dB A  A2 / 3  27  27 2 / 3 
For most stable nuclei  0  Z   1    1  
dZ Z Z  2  160  2  160 
1
 9  160
 Z   13.5 1    13.5   14.3  14
 160  151

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 A  1   mn  m p 
ac
Solution.3:  M Z1  M Z 
A1/3

14
 2.962 103  931.5  a c  0.0014  931.5
151/3

14 151/3
 2.7572  a c  1.3041  a c  4.06   0.76 MeV
151/3 14

  2.962 103  4   0.000844   a c

14
151/3

3.542
ac   0.58 MeV  1u=931.5MeV  .
6.08

Solution.4: (a) Since, M  Z, A   A  Z  Z2 for given A

For nuclear charge  Z 0  of “most stable” nuclei

 M  
   0    2Z0  0  Z0  .
 Z Z Z0 2

So mass of the “most stable” isobar is

M  Z0 , A   A  2Z0 Z0  Z0 2  =–2Z0   M  Z0 , A   A  Z0 2

Also, M  Z, A   A  2Z0 .Z  Z2

The difference in masses for odd A is:

M  Z, A   M  Z0 , A   2Z0 Z  Z2  Z02    Z  Z0 

2

(b) Q  M  Z, A   M  Z  1, A   M  Z, A   M  Z0 , A   M  Z  1, A   M  Z0 , A 

 1
Q    Z – Z0     Z  1  Z0      2  Z  Z0  –1  2   Z0  Z  
2 2

 2

 1
Thus Q  2   Z0  Z  
 2

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(c) Q  M  Z, A   M  Z  1, A   M  Z, A   M  Z0 , A   M  Z  1, A   M  Z0 , A 

 1
Q    Z – Z0     Z  1  Z0      2   Z  Z0  – 1  2   Z  Z0  
2 2

 2

 1
Thus Q  2   Z  Z0  
 2

1
Solution.5: Since M  Z , A    f  A   yZ  zZ 2 
c2 
 M 
For stable nucleus   0
 Z  Z  Z0

y 4aa A/ 2
 Z0    Z0 

2 z 2 ac A1/3  4aa A1   a 
1   c  A2/3
 4aa 
Solution.6:
Let 20 Ca 40  20 Ca 39  0 n1

So according to liquid drop model

20  19 19  40  40
2

B  20 Ca 40
  14.1  40  13  40 2/3
 0.595 
401/ 3

40
33.5
 3/ 4
40

B  20 
Ca 40  347.97 MeV

20 19 19  39  40 
2

Similarly B  20 Ca 39
  14.1 39  13  39 2/3
 0.595   0
391/ 3 39

B  20 
Ca39  333.24 MeV

Energy needed to extract neutron from 20 Ca 40 will be

 347.97  333.24  14.707 MeV

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Solution.7:
For  - decay 12 Mg 27 13 Al 27

So M  A, Z   M  A, Z  1  0.00084112  13 
0.0007668 2
271/ 3

12  132 
0.09966  27  
2 2
27  
 12    13   
27  2   2  

 M  A, Z   M  A, Z  1  0.000814  0.00639  0.00738

 M  A, Z   M  A, Z  1  0.0018 u  0.002 u

(Which also shows that 12 Mg 27 is heavier than 13 Al 27 and it is unstable in  -decay).

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Assignment-3
(Nuclear Shell Model)
Q1. The shell model energy levels are in the following way
1s1/ 2 ,1 p3/ 2 ,1 p3/ 2 ,1d5/ 2 , 2s1/ 21d3/ 2 ,1 f7 / 2 , 2 p3/ 2 ,1 f5/ 2 , 2 p1/ 2 ,1g9/ 2 , 2d5/ 2 , 2d3/ 2 ,3s1/ 2 ,1h11/ 2 …..

Assuming that the shells are filled in the order written, what spin and parities should be
expected for the ground state of the following nuclei?
7
3 Li, 168 O, 178 O, 19
39 45
K , 21 Sc, 147 N , 23
64
Cu .

125
Q2. According to the shell model find the spin and parity of the two nuclei 51 X and 89
38Y .

Q3. According to single particle shell model, find

(a) The ground state angular momentum and parity of 168O .
27
(b) The ground state angular momentum and parity of 13 Al .
33
(c) The ground state angular momentum and parity of 16 S.

(d) The configuration of neutrons in the ground state of 49 Be .

Q4. From the shell model predictions find the ground state spin and parity of the following
nuclides 32 He ; 10
20 27
Ne ; 13 41
Al ; 21Sc .

5
Q5. The ground state of radio isotope 17
9 F has spin parity I     and first excited state has
2
9
1
I    suggest the configuration for the excited state.
2

15 1
Q6. The spin parity of the 8 O in the ground state is   , the spin parity and energy in some
2
excited states is given by

1
  :  5.18 MeV
2

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
5
  :  5.24
2

3
  :  6.18
2
Use the shell model to find the reasonable configuration for the excited state.

Q7. Calculate the nuclear magnetic moment and electric quadrupole moment of the nuclei
17
8 
O . Take 1 barn  100 fm2 

Q8. Calculate the Nuclear magnetic moment and electric quadrupole moment of the nucleus
33
16 
S . Take 1 barn  100 fm2 

Q9. From shell model calculate the expected quadrupole moment of 209
Bi  9 / 2  .

 Take  209 
1/ 3
 5.934 and 1 barn  100 fm 2 
Q10. Given that the proton has a magnetic moment of 2.79 magnetons and a spin quantum
number of one half, what magnetic field strength would be required to a produce proton
resonance at a frequency of 60MHz in nuclear magnetic resonance spectrometer?

Q11. In a nuclear magnetic experiment for the nucleus 25 Mn55 of dipole moment 3.46 N , the
magnetic field employed is 0.8T . Find the resonance frequency. You may assume
J  7 / 2, N  3.15 1014 MeVT 1

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Solution
Assignment-3 Nuclear Shell Model

Solution.1: For 7
3 Li : Number of Neutrons N  7  3  4 (even); No contribution from

neutrons
Number of protons Z  3 (odd),
the proton configuration according to Shell model is

1s1/ 2  1 p3/ 2 
2 1

The spin is due to third proton in the p3/ 2 state

3
 spin  I  
2

Since l  1 , therefore the parity is;    1   1  1 (odd parity)

l 1


3
Thus, the spin-parity is written as; I    
2

For 16
8 O: N  16  8  8 (even); No contribution from neutrons

Z  8 (even); No contribution from protons

Since the spin is decided by odd number of nucleons (proton on nucleon), whereas the
16
8 O is an even-even nuclei i.e. it is a doubly magic nuclei and therefore I   0 

For 17
8 O : Z  8 (even); No contribution from protons
N  17  8  9 (odd), its configuration according to the Shell model is

: 1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2 

2 4 2 1

 I  5/ 2

and l  2 , thus the parity is    1   1  1

l 2


5

 Spin-parity is I  
2

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For 39
19 K : N  39 19  20 (even); No contribution from neutrons
Z  19 (odd), its configuration according to the Shell model is

: 1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2   2s1/ 2  1d3/ 2 

2 4 2 6 2 3

 the nuclear spin is I  3 / 2

and l  2 , thus the parity    1  1

l

 The spin-parity is I    3/ 2


For 45
21 Sc : N  45  21  24 (even); No contribution from neutrons

Z  21 (odd); its configuration according to the Shell model is

: 1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2   2s1/ 2  1d3/ 2  1 f7 / 2 

2 4 2 6 2 3 1

 the nuclear spin is, I  7 / 2

and l  3 , thus the parity is    1  1

l


7
 I  
2
For 14
7 N : In this nucleus the number of protons and number of neutrons are both odd,
Number of neutrons are N  7 (odd), its configuration according to the Shell

: 1/ s3/ 2  1 p3/ 2  1 p1/ 2 

2 4 1
model is

The spin due to neutron is I N  1/ 2, and lN  1

Number of protons are Z  7 (odd), its configuration according to the Shell

: 1s1/ 2  1 p3/ 2  1 p1/ 2 

2 4 1
model is

The spin due to proton is I p  1/ 2, and l p  1

1 1
Since I p  I N  lP  lN    1  1  3 (odd)
2 2
 The nuclear spin is I  I P  I N  1

and Parity is    1 N   1

l  lP 11
 1

 The spin-parity is I   1



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For 64
29 Cu : N  29  35 (odd), its configuration according to the Shell model is

:1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2   2s1/ 2  1d3/ 2  1 f7 / 2   2 p1/ 2  1g9/ 2 

2 4 2 6 2 4 3 1 5

 I N  9 / 2 and lN  4

and Z  29 (odd), its configuration according to the Shell mode is

: 1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2   2s1/ 2  1d3/ 2 1 f7 / 2   2 p1/ 2 

2 4 2 6 2 3 1

 I P  1/ 2 and lP  1

1 9
Now, I P  I N  lP  lN    1  4  10 (odd)
2 2
1 9
Thus, the nuclear spin is I  I P  I N   5
2 2

    1   1
l N  lP 41
and parity  1

 The Spin-parity is I    5


Solution.2:
125
51 X ; Z  51 and N  74

Z  51:
1s1/ 2  1p3/ 2  1p1/ 2  1d5/ 2   2s1/ 2  1d3/ 2  1 f7 / 2   2 p3/ 2  1 f5/ 2   2 p1/ 2  1g9/ 2  1g7 / 2 
2 4 2 6 2 4 8 4 6 2 10 1


7 7
I and l  4 . Thus, the spin-parity   
2 2
Y ; Z  38 and N  51
89
38

N  51:
1s1/ 2  1p3/ 2  1p1/ 2  1d5/ 2   2s1/ 2  1d3/ 2  1 f7 / 2   2 p3/ 2  1 f5/ 2   2 p1/ 2  1g9/ 2  1g7 / 2 
2 4 2 6 2 4 8 4 6 2 10 1


7 7
 I  and l  4 . Thus, the spin-parity   
2 2

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Solution.3:
(a) Z  8, N  8  even  even   Spin  0 and Parity  1 .

The spin-parity is I   0 
(b) Z  13, N  14  odd  even 

5
Z  13; 1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2   I  , l  2
2 4 2 5

2
5
Thus Spin  and Parity   1   1  1 .
l 2

2

5

Thus, the spin-parity is I   
2
(c) Z  16, N  17  even  odd 

3
N  13; 1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2   2s1/ 2  1d3/ 2   I  , l  2
2 4 2 6 2 1

2
3
Thus Spin  and Parity   1   1  1 .
l 2

2

3
Thus, the spin-parity is I    
2
(d) Z  4, N  5  even  odd 

3
N  5; 1s1/ 2  1 p3/ 2   I  , l  1
2 3

2
3
Thus Spin  and Parity   1   1  1 .
l 1

2

3

Thus, the spin-parity is I   
2

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Solution 4:
For 32 He : Z  2 (even), no contribution from protons

N  3  2  1 (odd) : 1s1/ 2 
1

Thus spin I  1/ 2 and l  0 . Parity     1  1

l


1 
 The spin-parity is I   
2

Z  N  10 (even). It is a doubly magic number therefore I    0 

20 
For 10 Ne :

For 27
13 Al : N  27 13  14 (even), no contribution from neutrons

Z  13 (odd), its configuration according to the Shell model is

: 1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2 

2 4 2 5

Thus, the spin and parity due to proton is I P  5 / 2 and lP  2;    1  1

2


5
 The nuclear spin-parity is I    
2
For 41
21 Sc : N  41  21  20 (even), no contribution from the neutrons

Z  21 (odd) : 1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2   2s1/ 2  1d3/ 2  1 f7 / 2 

2 4 2 6 2 4 1

Thus spin  I   7 / 2, l  3 and parity     1  1

3


 7
 I  
2
Solution 5:
For 17
F9 : N  17  9  8 (even), no contribution from neutrons

Z  9 (odd) : 1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2 

2 4 2 1

 I  5 / 2, and    1  1
l


5 
Thus, the spin-parity of ground state is I   
2

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
1 
Now, spin-parity of the excited state is I    .
2
This can be obtained by exciting a proton form 1p1/ 2 level to 1d5/ 2 level. le.

