Nuclear Physics

Single Correct Type Questions 5. For a nucleus ZA X having mass number A and atomic
number Z [8 April, 2023 (Shift-I)]
1. The ratio of the density of oxygen nucleus (168O) and
(A) The surface energy per nucleon (bs) = a1A2/3
helium nucleus (42He) is [25 Jan, 2023 (Shift-I)] (B) The Coulomb contribution to the binding energy
(a) 4 : 1 (b) 8 : 1 Z ( Z − 1)
bc = −a2
(c) 1 : 1 (d) 2 : 1 A4/3
2