1s1/ 2  1p3/ 2  1p1/ 2  1d5/ 2 

2 4 1 2

The spin parity of the nuclei is decided by the odd proton in the 1p1/ 2 level.

 I  1/ 2, l  1 thus    1   1  1

l 1


1
 I    .
2
Solution 6:
For 15
8 O : Z  8 (even), No contribution from protons

N  15  8  7 (odd) : 1s1/ 2 1 p3/ 2  1 p1/ 2 

4 1

 I  1/ 2, l  1,    1  1
l


1
Thus I   
2
For the excited state:

 1
I    : this is possible by promoting one nucleon from 1s1/ 2 level to the 1p1/ 2 level
2

 1s1/ 2  1 p3/ 2  1 p1/ 2 

1 4 2

I  1/ 2, l  0,    1  1
l
Thus

 1
 I  
2
For the excited state:

3
I P    : this is possible by promoting one nucleon from 1p3/ 2 level to the 1p1/ 2 level
2

 1s1/ 2  1p3/ 2  1p1/ 2 

2 3 2

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I  3/ 2, l  1,    1  1
l
Thus

 3
 I  
2
For the excited:

5
I P    : this is possible by promoting one nucleon from 1p1/ 2 level to the 1d5/ 2 level
2

 1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2 

1 4 0 1

I  5 / 2, l  0,    1  1
l
Thus

5
 I   
2

Solution 7:
17
8 O: Z  8 (even) no contribution from protons
N  9 (odd) nuclear contributes

1s1/ 2  1p3/ 2  1p1/ 2  1d5/ 2 

2 4 2 1

Thus, I  5 / 2, l  2, s  1/ 2
1 5
 J  l s  2 
2 2
Therefore, the nuclear magnetic moment is
 1 1   1 1 
   I   gl  g s   N   I   0   3.82    N
 2 2   2 
2  
   1.91 N

(where,  N is the nuclear magnetons)

The electric quadrupole moment is

3  2J 1  2
Q  R
5  2J  2 

where R  1 2 A1/ 3 fm

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3  2  5 / 2 1 
Q      1  2   17   fm 
2 2/3 2

5  2  5 / 2 1 

 3 19 1030 m 2  0.032 1028 m2

 Q  0.032b

Solution 8:
For 33
16 S , Z  16 (even) No contribution due to proton

N  17 (odd) Neutrons Contribute

 1s1/ 2  1 p3/ 2  1 p1/ 2  1d5/ 2   2s1/ 2  1d3/ 2 

2 4 2 6 2 1

Thus, I  3 / 2, l  2, s  1/ 2
1 3
 J  l s  2 
2 2
Therefore, the nuclear magnetic moment is
 I   3 1   I  1 
    I   g l  g s   N     I  3 /1  0   3.82   N
 I  1   2 2   I 1  2 
I 3/ 2 3/ 2
  1.91  N  1 91  N  1 91 N
I 1 3 / 2 1 5/ 2
   1.148 N

 3 
 
3  R2   3
2 I  1  2 1  2 3 2 2
2
and Q    R    R
5  2I  2 3
5  2  2  5 5/ 2
 2 

where R2  1.2 A1/ 3  ln   1.2  33  fm

2/ 3

3 2
 Q     1.2   33  fm 
2 2/3 2

5 5
 3.556 1030 m 2  0.036 1028 m 2
Q  0.036b

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Solution.9:
Quadrupole Moment of Nuclei using Shell Model
2 j 1 3
 r 2  barn where  r 2  R 2 ; R  R0  A .
1/ 3
Q
2  j  1 5
3 3
 r 2  R02  A   1.2   209   30.4 fm 2  0.3 barn
2/3 2 2/3

5 5
2 j 1 9 1
Q   r 2  barn   0.3 barn  0.22 barn
2  j  1 2  9 / 2  1

Solution 10: The resonance frequency v is given by


v
Ih
where  is the magnetic moment, B the magnetic induction, I the nuclear spin in units

of . A nuclear magneton N  5.05 1027 JT 1 .

vIh 60 1016  1/ 2   6.625 1034

B   1.4T
 2.79  5.05 1027

B 3.46  3.15 1014 1.6 1013  0.8

Solution 11: f  
Ih 7 34
   6.63 10
 
2

 6.012 106 Hz  6.012MHz

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Assignment-4
(Radioactivity)
Q1. The half-life of radioactive radon is 3.8 day. Find the time at the end of which 1 / 20 
th
of
the radon sample will remain undecayed. (Given log10 e  0.4343 )

Q2. A radioactive material decays by simultaneous emission of two particles with respective
half-lives 1620 and 810 year. Find the time, in year, after which one-fourth of the
material remains.

Q3. A 280 days old radioactive substance shows an activity of 6000 dps, 140 day later its
activity becomes 3000 dps. What was its initial activity?

Q4. A small quantity of containing Na 24 radio nuclide (half-life = 15 hour)of activity 1.0
micro curie is injected into the blood of a person A sample of the blood of volume 1cm 3
taken after 5 hour shows an activity of 296 disintegration per minute. Determine the total
volume blood in the body of the person. Assume that radioactive solution mixes
uniformly in the blood of the person.
( 1curie  3.7  1010 disintegration per second)

Q5. A radioactive sample emits n -particles in 2 sec. In next 2 sec it emits 0.75n -particles.
What is the mean life of the sample? Given ln 2  0.6931 and ln 3  1.0986

238
Q6. A radioactive sample of U decays to Pb through a process for which the half-life is
4.5  10 9 year. Find the ratio of number of nuclei of Pb to 238
U after a time of 1.5  109

year. Given 2  1.26 .

1/ 3

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Q7. In an ore containing uranium, the ratio of U 238 to Pb 206 nuclei is 3. Calculate the age of
the ore, assuming that all the lead present in the ore is the final stable product of U 238
Take the half-life of U 238 to be 4.5  10 9 year.

238
Q8. In the mixture of isotopes normally found on the earth at the present time, 92 U has an
235
abundance of 99.3% and 92 U has an abundance of 0.7%. The measured lifetimes of

these isotopes are 6.52  109 years and 1.02  109 years, respectively. Assuming that they
were equally abundant when the earth was formed, the estimate age of the earth, in years.

Q9. Nuclei of a radioactive element A are being produced at a constant rate  . The element
has a decay constant  . At time t  0 , there are N 0 nuclei of the element.

(a) Calculate the number N of nuclei of A at time

(b) If   2 N 0  , calculate the number of nuclei of after one half-life of A , and also the

limiting value of N as t  

Q10. A radioactive nucleus X decays to a nucleus Y with a decay constant  X  0.1s 1 .

 1 
Y further decays to a stable nucleus Z with a decay constant Y    s 1 Initially there
 30 
are only X s nuclei and their number is N 0  10 20 .

Set, up the rate equations for the populations of X , Y and. Z . The. Populations of
Y nucleus as a function ‘of time is given by

 N 0X 
N Y t    exp Y t   exp  X t . Find the time at which N Y is maximum

 X   
Y 

and determine the populations of X and Z at that instant.

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Solution
Assignment-4 Radioactivity
0.693
Solution.1: N  N 0e / t where    0.18
3.8
N0
 N 0 e 018t or log10 20  0.18  t  log10 e
20
1.3
or 1.3  0.18  0.4343 t or t  or t  16.5 days
0.18  0.4343
N
Solution.2:  e t
N0

There is a simultaneous emission of two particles

N N
  e  1  2 t  0  e  1  2 t or log 4   1  2  t log e
N0 4 N0

0.693 0.693
Now 1  2 
1620 810
 1 1  2.303  0.6  1620
 2.303  2  0.3  0.693    t or t  or t  1080 years
1620 810  0.693  3
Solution.3:
1 N 1 A  1  A 
For radioactive disintegration,   ln 0  ln  0  or   ln  0 
t N t  A 280  6000 

1  A 
also   ln  0 
 280  140   3000 
1  A  1  A   A   A 
 ln  0   ln  0  or 3ln  0   2 ln  0 
280  6000  420  3000   6000   3000 

A3  6000 
3 2 3
 A   A 
 0    0  or 02 
 6000   3000  A0  3000 2

6  6  6 109
or A0  or A0  24  103 or A0  24000
3  3 106
Initial activity =24000 dps

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Solution.4:

For radioactive decay, R  R0 e  t or   t   ln

R
R0

2.303 R  0.693
or t  log10  0  , where   per hour
  R  15

Half-life T  of the radio nuclide Na 24  15hours

Volume of blood  1 cm3 taken as a sample

After 5 hours, R  296 dpm in the blood sample.

2.303  R   R  5  0.693 R0
t log10  0  or log10  0   or  0.10033
0.693  296   296  2.303 15 296
15
373
or R0  296  1.26  373dpm or R0  dps
60
Activity of one micro curie 106 curie  3.7  104 dps

 373 
If activity is   , volume of blood 1 cm
3

 60 

3.7  104  60 3
If activity is 3.7  104 dps volume  cm
373
 Volume of blood  5951.7 cm 3 or Volume of blood  5.95 litre
 Total volume of blood in the body of person is 5.95 liter.
Solution.5:
Let N 0 be the initial number of nuclei. So, after 2 seconds amount remaining  N 0 e 2 

 Amount dissociated, n  N 0 1  e 2   (i)

Again after 2 seconds amount remaining  N 0 e 2 e 2

 Amount dissociated, 0.75n  N 0 e 2  N 0 e 2 e 2


 0.75  N 0 e 2 1  e 2  (ii)
Solving equations (i) and (ii)

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

0.75n N 0 e 2  1  e 2   3 4
 
or, e  2    e 2  or, ln e 2  ln 4  ln 3
n N0 1 e 
2
 4 3

or, 2  2 ln 2  ln 3 or, 2  2  0.6931 1.0986 or,   0.1438sec1

1
 Mean life   6.954second or mean life  7 seconds

Solution.6:
Radio of number of nuclei of Pb to U 238 .
U 238 decays and Pb is formed through the radioactive process.
n
1
N  N0   where N 0  Initial number of radioactive materials
2
N  Number of radioactive materials left, n  Number of half lives
time of decay 1.5  109 1
or n   
half life 4.5  109 3
1/ 3
1
NU  N 0  
2

1
1/ 3
  1 1 / 3 
N Pb  N 0  NU  N 0  N 0    N 0 1    
2   2  

  1 1 / 3 
N 0 1    
N Pb   2   1
 1/ 3
 1/ 3
 1  21 / 3  1  1.26  1  0.26
NU 1 1
N0    
2 2
Solution.7:
Initial number of U 238 4 N 4
238
  0 
Final number of U 3 N 3
N0  4.5  109 
 log10  15 109 log10 4  log10 3
2.303 4
t  log10  2.303 
 N  0.693  3

 15  109 0.6021 0.4771  15  109  0.1250  1.875  109 years

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Solution.8:
If the number of 92 U 238 nuclei originally formed is N , the number present now is

N238  Net / T  Net / 6.52

where t is elapsed time in units of 109 year and T is life time of U . Since the number of

92 U 235 nuclei originally formed. The number now present is

N235  Net /1.02

The present abundance of 92 U 235 is

N 235 N Ne t /1.02 1 4.96

7 103   235   t / 6.52
 e0.827 t  3
 143  t   6.0
N 238  N 235 N 238 Ne 7 10 0.827

That is, the elapsed time is t  6.0 109 years.

Solution.9:
Nuclei of a radioactive element A are being produced at a constant rate  .
Decay constant of element =  , At t  0 , nuclei of element present N 0

(a) Number N of nuclei of A at time t :

dN
Net rate of formation of nuclei of element A 
dt
N t
dN dN dN 1
N   N  0 dt ln    N   N  t
N
    N or  dt or or 
dt   N 0
 0

   N    N
ln   t or  e  t
   N 0    N 0

or   N  e  t   N 0  or N 

1
    N  e 
0
 t
(i)

ln2
(b) (i) If   2N 0 , t  half life 

1
N   2 N 0   2 N 0   N 0  e   t 


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N 0
or N 

2  e ln2  

 Here e
 ln  2  1
 
2
N 0  1  3N 0 3
or N  2  2   2 or N  N0
   2
(ii) when t   and   2N 0

 2N 0
N   2N 0 or N  2N 0
 
Solution.10:
Radioactive nucleus X decays to a nucleus Y
Y further decays to a stable nucleus Z
Initially there are only X nuclei and their number are N 0  10

(i) Rate equations:

At an instant t ,
Let number of nuclei of Y  NY and number of nuclei of  N 2
 Rate equations of populations of X , Y and Z are

 dN X 
    X N Y
 dt 

 dNY 
    X N X  Y N Y
 dt 

 dN Z 
   Y N Y
 dt 
(ii) Time t at which N Y is maximum:
N0X
Given: N Y t  
 X  Y

e Y t  e  X t 

For NY t  to be maximum, N Y t   0
d
dt
dNY
From (ii),   X N X  Y N Y   X N X  Y N Y
dt

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N0X
 
or  X N 0 e  X t  Y
 X  Y 

e Y t  e  X t 
 X  Y e   t  e  
Y Xt
 X
or  or  1  e  X  Y t
1
Y e  t X
Y

 
     0.1 
or ln X    X  Y t or t 
1 1 
ln X  or t  ln
 Y   X  Y  Y  0.1 
1  1 
 
30  30 
or t  15 ln3 or t  16.48 sec
(iii) Populations of X and Z when N Y is maximum:

N X  N 0 e X t or N x  10 e 0.116.48

20

or N X  1.92  1019
Since  X N X  Y N Y according to equation (iv),

 
N X X  0.1 
N Y  or N Y  1.92  1019   
Y  1 
 
 30 
or N Y  5.76  1019
Again N Z  N 0  N X  N Y

or N Z  1020  1.92  1019   5.67  1019 

or N Z  1019 10  1.92  5.76

N Z  2.32  1019
Hence (i) Rate equations for populations X , Y , Z are given in equations (i), (ii) and (iii)
(ii) Time at which N Y is maximum  16.48 sec

(iii) Population of X  N X  1.92  1019

Population of Z  N Z  2.32  1019

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Assignment-5
(Nuclear Reaction (Fission and Fusion)
Q1. A nucleus with mass number 220 initially at rest emits an  - particle. If the Q value of
the reaction is 5.5 MeV , calculate the kinetic energy of the  - particle
Q2. If a star can convert all the He nuclei completely into oxygen nuclei, then find the energy
released per oxygen nuclei. [Mass of He nucleus is 4.0026 amu and mass of oxygen
nucleus is 15.9994 amu]
Q3. A star initially has 10 40 deuterons. It produces energy via the processes

1 H 2 1 H 2 1 H 3  p , and 1 H 2 1 H 2 2 He4  n .

If the average power radiated by the star is 1016 W , then the time required for the deuteron
supply of the star to be exhausted.
The masses of the nuclei are as follow:

 
M H 2  2.014 amu ; M  p   1.007 amu; M  n   1.008 amu; M He4  4.001amu .  
Q4. It is proposed to use the nuclear fusion reaction
2
1 H 12 H  42 He
in a nuclear reactor of 200 MW rating. If the energy from the above reaction is used with
25 percent efficiency in the reactor, how many grams of deuterium fuel will be needed
2 4
per day. (The masses 1 H and 2 He are 2.0141 atomic mass unit and 4.0026 atomic
mass unit respectively).
Q5. A nucleus X , initially at rest, undergoes alpha decay according to the equation,

Z Y 
X 228
A
92

(a) Find the values of A and Z in the above process.

(b) The alpha particle produced in the above process is found: to move in a circular track.
of radius 0.11 m in a uniform magnetic field of 3 tesla. Find the energy (in MeV )
released during the process and the binding energy of the parent nucleus X .
 
Given that: mY   228.03u; m 10 n  1.009u ,  4
2   
He  4.003u; m 11 H  1.008u.

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248
Q6. The element Curium 96 Cm has a mean life of 1013 second, its primary decay modes are

spontaneous fission and  -decay, the former with a probability of 8% and the latter with
a probability of 92%. Each fission releases 200 MeV of energy. The masses involved in
 -decay are as follows:
248
96 Cm  248.072220u, 94
244
Pu  244.064100u and 42 He  4.002603u . Calculate the power
MeV
output from a sample of 1020 Cm atoms. ( 1 u  931 ).
c2

235
Q.7. In a nuclear reaction U undergoes fission liberating 200 MeV of energy. The reactor
has 10% efficiency and produces 1000 MW power. If the reactor is to function for 10
year, find the total mass of uranium required.

Q8. A nucleus at rest undergoes a decay emitting an   particle of de-Broglie

wavelength   5.76  1015 m . If the mass of the daughter nucleus is 223.610 amu and
that of the   particle is 4.002 amu , determine the total kinetic energy in the final state.
MeV
Hence, obtain the mass of the parent nucleus in amu. ( 1 amu  931.470 )
c2

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Solution
Assignment-5 Nuclear Reaction & Fission and Fusion
Solution.1:
Linear momentum is conserved.  p1  p2

but p  2mK where K  kinetic energy

 2  216m  K1  2  4m  K 2

where K 2 is for   particle and K1 is for nucleus.

or 216 K1  4 K 2 or K 2  54 K1 (i)

Given: K1  K 2  5.5MeV (ii)

From (i) and (ii),

K1  54K1  5.5  MeV 

5.5 1
or 55K1  5.5( MeV ) or K1  MeV or K1  MeV (iii)
55 10
54
 K 2 54  K1 or K 2  MeV or K 2  5.4 MeV (iv)
10
 Kinetic energy of   particle  5.4 MeV

Solution.2: 4  2 He4  8 O16

Binding energy m  931.5 MeV   4  4.0026  15.9994   931.5 MeV  10.24MeV

 Energy released per oxygen nuclei  10.24 MeV

Solution.3:

1 H 2 1 H 2 1 H 3  p ; 1 H 2 1 H 3 2 He4  n

by adding 3  1 H 2  2 He4  p  n

m  3  2.014    4.001  1.007  1.008 or m  0.026 amu

Mass is converted into energy

1 amu  931.5 MeV or 1 amu  931.5 106 1.6 1019 J

 Energy from  m   0.026  931.5 1.6 1013 J  3.87 1012 J

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 Energy released by 3 deuterons  3.87  1012 J

3.87 1012 1040
 Energy for 10 deuterons 
40
 E  1.29 1028 J
3
Energy Energy
Power= or time =
Time Power
1.29 1028
Time = sec  1.29 1012 sec
1016
Time is of the order of 1012 sec.
Solution.4:
Deuterium fuel needed per day in the reactor:
Mass defect provides the energy in the reactor:
m  22.0141  4.0026  4.0282  4.0026  0.0256amu

 E  m931.5 MeV   0.0256931.5  1.6  1013 J  3.82  1012 J

Efficiency = 25%
25
 Energy available due to fusion of two deuterium nuclei   3.82  1012 J
100
 Energy available  9.55  1013 J

9.55  1013
Energy available per deuterium nuclei  J
2
 
Total energy needed  power  time  200  106  24  60  60  1.728 1013 J

1.728  1013
Number of deuterium nuclei required   0.362  1026
9.55
 1013
2
 Mass of deuterium required  m
m  Avogardo number
  0.362  10 26
2
0.362  1026  2
or m  or Mass = 120.26gram
6.02  1023
Hence 120.26gram of deuterium fuel will be needed per day in the nuc1ear reactor

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Solution.5:
(a) A  228  4  232, Z  92  2  90

mv 2
(b) The magnetic force qvB provides centripetal force to   particle for its circular
r
motion.
mv 2 qr B
  qvB or v 
r m

or v 
2  1.6  10  0.11 3
19
 v  1.58  107 ms 1 (i)
4.003 1.67  10  27
Linear momentum is conserved in the process.
mY vY  m v (ii)

Nucleus X is initially at rest

Total kinetic energy  K   K r
2
1 1  m v  1 1 m2 v2 1  m 
 m v  mY 
2
  m v 
2
 m v2 1   
2 2  mY  2 2 mY 2  mY 


1
   2
 4.003  1.67  10  27  1.58  10 7 1  
4.003 
J
2  228.03 

13 8.49  1013

 8.49  10 J  5.31MeV (iii)
1.6  1019
5.31
Equivalent mass   0.0057 amu
931.5
X is parent nucleus, Y is daughter nucleus
Also m X  mY  m  equivalent mass
 228.03  4.003  0.0057  232.0387 amu
Now, Nucleons = 92 protons + 140 neutron
Binding energy  92 1.008  1401.009  232.0387
B.E.  1.9573amu or B.E.  1.9573 931.5 MeV
 B.E. of nucleus X  1823MeV (iv)

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Solution.6:
The primary decay modes of 96 Cm 248 are
(i) Spontaneous fission (probability 8%)
(ii)   decay (probability 92%).
The nuclear reaction is given below:

96 Cm 248  94 Pu 244  2 He 4

 Mass defect in the reaction  m

 m  mass of Cm  mass of Pu  mass of He
or m  248.072220 244.064100 4.002603 or m  0.005517u

Energy released  0.00515 931 MeV

E  5.136 MeV

Given: E f  Each fission released 200 meV of energy

Mean life of Cm  1013 sec

1
   10 13 s 1
mean life
dN
again  N where N is given to be 10 20
dt
 Rate of decay  1013 1020  or Rate of decay 107 dps

of these 107 dps , 8% are in fission and 92% are in   decay process.
8
Energy released per second due to fission   107  200  16  107 MeV
100
92
Energy released per sec due to   decay   107  5.136  4.725  107 MeV
100
Total energy released per second  16  4.725 107 MeV  20.725 107 MeV

 Power output = Energy per second

   
 20.725  107  1.6  1013 J / s  3.316  105 W  3.32  105 W

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Solution.7:
Power output
Efficiencyof reactor 
Power input

1000 106
Power input   1010 W
10 / 100
 Equivalent energy  Power input  time  1010  10  365  24  60  60 J

 31.1536 1018 J (i)

Now 1 fission releases 200 MeV energy

 1 fission releases energy  200  1.6  10 13 J (ii)

From (i) and (ii),
3.1536 1018
Number of fissions  or Number of fissions  9.85  10 28
200  1.6  1013
 Number of U 235 atoms  9.85  1028

Avogadro number  6.02  1026 per kg

 6.02  106 atoms are contained in 235kg of U 235

235 9.85  1028

 9.85  1028 atoms are contained in mass  kg
6.02  1026
 Mass of U 235  38451kg
Solution.8:
Let Z X A represent the parent nucleus. The parent nucleus decays emitting   particles

 Z X A  Z 2 Y A4  2 He 4
Mass of daughter nucleus Y  223.610 amu , Mass of   particle  4.002 amu

De-Broglie wavelength of   particle,   5.76  1015 m

So, momentum of   particle would be
h 6.63 1034
p    p  1.1511019 kg  m / s
 5.76 1015

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From law of conservation of linear momentum, this should also be equal to the linear
momentum the daughter nucleus thus p  pY

Let K and K Y be the kinetic energies of   particle and daughter nucleus. Then total

kinetic energy in the final state is

p2 p2 p2  1 1  p2  M Y  m 
K  K  K Y   Y      
2m 2 M Y 2  m M Y  2  M Y m 

K 
1.15110  19 2
 4.002  223.610  1.67 1027 
2  4.002 1.67 10  223.611.67 10 
27 27

1 amu  1.67 1027 kg

1012
 K  1012 J  K  MeV  6.25 MeV
1.6 1013
6.25
Mass equivalent to energy K  amu
931.470
or m  0.0067amu  m X  mY  m  m

or mX  223.610  4.002  0.0067 amu or m X  227.67 amu

 Mass of parent nucleus  227.62 amu

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Assignment-6
(Particle Physics)
Q1. Indicate with an explanation, whether the following interactions proceed through the
strong, electromagnetic or weak interaction or whether they do not occur.
(i)       

(ii)      

(iii)  0    r
(iv) p  n  e  e

(v)    p   0  0

(vi)    p  K 0  0

(vii) e  e      

Q2. Particles X and Y can be produced by strong interaction

K  p  K  X

K  p  0 Y

identify the particles X 1321 MeV  and Y 1192MeV  and deduce their quark content.

Q3. Which of the following processes are allowed in electromagnetic interaction and which
one allowed in weak interaction via the exchange of a single W  or Z 0 ?
(a) K    0  e  e (b)  0     e   e

Q4. A particle X decays at rest weakly as follows X   0      identify the particle X

and give its quark content.

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Q5. (a) What are the quark constituents of the states   , 0 ,   ,  

(b) Assuming the quarks are in the states of zero angular momentum, what fundamental
difficulty appears to be associated with the  state, which has I  3/ 2 and how it is
resolved?

Q6. Use the quark model to determine the quark composition of

(a)   ,   , n and p (b) K  , K  ,   ,  

Q7. Which of the following reaction are allowed and which forbidden as under weak
interaction?
(a)    p      (b)  e  p  e     p

(c) K    0      (d)      e  e

Q8. Which of the following interaction is allowed or forbidden? Also mention the interaction
type.
(a) K   p       (b)       K   K 

(c)    p   0 (d)          

Q9. (a) Write the Gellmann’s equation for quarks. Are quarks observed as free particle?
(b) Write the quark content of D  - meson and  0 -meson.

Q10. Indicate how the following quantities will transform under the P (space inversion) and
T (time reversal) operation:
(a) Position coordinate  r  (b) Momentum vector  p 

(c) Electric field E  (d) Magnetic field B  

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Solution
Assignment-6 Particle Physics

Solution 1: (i)        

q: 1 1 0 : Conserved
Spin: 0 1/2 1/2 : Conserved
L : 0 1 1 : Conserved

This is a allowed interaction since neutrino is involved, Parity is violated, therefore this is
allowed through weak interaction
(ii) z      z

q: 1 1 0 : Conserved
q: 1 1 0 : Conserved
Spin: 1/ 2 1/ 2 1/ 2 : Conserved
Lz : 1 0 1 : Conserved

L : 0 1 0 : Not conserved

This is not a allowed interaction because lepton number is violated.

(iii) 0    r
q: 0 0 0 : Conserved
Spin: 1/ 2 1/ 2 1: Conserved
B : 1 1 0 : Conserved
I: 1 0 0 : Not Conserved

I3 : 0 0 0 : Conserved

S: 1 1 0 : Conserved
This is allowed interaction through electromagnetic interaction

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(iv) p  n  e  e
q: 1 0 1 0 : Conserved

1 1 1 1
Spin: : Conserved
2 2 2 2
Le : 0 0  1  1 : Conserved

B  1 1 0 0 : Conserved
This is an allowed interaction. Since neutrino is involved, this is allowed through weak
interaction.
Note: This interaction is allowed if proton is bound but forbidden when proton is free
because proton is lighter than the sum of the masses of the product particles.
(v)    p   0  0
q: 1 1 0 0 : Conserved
1 1
Spin: 0 0 : Conserved
2 2
B: 0 1 0  1 : Conserved

1 1
I: 1 1 : Conserved
2 2
1
I3 : 1  0 0 : Not Conserved
2
S: 0 0 0  1 : Not Conserved

This is not allowed interaction become I 3  0 and S  0

(vi)    p  K 0  0
q: 1 1 0 0 : Conserved

1 1
Spin: 0 0 : Conserved
2 2
B: 0 1 0  1 : Conserved
1 1
I: 1 1 : Conserved
2 2

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1 1
I3 :  1   0 : Conserved
2 2
S: 0 0 1  1 : Conserved
This is allowed through strong interaction.
(vii) e  e      
q: 1 1 1  1 : Conserved

1 1 1 1
Spin: : Conserved
2 2 2 2
Le : 1 1 0 0 : Conserved

L : 0 0 1  1: Conserved

This is a allowed interaction. Since a lepton – anti lepton pair is involved, this is allowed
through weak interaction.
Solution 2: (i) K   p  K   X

q : 1 1 1  1
1 1
Spin: 0 0
2 2
B: 0 1 0  1
S : 1 0 1  2 
The particle X is Xi -hyper on    . The quark content of  is dss

(ii) K  p  0 Y

q : 1 1 0 0
1 1
Spin: 0 0
2 2
B: 0 1 0  1
S : 1 0 0  1
Therefore the particle Y is sigma hyperon  0  . The quark content of  0  is uds .

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Solution 3: (a) K    0  e  e

q: 1 0 1 0 : Conserved

1 1
Spin: 0 0 : Conserved
2 2
1
I: 1 0 0 : Not Conserved
2
1
I3 : 0 0 0 : Not Conserved
2
Str S :  1 0 0 0 : Not Conserved
It is forbidden as electromagnetic interaction because S  0 and also forbidden as weak
interaction because there is no strangeness changing current.
(b) 0    e  e

S: 0 0 0 0 : Conserved
1 1 1 1
Spin: : Conserved
2 2 2 2
Le : 0 0  1  1 : Conserved

B: 1 1 0 0 : Conserved
I: 1 0 0 0 : Conserved

I3 : 1 0 0 0 : Conserved

S: 1 1 0 0 : Conserved
It is allowed as electromagnetic interaction.
Solution 4: X   0     

q:  1 0 1 0 : Conserved

1 1
Spin:  0  : 0 : Conserved
2 2
L:  0 0 1  1 : Conserved

B:  0 0 0 0 : Conserved

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I: 1/ 2  1 0 0 : Must violate

I3 :  1/ 2  0 0 0 : Must violate

S:  1 0 0 0 : Must change by S  1

Therefore, the particle X is K  - meson. The quark content of K  - meson is us

Solution 5: (a)   : ddd , 0 : ddu ,   : uud ,   : uuu
(b) The fundamental difficulty is that Pauli’s principle is violated. For example consider
1
the spin at  . All the three d -quarks have to be aligned with the same jz   . The
2
same difficulty arises for   . The difficulty is removed by endowing a new intrinsic
quantum number (color) to the quarks. Thus, the three d -quarks in  or the three u -
quarks in   differ in color, Red (R), Green (G), and Blue (B).
Solution 6: (a)  : uus,  : dds, n : udd , p : uud

(b) K  : us , K  : ds,   : ud ,   : ud
Solution 7: (a) Forbidden because Lepton number is not conserved
(b) Forbidden because charge is not conserved
(c) Allowed
(d) Allowed
Solution 8: (a) K   p     
q: 1 1 1  1 : Conserved

1 1
Spin: 0 0 : Conserved
2 2
B: 0 1 1 0 : Conserved
1 1
I: 1 1 : Conserved
2 2
1 1
I3 :   1 1 : Conserved
2 2
S: 1 0 1 0 : Conserved
This (a) allowed through strong interaction

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(b)     K   K 
q:  1  1  1  1 : Conserved
1 1
Spin: 0 0 : Conserved
2 2
L :  1 1 0 0 : Conserved

1 1
I: 0 0 : Conserved
2 2
1 1
I3 : 0 0 : Conserved
2 2
S: 0 0 1  1 : Conserved
Since Leptons are involved, therefore this interaction cannot be strong but it is allowed
through electromagnetic interaction.
(c)   p   0
q: 1 1 0 : Conserved
Spin: 1/ 2 1/ 2 0 : Conserved
B: 1 1 0 : Conserved
I: 1 1/ 2 1: Not Conserved
I3 : 1 1/ 2 0 : Not Conserved

S: 1 0 0 : Not Conserved  S  1

Therefore, this is allowed through weak interaction.
Solution 9: (a) Gellmann’s equation is generalized as
Q
e
1

 I 3  B  S  C  B*  T
2

where B denotes Baryon number, S the strangeness , C the charm, B* the beauty of
bottom and T the top.
Quarks are not observed as free particle as they are confined in Hardons
(b) D  : cd

uu  d d
0 :
2

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Solution 10: (a) p : r  r

T: r r
dr
(b) Since p  mv  m
dt
 p: p  p  p : r  r  and T : p   p  T : t  t 

V
(c) Since E  
r
 P : E   E and T : E  E
(d) Since B  I  r
P:B  B ( p : r  r and p : I   I )
T : B  B ( T : r  r and T : I  I )

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Nuclear Physics (Objective Questions)

Q1. The radius of Ge nucleus is measured to be twice the radius of 49 Be . How many

nucleons are there in Ge nucleus?

(a) 64 (b) 72 (c) 82 (d) 86

208 26
Q2. The ratio of the sizes of 82 Pb and 12 Mg nuclei is approximately

(a) 2 (b) 4 (c) 8 (d) 16

ks
27
Q3. If the nuclear radius of Al is 3.6 Fermi, the approximate nuclear radius of 64Cu in
Fermi is
(a) 4.8 (b) 3.6 (c) 2.4 (d) 1.2

16
Q4. According to the empirical observations of charge radii, an 8 X nucleus is spherical

4
and has charge radius R and a volume V   R 3 . Then the volume of the 128
54Y nucleus,
3
zi
is
(a) 1.5V (b) 2V (c) 6.5V (d) 8V

Q5. Assume spherical symmetry of the nucleus ZA X , where Z is atomic number and A is
mass number of the nucleus. Then the nuclear density and nuclear particle density of


nucleus is of the order of: m p  mn  1.67  1027 kg and R0  1.2 fermi 
fi
(a) 1015 kg / m3 and 1040 nucleons / m3

(b) 1017 kg / m3 and 1044 nucleons / m3

(c) 1017 kg / m3 and 1040 nucleons / m3

(d) 1015 kg / m3 and 1044 nucleons / m3

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Q6. Which of the following statement is not correct?

(a) Coulomb barrier prevent the elements heavier than Iron but lighter than lead from
fissioning spontaneously.
(b) The amount of energy required to add a proton in nucleus is more than that of
neutron.
(c) Mirror nuclei have the same total Isotopic spin  I  but opposite third component of

isotopic spin  I Z  .

ks
(d) Heavy nuclei are radioactive but stable against neutron emission.
The amount of energy required to add a neutron in nucleus is more than that of proton.

Q7. Consider a nucleus with N neutrons and Z protons. If m p , mn and B.E . represents

the mass of the proton, the mass of the neutron and binding energy of the nucleus
respectively. Then mass of the nucleus is given by (and c is the velocity of light in free
space)
(a) Nmn  Zm p (b) Nm p  Zmn
zi
B.E. B.E.
(c) Nmn  Zm p  (d) Nm p  Zmn 
c2 c2

20
Q8. Let m p and mn be the mass of proton and neutron. M 1 is the mass of 10 Ne nucleus
40
and M 2 is the mass of a Ca nucleus. Then find the correct relation:
20

(a) M 1  10  m p  mn  , M 2 = 20  m p  mn  and M 2 = 2 M1
fi
(b) M 1  10  m p  mn  , M 2  20  m p  mn  and M 2 = 2M 1

(c) M 1  10  m p  mn  , M 2  20  m p  mn  and M 2  2M 1

(d) M 1  10  m p  mn  , M 2  20  m p  mn  and M 2  2M 1

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 
Q9. The measured mass of deuteron atom 12 H , Hydrogen atom 11H , proton  p  and  
neutrons  n  are 2.0141 u , 1.0078 u , 1.0073 u and 1.0087 u . Then the binding energy per

nucleon of the deuteron nucleus is:

(a) 1.11 MeV / nucleons (b) 2.22 MeV / nucleons
(c) 3.33 MeV / nucleons (d) 4.44 MeV / nucleons

238
Q10. The masses of a hydrogen atom, neutron and U atom are given by 1.0078 u ,

ks
92

238
1.0087 u and 238.0508 u respectively. The binding energy of U nucleus is therefore
92

approximately equal to
(a) 2200 MeV (b) 2000 MeV
(c) 1800 MeV (d) 1600 MeV

Q11. The atomic masses of 42

20 Ca  41.958622 u ,
41
20 Ca  40.962278 u and mass

of 10 n  1.008665 u , 11 p  1.007276 u . Then the energy needed to remove a neutron from

zi
the nucleus of the calcium isotope 42
20 Ca is:

(a) 11.5 MeV (b) 12.5 MeV

(c) 13.5 MeV (d) 14.5 MeV

Q12. The atomic masses of 42 41

20 Ca  41.958622 u , 19 K  40.961825 u and mass of
fi
1
0 n  1.008665 u , 11 p  1.007276 u . Then the energy needed to remove a proton from the

nucleus of the calcium isotope 42

20 Ca is:

(a) 11.3 MeV (b) 10.3 MeV

(c) 9.3 MeV (d) 8.3MeV

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 B.E. 
Q13. The following histogram represents the binding energy per particle   in
 A 
MeV as a function of the mass number  A  of a nucleus. A nucleus with mass number

A  180 fission into two nuclei of

equal masses.
8
In the process
6
(a) 180 MeV of energy is released B.E.
A 4

ks
(b) 180 MeV of energy is absorbed
(c) 360 MeV of energy is released 2
(d) 360 MeV of energy is absorbed
40
160 80
200 120
A
Q14. Binding energy per nucleon Vs mass number curve for nuclei W , X , Y and Z is
indicated on the curve. The process
8 .5 Y
that would release energy is:
(a) Y  2 Z  /  8 X
MeV 
zi
(b) W  X  Z 7. 5 W
(c) W  2Y
5 Z
(d) X  Y  Z

30 60 90 120


Q15. According to liquid drop model consider binding energy formula. Then the value of
fi
atomic number  Z 0  for the most stable nuclei among members of an isobaric

family  A
Z X  is: (where symbols have their usual meanings)
4aa  ac A1/3 4aa  ac A1/3
(a) Z 0  (b) Z 0 
2 a c  8 aa A 2ac A1/3  8aa A1

4aa  ac A1/3 4aa  ac A1/3

(c) Z 0  (d) Z 0 
2ac  8aa A1 2ac  8aa A1/3

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Q16. According to liquid drop model, which of the following equation represents the
nuclear mass  M  Z, A   for members of an isobaric family  A
Z X  (Here ,  and  are

some constant with suitable dimensions and  is pairing energy term).

(a) M  Z, A   A   Z  Z3  

(b) M  Z, A   A   Z2  Z3  

(c) M  Z, A   A   Z  Z2  

ks
(d) M  Z, A   AZ   Z2  Z3  

Q17. Consider the following expression for the mass of a nucleus with Z proton and A
nucleons: (where A is some odd number)
M  Z, A   A   Z  Z2

where ,  and  are some constant with suitable dimensions. If M  Z0 , A  is nuclear

mass of most stable nuclei of the isobaric family, then the difference in mass
 M  Z, A   M  Z0 , A   will be:
zi
2
(a)   Z – Z 0  (b)   Z – Z 0 
2
(c)   Z – Z 0  (d)   Z – Z 0 

Q18. Consider the following expression for the mass of a nucleus with Z proton and A
nucleons: (where A is some odd number)
M  Z, A   A   Z  Z2

where ,  and  are some constant with suitable dimensions. If M  Z0 , A  is nuclear

fi
mass of most stable isobar after emitting   -particle, then the energy released in this
process will be:
 1  1
(a) Q   2  Z 0  Z   (b) Q   2  Z  Z 0  
 2  2

 1  1
(c) Q   2  Z 0  Z   (d) Q   2  Z  Z 0  
 2  2

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Q19. Consider the following expression for the mass of a nucleus with Z proton and A
nucleons: (where A is some odd number)
M  Z, A   A   Z  Z2

where ,  and  are some constant with suitable dimensions. If M  Z0 , A  is nuclear

mass of most stable isobar after emitting   -particle, then the energy released in this
process will be:

ks
 1  1
(a) Q   2  Z 0  Z   (b) Q   2  Z  Z 0  
 2  2

 1  1
(c) Q   2  Z 0  Z   (d) Q   2  Z  Z 0  
 2  2

Q20. The mass of a nucleus with Z protons and A nucleons is

1
M  Z , A   f A  yZ  zZ 2 
2   
c
zi
1
Here f  A  is function of A , y  4aa and z  ac A 3
 4aa A1

aa and ac are constants of suitable dimensions. For fixed A , the expression of Z for the

most stable nucleus is

A/ 2 A/ 2
(a) Z  (b) Z 
a   a 
1   c  A2/3 1   c  A2/3
 aa   4 aa 
fi
A/ 2 A
(c) Z  (d) Z 
 a  1  A2/3
1   c  A2/3
 4aa 

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Q21. “Mirror Nuclei” are nuclei having same odd value of mass number A and have
interchangeable N and Z differing by one unit. According to the semi-empirical mass
A A
formula, the mass difference between the pair of mirror nuclei Z X and Y will
Z 1

approximately be: (where constants have their usual meaning)

ac aS
(a)  A  1   mn  m p  (b)  A  1   mn  m p 
A1/3 A1/3
ac aS
(c)  A  1   mn  m p  (d)  A  1   mn  m p 
A A

ks
Q22. According to nuclear shell model which includes spin orbit coupling, the spin and
parity of the ground state of 115B is
   
3 3 1 1
(a)   (b)   (c)   (d)  
2 2 2 2
41
Q23. The nucleus of 20Ca can be described by the single particle shell mode. Then

which of the following statement is true?

(a) The single particle states occupied by the last proton and the last neutron respectively
zi
are d3/2 and p3/2

41 7
(b) The ground state angular momentum and parity of 20 Ca is  
2
(c) The single particle state occupied by the last proton and the last neutron, respectively
are d3/2 and f5/2

41 7
(d) The ground state angular momentum and parity of Ca is  
fi
20
2
Q24. According to single particle shell model which of the following represents ground
state angular momentum and parity of 167 N

(a) J  2 and even parity (b) J  3 and even parity

(c) J  5 / 2 and odd parity (d) J  2 and odd parity

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Q25. According to single particle shell model, which of the following statements is
not true?

(a) The ground state angular momentum and parity of 168O is  0 

27 5
(b) The ground state angular momentum and parity of 13 Al is  
2

33 3
(c) The ground state angular momentum and parity of 16 S is  
2

ks

3 9
(d) The configuration of neutrons in the ground state of Be is   4
2
Q26. In the nuclear shell model, orbitals are filled in the order
2 4 2 6 2 4 8
1S1/2   2 P3/2   2 P1/2   3d5/2   2S1/2   3d3/2   4 f7/2  etc.

For an odd-odd nucleus a range of J p (spin-parity) is allowed. Which of the following is

not the allowed value for 189 F ?

(a) 0 , 2 , 4  (b) 1 , 3 , 5
zi
(c) 2 , 4  , 6  (d) 1 , 2 , 3

Q27. The neutron and proton form a deuteron bound state which is stable while there is
no bound state for two neutrons because?
(a) Nuclear forces are saturated
(b) Nuclear forces are spin dependent
fi
(c) Nuclear forces are charge independent
(d) Nuclear forces depends upon magnetic moment

Q28. The ground state of the deuteron is a

(a) pure 3S1 state (b) pure 3 P1 state

(c) mixture of 3S1 and 3 P1 states (d) mixture of 3S1 and 3 D1 states

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Q29. The ground state wavefunction of deuteron is in a superposition of s and d states.

Which of the following is NOT true as a consequence?
(a) It has a non-zero quadruple moment
(b) The neutron-proton potential is non-central
(c) The orbital wavefunction is not spherically symmetric
(d) The Hamiltonian does not conserve the total angular momentum
Q30. Deuteron has only one bound state with spin parity 1+, isospin 0 and electric
quadrupole moment 0.286 efm2. These data suggest that the nuclear forces are having

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(a) only spin and isospin dependence
(b) no spin dependence and no tensor components
(c) spin dependence but no tensor components
(d) spin dependence along with tensor components
238 234
Q31. The atomic ratio between the uranium isotopes U and U in a mineral sample
is found to be 1.8 10 4 . The half life of 238
U is 4.5  109 years , then half life of 234
U is:

(a) 2.5  105 years (b) 3.5  105 years

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(c) 2.5  10 4 years (d) 3.5  10 4 years

Q32. A radioactive sample contains 100 mg of radon 222 Rn , whose atomic mass is 222 u .
The half life of the radon is 3.8 day . Then the activity of the radon sample is

(a) 5.7  108 decay / sec (b) 5.7  1010 decay / sec

(c) 5.7  1012 decay / sec (d) 5.7  1014 decay / sec

Q33. The radio isotope 14C maintains a fixed proportion in a living entity by exchanging
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14
carbon with the atmosphere. After it dies exchange ceases and proportion of C
decreases continuously as 14C beta decays with half life of 5500 years . Estimate the age
of the dead tree whose present activity is 1 / 3 of initial activity.
(a) 8717 years (b) 6520 years
(c) 5500 years (d) 4500 years

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Q34. A radioactive sample contains 3  109 kg of active gold 200 Au , whose half life
is 48 min . Then the activity of the radon sample is
(a) 55 Ci (b) 57 Ci (c) 59 Ci (d) 61 Ci

Q35. In an ore containing uranium, the ratio of U 238 to Pb 206 nuclei is 3. Calculate the
age of the ore, assuming that all the lead present in the ore is the final stable product of
U 238 Take the half-life of U 238 to be 4.5  10 9 year.
(a) 2  109 Years (b) 3  109 Years

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(c) 4  109 Years (d) 5  109 Years

Q36. In an ore containing uranium, the ratio of U 238 to Pb 206 nuclei is 2. Calculate the
age of the ore, assuming that all the lead present in the ore is the final stable product of
U 238 Take the half-life of U 238 to be 4.5  10 9 year.
(a) 2  109 Years (b) 3  109 Years

(c) 4  109 Years (d) 5  109 Years

Q37. In a successive growth/decay process A  B  C , element C is a stable nucleus.
zi
The following parameters are given for the process:
1 2
A B C
t  0 N0 0 0
t t N1 N2 N3

where 1 and  2 are decay constant, N 0 is the concentration of A at t  0 and N 1 , N 2 ,

N 3 are concentration of A, B, C at any time t . Then the concentration of intermediate

fi
member B  will be:
1 N 0 1t 1 N 0  1t
(a) N 2 
 2  1
e  e  2t  (b) N 2 
 2  1
e  e  2t 
 2 N 0 1t  2 N 0 1t
(c) N 2 
 2  1
e  e  2t  (d) N 2 
 2  1
e  e  2t 

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Q38. In a successive growth/decay process A  B  C , element C is a stable nucleus.

The following parameters are given for the process:
1 2
A B C
t  0 N0 0 0
t t N1 N2 N3

where 1 and  2 are decay constant, N 0 is the concentration of A at t  0 and N 1 , N 2 ,

N 3 are concentration of A, B, C at time t . Then the time at which concentration of

ks
intermediate member B  will reach maxima is:

ln 1 /  2  ln  2 / 1 
(a) t '  (b) t ' 
1  2  2  1 
ln 1 2  ln 1 2 
(c) t '  (d) t ' 
1  2  2  1 
Q39. Nuclei of a radioactive element A are being produced at a constant rate  . The
element has decay constant  . At time t  0 , there are N 0 nuclei of the element. Then the

number N of nuclei of A at time

zi
1 1
(a)    1  N 0   e   t  (b)   N 0   e   t 
 
1 1
(c)     N 0   e   t  (d)   1  N 0   e   t 
 
Q40. The disintegration energy is defined to be the difference in the rest energy between
the initial and final states. Consider the following process:
240 236
Pu  U  24 He
fi
94 92

The emitted α-particle has a kinetic energy 5.17 MeV . The value of the disintegration
energy is
(a) 5.26 MeV (b) 5.17 MeV (c) 5.08 MeV (d) 2.59 MeV

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210
Q41. The polonium isotope 84 Po is unstable and emits a 5.30 MeV α-particle. The
210
atomic mass of 84 Po is 209.9829 u and that of 24 He is 4.0026 u , then the atomic mass

of its daughter nuclei is

(a) 203.9723 u (b) 204.9052 u
(c) 205.9754 u (d) 206.1053 u

Q42. In the β-decay of neutron n→ p + e- + e , the anti-neutrino e , escapes detection. Its

ks
existence is inferred from the measurement of
(a) energy distribution of electrons (b) angular distribution of electrons
(c) helicity distribution of electrons (d) forward-backward asymmetry of electrons

Q43. The Thermal Neutrons are captured by 5 B10 to form 5 B11 which further decays into

 - particle and 3 Li 7 , then the Q - value will be

(Given mn  1.008665 u , mB  10.01611 u, m  4.003879 u, mLi  7.01823 u )

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(a) 4 MeV (b) 3.8 MeV (c) 2.8 MeV (d) 1.7 MeV

Q44. The Thermal Neutrons are captured by 5 B10 to form 5 B11 which further decays into

 - particle and 3 Li 7 , the kinetic energy of Li is

(Given mn  1.008665 u , mB  10.01611 u, m  4.003879 u, mLi  7.01823 u )

(a) 1.8 MeV (b) 2.8 MeV (c) 1.0 MeV (d) 4.0 MeV
fi
Q45. Consider the nuclear reaction 7 N 14  , p  8O17 which occurred in Rutherford’s

 -range in nitrogen experiment. The mass of N 14  14.0031 u , He 4  4.0026 u ,

O17  16.9994 u and p  1.0078 u p. Then the Q -value of the reaction is

(a) 1.22 MeV (b) 1.32 MeV (c) 1.42 MeV (d) 1.49 MeV

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Q46. Consider the nuclear reaction 7 N 14  , p  8O17 which occurred in Rutherford’s

 -range in nitrogen experiment. The mass of N 14  14.0031 u , He 4  4.0026 u ,

O17  16.9994 u and p  1.0078 u p. Then the lowest  energy which will make this
reaction possible is:
(a) 1.52 MeV (b) 1.72 MeV (c) 1.92 MeV (d) 2.92 MeV

Q47. A neutron beam is incident on a stationary target of fluorine atoms. The

ks
reaction F 17  n, p  O19 has a Q -value of 4.0 MeV . Then the lowest neutron energy

which will make this reaction possible is:

(a) 2.2 MeV (b) 3.2 MeV (c) 4.2 MeV (d) 5.2 MeV

Q48. Neutrons are observed in a nuclear reaction Li 7  p, n  Be 7 . Then the bombarding

energy of proton at which neutrons of zero energy is obtained, will be ( Q -value of

reaction is 1.65 MeV ):
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(a) 1.7 MeV (b) 1.9 MeV (c) 2.1 MeV (d) 5.2 MeV

Q49. If 92 U 235 captures a thermal neutron a releases 160 MeV and if the resulting fission

fragments have mass numbers 138 and 95 , the kinetic energy of the lighter fragment is:
(a) 85 MeV (b) 95 MeV (c) 105 MeV (d) 115 MeV
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Q50. A nuclear decay process is given

Z XA  Z 1Y A  e

The atomic masses of X and X are 51.9648 u and 51.9571 u . Then the Q -value of
the reaction is:
(a) 2.7 MeV (b) 3.7 MeV (c) 4.7 MeV (d) 6.2 MeV

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Q51. If a star can convert all the He nuclei completely into oxygen nuclei, the energy
released per oxygen nucici is [Mass of He nucleus is 4.0026 amu and mass of oxygen
nucleus is 15.9994 amu]
(a) 7.6 MeV (b) 56.12 MeV (c) 10.24 MeV (d) 23.9 MeV
Q52. In the options given below, let E denote the rest mass energy of a nucleus and n a
neutron. The correct option is
(a) E  92
236
U   E 137
53 I   E  39 Y   2 E n 
97
(b) E  92
236
U   E 137
53 I   E  39 Y   2 E n 
97

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(c) E  92
236
U   E 140
56 Ba   E  36 Kr   2 E n 
94
(d) E  92
236
U   E 140
56 Ba   E  36 Kr   2 E n 
94

Q53. Consider nuclear fission process as given below

235
92 U  0 n1 92
236
U * 36
92
Kr 141 1
56 Ba  30 n  Q

235 92
(Mass of 92 U  235.045733 u , Mass of 141
56 Ba  140.09177 u , Mass of 36 Kr  91.8854 u
235
and Mass of n  1.008665 u ). Then the energy released by one kilogram of 92 U will be

(a) 5  1023 MeV (b) 5  1026 MeV

(c) 5  1027 MeV (d) 5  1028 MeV

zi
Q54. A Uranium nucleus decays at rest into a Thorium nucleus and a Helium nucleus as
shown below U 235  Th 231  He 4
Which of the following is true?
(a) Each decay product has the same kinetic energy
(b) Each decay product has the same speed
(c) The Thorium nucleus has more momentum than the Helium nucleus
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(d) The Helium nucleus has more kinetic energy than the Thorium nucleus
Q55. For nuclear fusion reaction to take place, which one of the following is true?
(a) Only very high temperature is required
(b) Normal temperature and comparatively high pressure is required.
(c) Very high temperature and comparatively high pressure is required.
(d) Very high temperature and very low pressure is required.

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Q56. Which of the following fusion reaction give more energy? The nuclear mass of the
different nuclei is as follows

M  1 H 1   1.00783 u , M  1 H 2   2.01410 u , M  1 H 3   3.01605 u ,

M  2 He3   3.01603 u , M  2 He 4   4.02603 u , M  3 Li 6   6.01512 u ,

M  e    0.00055 u

(a) 1 H 1  1 H 1  1 H 2  e    (b) 1 H 2  1 H 1  2 He3

ks
(c) 1 H 3  2 He3  3 Li 6 (d) 1 H 2  2 He4  3 Li 6

Q57. Consider the four processes

(a) p   n  e   ve (Within nucleus) (b)  0  p   e    e

(c)    e    e (d)  0    

Which of the above is forbidden?

Q58. Match the reactions on the left with the associated interactions on the right.
zi
(1) π+ → μ+ +  (i) Strong

(2) π0 → γ + γ (ii) Electromagnetic

(3) π0 + n → π- + p (iii) Weak
(a) (1, iii), (2, ii), (3, i) (b) (1, i), (2, ii), (3, iii)
(c) (1, ii), (2, i), (3, iii) (d) (1, iii), (2, i), (3, ii)
fi
Q59. The isospin and the strangeness of   baryon are
(a) 1, -3 (b) 0, -3 (c) 1, 3 (d) 0, 3

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Q60. The quark content of   , K  ,   and p is indicated:

   uus ; K   su ;    u d ; p  uud .

In the process,    p  K     , considering strong interactions only, which of the

following statements is true?
(a) The process, is allowed because ∆S = 0
(b) The process is allowed because ∆I3 =0
(c) The process is not allowed because ∆S ≠ 0 and ∆I3 ≠ 0

ks
(d) The process is not allowed because the baryon number is violated
Q61. Which one of the following sets corresponds to fundamental particles?
(a) proton, electron and neutron (b) proton, electron and photon
(c) electron, photon and neutrino (d) quark, electron and meson
Q62. Choose the CORRECT statement from the following
(a) Neutron interacts through electromagnetic interaction
(b) Electron does not interact through weak interaction
(c) Neutrino interacts through weak and electromagnetic interaction
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(d) Quark interacts through strong interaction but not through weak interaction
Q63. The decay process n  p   e   v e violates

(a) Baryon number (b) lepton number (c) isospin (d) strangeness
Q64. The isospin I  and baryon number B  of the up quark is
(a) I  1, B  1 (b) I  1, B  1 / 3
(c) I  1 / 2, B  1 (d) I  1 / 2, B  1 / 3
fi
Q65. In the  decay process, the transition 2   3  , is
(a) allowed both by Fermi and Gamow-Teller selection rule
(b) allowed by Fermi and but not by Gamow-Teller selection rule
(c) not allowed by Fermi but allowed by Gamow-Teller selection rule
(d) not allowed both by Fermi and Gamow-Teller selection rule

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Q66. Which one of the following is a fermions’?

(a)  -particle (b) 4 Be7 nucleus

(c) Hydrogen atom (d) deuteron

Q67. Which one of the following three-quark states  qqq  denoted by X CANNOT be a

possible baryon? The corresponding electric charge is indicated in the superscript.

(a) X   (b) X  (c) X  (d) X  
Q68. The decay    e    is forbidden, because it violates

ks
(a) momentum and lepton number conservations
(b) baryon and lepton number conservations
(c) angular momentum conservation
(d) lepton number conservation

Q69. Consider the reaction 54 54
25 Mn  e  24 Cr  X . The particle X is

(a)  (b)  e (c) n (d)  0

Q70. A beam of pions (π+) is incident on a proton target, giving rise to the process
π+p → n + π+ + π+
zi
Assuming that the decay proceeds through strong interactions, the total isospin I and its
third component I3 for the decay products, are
3 3 5 5
(a) I  , I 3  (b) I  , I 3 
2 2 2 2
5 3 1 1
(c) I  , I 3  (d) I  , I 3  
2 2 2 2
Q71. The dominant interactions underlying the following processes
fi
A. K   p      , B.       K   K  , C.    p   0 are

(a) A: strong, B: electromagnetic and; C: weak

(b) A: strong, B: weak and; C: weak
(c) A: weak, B: electromagnetic and; C: strong
(d) A: weak, B: electromagnetic and; C: weak

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Q72. A spin-1/2 particle A undergoes the delay

A  B C  D
where it is known that B and C are also spin-1/2 particles. The complete set of allowed
values of the spin of the particle D is
1 3 5
(a) ,1, , 2, , 3, ... (b) 0, 1
2 2 2
1 1 3 5 7
(c) only (d) , , , ,....
2 2 2 2 2

ks
Q73. Consider the four processes
(i) p   n  e   ve (ii) 0  p   e   v e

(iii)    e   ve (iv)  0    

which of the above is/are forbidden for free particles?

(a) only (ii) (b) (ii) and (iv) (c) (i) and (iv) (d) (i) and (ii)
Q74. The charm quark s assigned a charm quantum number C  1 . How should the
Gellmann-Nishijima formula for electric charge be modified for four flavors of quarks?
1 1
zi
(a) I 3  B  S  C (b) I 3  B  S  C
2 2
1 1
(c) I 3  B  S C (d) I 3  B  S  C
2 2
Q75. Consider the following processes involving free particles
(i) n  p  e   ve (ii) p  n   

(iii) p  n      0   0 (iv) p  ve  n  e
fi
Which of the following statements is true?
(a) Process (i) obeys all conservation laws
(b) Process (ii) conserves baryon number, but violates energy-momentum conservation
(c) process (iii) is not allowed by strong interaction but is allowed by weak interactions
(d) Process (iv) conserves baryon number, but violates lepton number conservation

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Nuclear Physics (Objective Questions-Solution)

Ans1: (b)
1/3 1/3 1/3
 R  Ro  A and given that RGe  2 RBe  Ro  AGe   2 Ro  9   AGe  72 .

Ans2: (a)
1/3 1/3
1/3 R A   208  1/3
Since R  R0  A   Pb   Pb     8 2
RMg  AMg   26 

Ans3: (d)

ks
1/3 1/3
1/3 R A   64  4
Since R  R0  A   Cu   Cu      1.33
RAl  AAl   27  3

Ans4: (d)
4 4 4 4
V   R 3   R03 A   R03  16; V '   R03  128  8V .
3 3 3 3
Ans5: (b)
Ans6: (b)
The amount of energy required to add a neutron in nucleus is more than that of proton.
zi
Ans7: (c)
Ans8: (d)
Since nuclear mass is always less than their constituent particles so
M 1  10  m p  mn  and M 2  20  m p  mn 
40 20
Since B.E. of 20 Ca  B.E. of 10 Ne

  20  m p  mn  – M 2  c 2  10  m p + mn  – M1  c 2  M 2  10  m p + mn   M1   2 M1
fi
 
Ans9: (a) B.E .   ZmH  Nmn  m 12 H   931.5 MeV

 B.E.  1 1.0078  1 1.0087  2.0141  931.5 MeV

 B.E.  0.0024  931.5 MeV  2.2356 MeV

B.E. 2.2356
   1.1178 MeV / nucleons
A 2

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Ans10: (c)

B.E .   ZmH  Nmn  m  238


U   931.5 MeV
92

 B.E.  92  1.0078  146  1.0087  238.0508  931.5 MeV

 B.E.  1.937  931.5 MeV  1804 MeV

Ans11: (a)
42
20 Ca  41 1
20 Ca  0 n ;

ks
41
Total mass of the 20 Ca and 10 n  41.970943 u .

Mass defect m  41.970943  41.958622  0.012321 u

So, B.E. of missing neutron= m  931.5  11.48 MeV

Ans12: (b)
42 41
20 Ca 19 K 11 p ;
41
Total mass of the 19 K and 11 p  41.969101 u .
zi
Mass defect m  41.969101  41.958622  0.010479 u
So, B.E. of missing proton= m  931.5  10.27 MeV .

Ans13: (c)

Since a nucleus with mass number A  180 fission into two nuclei of equal masses thus
fi
180  90  90 .
So B.E . of the heavier nucleus is  180  4  720 MeV .
Total B.E . of the lighter nuclei is  90  6  90  6  1080 MeV .
Since product nuclei have higher B.E . so in this process energy is released
i.e  1080  720  360MeV .

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Ans14: (c)
Let us write B.E . in MeV of both sides.
(a) Y  2 Z ; 8.5  60  510  2  5  30  300
(b) W  X  Z ; 7.5  120  900  8  90  5  30  870
(c) W  2Y ; 7.5  120  900  2  8.5  60  1020
(d) X  Y  Z ; 8  90  720  8.5  60  5  30  660
In W  2Y , product have higher B.E. than reactant. So energy will release.
Ans15: (b)

ks
2/3 Z(Z  1) (A  2Z)2 ap
Since B  a v A  a s A  ac  a a (  , 0)
A1/3 A A 3/ 4

 dB  ac a
    1/3  2Z0  1  a 2(A  2Z0 )(2)  0
 dZ  Z Z0 A A

ac 4a  ac 4a  ac
1/3 
 2Z0  1  a  A  2Z 0   0  2Z0  1/3  a    A1/3  4a a
A A A A 
 ac 
zi
 4a a  A1/3  4a a  a c A 1/3
 Z0     Z0 
 ac 4a a  2a c A 1/3  8a a A 1
2  1/3  
A A 

Ans16: (c)
Ans17: (a)

Since, M  Z, A   A   Z  Z2 for given A

fi
For nuclear charge  Z 0  of “most stable” nuclei

 M  
   0    2 Z0  0  Z0  .
 Z  Z Z0 2

So mass of the “most stable” isobar is

M  Z0 , A   A  2Z0 Z0  Z0 2   =–2Z0   M  Z0 , A   A  Z0 2

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Also, M  Z, A   A  2 Z0 .Z  Z2

The difference in masses for odd A is:

2
M  Z, A   M  Z0 , A   2Z0 Z  Z2  Z0 2    Z  Z0 

Ans18: (a)

Q  M  Z, A   M  Z  1, A    M  Z, A   M  Z0 , A     M  Z  1, A   M  Z0 , A  

ks
2 2  1
Q    Z – Z0     Z  1  Z 0      2  Z  Z0  –1  2  Z 0  Z  
 2

 1
Thus Q  2  Z0  Z  
 2

Ans19: (b)

Q  M  Z, A   M  Z  1, A    M  Z, A   M  Z0 , A     M  Z  1, A   M  Z0 , A  

 1
zi
2 2
Q    Z – Z0     Z  1  Z0      2  Z  Z0  – 1  2  Z  Z0  
 2

 1
Thus Q  2   Z  Z0  
 2

Ans20: (c)
1
Since M  Z , A    f A  yZ  zZ 2 
2   
c
fi
 M 
For stable nucleus   0
 Z  Z  Z0

y 4 aa A/ 2
 Z0    Z0 

2 z 2 ac A  4aa A1
1/3
  a 
1   c  A2/3
 4aa 

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Ans21: (a)

Mirror nuclei to be considered have the same odd value of A but the values of N and Z
are interchanged such that they differ by one unit  N – Z  1 .

Now, from semi empirical mass formula we know

2
Z  Z  1  A  2Z  1
M  Z, A   M N A   M N  M P  Z  a v A  a s A 2/3
 ac 1/3
 aa  , 0  a p .
A A A3/4

ks
Now to find mass difference between pair Mirror Nuclei are

M Z1  M Z  M  Z  1, A  – M  Z, A 

But A – 2Z  N  Z – 2Z  N – Z  1  Let N  Z 

ac
 M Z1  M Z    M N  M P   Z  1  Z    Z  Z  1  Z  Z  1
A1/3 
ac ac
1/3 
 M Z1  M Z   M P  M N   A  1  1/3  A  1   M N  M P 
A A
zi
Ans22: (a)
Z  5, N  6  odd  even 

2 3 3
N  5; 1S1/2   2 P3/2   j  , l  1
2
3 l 1
Thus Spin  and Parity   1   1  1 .
2
Ans23: (b)
fi
Z  20, N  21  even  odd 

2 4 2 6 2 4 1 7
N  21; 1S1/2   2 P3/2   2 P1/2   3d5/2   2 S1/2   3d3/2   4 f 7/2   j  , l  3
2
7 l 3
Thus Spin  and Parity   1   1  1 .
2

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Ans24: (d)
Z  7, N  9  odd  odd 

2 4 1 1
Z  7; 1S1/2   2 P3/2   2 P1/2   j p  , l p  1
2
2 4 2 1 5
N  9; 1S1/2   2 P3/2   2 P1/2   3d 5/2   jn  , ln  2
2
1 5
Let us find j p  jn  l p  ln    1  2  6  even number 
2 2

ks
l  ln 3
 J  j p  jn  2 and Parity   1 p   1  1 .

Ans25: (d)
(a) Z  8, N  8  even  even   Spin  0 and Parity  1 .

(b) Z  13, N  14  odd  even 

2 4 2 5 5
Z  13; 1S1/2   2 P3/2   2 P1/2   3d 5/2   j  , l  2
2
5 l 2
Thus Spin  and Parity   1   1  1 .
zi
2
(c) Z  16, N  17  even  odd 

2 4 2 6 2 1 3
N  13; 1S1/2   2 P3/2   2 P1/2   3d 5/2   2S1/2   3d 3/2   j  , l  2
2
3 l 2
Thus Spin  and Parity   1   1  1 .
2
(d) Z  4, N  5  even  odd 
fi
2 3 3
N  5; 1S1/2   2 P3/2   j  , l  1
2
3 l 1
Thus Spin  and Parity   1   1  1 .
2

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Ans26: (c)
Z  9, N  9  odd  odd 

2 4 2 1 5
Z  9; 1S1/2   2 P3/2   2 P1/2   3d5/2   j p  , l p  2
2
2 4 2 1 5
N  9; 1S1/2   2 P3/2   2 P1/2   3d 5/2   jn  , ln  2
2
5 5 l l 4
 j p  , jn   J  0,1, 2, 3, 4, 5 and Parity   1 p n   1  1 .
2 2

ks
Ans27: (b)
The energy of the singlet state is higher than the triplet state. A combination of proton
and neutron form triplet state while combination of neutron-neutron forms singlet state.
Which is higher in energy and no bound state is possible. This is confirming the spin
dependency of the nuclear forces.
Ans28: (d)
Ans29: (d)
Ans30: (d)
zi
Ans.31: (a)
N A B T1/2 A
N A  A  N B B   
N B  A T1/2 B
NB 1
 T1/2 B  T1/2 A  4
4.5  109  2.5  105 years
NA 1.8  10
Ans32 (d)
0.693 0.693
fi
Decay constant     2.1 106 sec 1
T1 3.8  24  60  60
2

100  10 6 kg
Number of atoms in 1.00 mg is N  27
 2.7  1020 atoms
 222u   1.66  10 kg / u

101
or N   6.023  1023 =2.7  1020 atoms
222
Hence, activity R   N  2.110 6  2.7  1020  5.7 1014 decay / sec

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Ans33: (a)  R  R0e  t

The age of the dead tree is

1  R0  T1/2  R  5500  3 
t ln    ln  0   ln    8717 years
  R  0.693  R  0.693  1 
Ans.34: (a)
0.693 0.693
Decay constant     2.406  104 sec 1
T1 48  60
2

ks
3 106 g
Number of atoms in 3  10 9
kg is N   6.023 10 23 =9.04  1015 atoms
200
Hence, activity R   N  2.406 104  9.04 1015  2.18 1012 decay / sec  59 Ci

1.0 Ci  3.7 1010 decay / sec

Ans35. (a)
NU N N N N 4
Solution:  3 where N Pb  N 0U  NU . Thus Pb  0U  1  0U  Pb  1 
N Pb NU NU NU NU 3

Initial number of U 238 4 N 4

  0 
zi
238
Final number of U 3 N 3
2.303 N  4.5  10 9  4
t  log 10 0  2.303    log 10  15  10 9 log 10 4  log 10 3
 N  0.693  3

 15  10 9 0.6021  0.4771  15  10 9  0.1250  1.875  10 9 year

Ans36: (b)
NU N N N N 3
Solution:  2 where N Pb  N 0U  NU . Thus Pb  0U  1  0U  Pb  1 
fi
N Pb NU NU NU NU 2

Initial number of U 238 3 N 0 3

  
Final number of U238 2 N 2

2.303 N0  4.5  109  3 9

t  log10  2.303    log10  15  10  log10 3  log10 2 
 N  0.693  2

 15  109  0.4771  0.3010   15  109  0.1761  2.641109 years

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dN 2 dN 2
Ans37: (a)  1 N1  2 N 2   2 N 2  0 N 0e  1t
dt dt

t 1 N o e 2 1 
Multiply both side by e 2 dt and then integrate  N 2 e 2t  K.
 2  1
1 N 0 N
At t  0, N 2  0  K  
 2  1  2  1

, thus N 2  1 o e 1t  e 2t . 
Ans38: (b)
dN 2 dN 2

ks
 1 N1  2 N 2   2 N 2  0 N 0e  1t
dt dt

t 1 N o e 2 1 
Multiply both side by e 2 dt and then integrate  N 2 e 2t  K.
 2  1
1 N 0 N
At t  0, N 2  0  K  
 2  1  2  1

, thus N 2  1 o e 1t  e 2t 
 dN 2  ln  2 / 1 
  0t' .
 dt t t ' 2  1
Ans39: (c)
zi
Solution: Nuclei of a radioactive element A are being produced at a constant rate  .
Decay constant of element =  . At t  0 , nuclei of element present N 0
(a) Number N of nuclei of A at time t :
dN
Net rate of formation of nuclei of element A 
dt
dN dN
    N or  dt
fi
dt   N
N t
dN 1 N
or   dt or   ln    N   N  t
N0
  N 0  0

   N    N
ln    t or  e  t
   N 0    N0

1
or   N  e   t   N 0  or N 


    N 0  e  t 
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Ans40: (a)
 A 4   A  240
K .E   Q  Q    KE   5.17  5.26 MeV
 A   A 4  236
Ans41: (c)
 A  210
Q  KE   5.30  5.40 MeV
 A4 206
5.40MeV
The mass equivalent of this Q-value, mQ   0.0058 u
931.5MeV / u

ks
Hence m f  mi – m - mx  209.9829 – 4.026 – 0.0058  m f  205.9754 u

Ans42: (a)
Ans43: (c)
Solution:

5 B10  0 n1 5 B11  2 He 4     3 Li 7  Q

Q   M B  M N  M   M Li   931.5 MeV

Q  10.01611  1.008665  4.003879  7.018231  931.5  2.78 MeV  2.8 MeV

zi
Ans44: (c)
Solution:

5 B10  0 n1 5 B11  2 He 4     3 Li 7  Q

Q   M B  M N  M   M Li   931.5 MeV

Q  10.01611  1.008665  4.003879  7.018231  931.5  2.78 MeV  2.8 MeV

Energy released in the process for  - particle ( Q  0 and it is an exothermic reaction).

fi
Q  M Li 2.78  7.018221
E    E  1.78 MeV
M Li  M  4.003879  7.018221

We know that ELi  E  Q  ELi  Q  E  2.78  1.78  1.00 MeV

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Ans45: (d)
a A Bb;   7 N 14  8O17  p

Q   M A  M a    M B  M b    931.5 MeV

Q  14.003  4.0026   16.9994  1.0078   931.5 MeV

Q  0.0013  931.5 MeV  1.49 MeV

Ans46: (c)

ks
a A Bb;   7 N 14  8O17  p

Q   M A  M a    M B  M b    931.5 MeV

Q  14.003  4.0026   16.9994  1.0078   931.5 MeV

Q  0.0013  931.5 MeV  1.49 MeV

 M  Ma   14  4 
Eth  Q  A   1.49    1.92 MeV
 MA   14 
Ans47: (c)
zi
a A Bb; n  F 17  O19  p

 M  Ma   19  1 
Eth  Q  A   4   4.2 MeV
 MA   19 
Ans48: (b)
a A Bb; p  Li 7  Be 7  n

 M  Ma   7 1 
Eth  Q  A   1.65    1.9 MeV
fi
 MA   7 
Ans49: (b)
Since the reaction is due to thermal neutron Ea  0

a A Bb;
MB 138
 Eb  Q  160   95MeV
M B  mb 138  95

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Ans50: (d)
Q  X  Y  2me    X  Y   2me   51.9648 u  51.9571 u   931.5MeV  1.02 MeV

Q  7.17 MeV  1.02 MeV  6.2 MeV

Ans51: (c)
Solution: 4  2 He4  8 O16

Binding energy m  931.5 MeV   4  4.0026  15.9994   931.5 MeV  10.24MeV

ks
 Energy released per oxygen nuclei  10.24 MeV
Ans52: (a)
Solution: Rest mass energy of U will be greater than the rest mass energy of the nuclei
into which it breaks. The constituent nuclei and neutrons will have kinetic energy also, as
a result of conservation of linear momentum.
Ans53: (b)
Ans54: (d)
For momentum conservation momentum of Th and He must be same and opposite.
mTh vTh  mHe vHe  vTh  vHe  mTh  mHe
zi
1 1
Also mTh vTh2  mHe v 2He
2 2
Ans55: (c)

Ans56: (d)
The energy  Q  released in all fusion reaction
fi
(a) 1 H 1  1 H 1  1 H 2  e   

Q   M  1 H 1   M  1 H 1   M  1 H 2   M  e     931.5 MeV

 Q  1.00783 u  1.00783 u  2.01410 u  0.00055 u   931.5 MeV

 Q  1.07 10 3   931.5 MeV  Q  0.99 MeV

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(b) 1 H 2  1 H 1  2 He3

Q   M  1 H 2   M  1 H 1   M  2 He3    931.5 MeV

 Q   2.01410 u  1.00783 u  3.01603 u   931.5 MeV

 Q   5.9  103   931.5 MeV  Q  5.5 MeV

(c) 1 H 3  2 He3  3 Li 6

Q   M  1 H 3   M  2 He3   M  3 Li 6    931.5 MeV

ks
 Q  3.01605 u  3.01603 u  6.01512 u   931.5 MeV

 Q  .01696 u   931.5 MeV  Q  15.8 MeV

(d) 1 H 2  2 He4  3 Li 6

Q   M  1 H 2   M  2 He4   M  3 Li 6    931.5 MeV

 Q   2.01410 u  4.02603 u  6.01512 u   931.5 MeV

 Q   0.02501 u   931.5 MeV  Q  23.3 MeV

zi
Ans57: (b)
Solution:
(a) p   n  e    e [Allowed within nucleus]

(ii)  0  p   e    e [Not allowed]. In this decay charge is not conserved

(iii)    e    e [Allowed through Weak interaction]

fi
(iv)  0     [Allowed through Electromagnetic interaction]
Ans58: (a)

Ans59: (b)

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Ans60: (c)
Solution:    p  k    

S: 0 0 1 1 (not conserved)
1 1
I3 : 1   1 (not conserved)
2 2
For strong interaction S and I3 must conserve. Therefore this process is not allowed under
strong interaction

ks
Ans61: (a)
Ans62: (d)
Ans63: (c)
Ans64: (d)
Ans65: (c)
Solution: According to Fermi Selection Rule:
I  0, Parity  No Change
According to Gammow-Teller Selection Rule:
I  0,1, Parity  No Change
zi
In the  decay process, the transition 2   3  ,
I  1, Parity  No Change .
Ans66: (b)
Ans67: (d)
Solution: X  qqq
2 2 2 6
X   uuu      2  two unit positive charge 
fi
3 3 3 3
2 2 1 4 1
X   uud       1 single unit positive charge 
3 3 3 3 3
1 1 1
X   ddd       1 single unit negative charge 
3 3 3
X   Not possible with qqq  . So the correct option is (d)

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Ans68: (d)
Solution:    e    . In this decay lepton number is not conserved.

Ans69: (b)

Ans70: (c)
1 5 1 3
Solution:    p  n       ; I : 11  , I3 :  1 1 
2 2 2 2

ks
Ans71: (a)
(A) K   p      (Strong interaction)

1 1
I 3 :    1  1 (Conserved)
2 2
(B)       K   K  (Electromagnetic interaction)

(C)    p   0 (Weak interaction)

1
I3 :1  0 (Not conserved)
2
zi
Ans72: (c)
Solution: Spin of the left side and combined spin of the products must be same to
conserve the spin angular momentum conservation law.
Ans73: (d)
Solution: (i) p   n  e    e [Not allowed]
It violates energy conservation. The mass of proton is less than mass of neutron. Free
fi
proton is stable and can not decay to neutron. Proton can decay to neutron only inside the
nucleus, where energy violation is taken care by Heisenberg uncertainty principle.
(ii)  0  p   e    e [Not allowed]. In this decay charge is not conserved

(iii)    e    e [allowed through Weak interaction]

(iv)  0     [allowed through Electromagnetic interaction]

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Ans74: (d)
1
Solution: From Gell-Mann-Nishijima formula Q  I 3  B  S
2
1
For Quark it is generalized as Q  I 3  B  S  C
2

Ans75: (b)

ks
Solution:
(i) n  p  e   ve

q 0 1 1 0 (conserved)
1 1 1 1
spin     (not conserved)
2 2 2 2
Le 0 0  1  1 (not conserved)
(ii) Baryon number is conserved but energy and momentum conservation violated.
(iii) spin is not conserved
zi
(iv) obeys all conservation laws.
fi

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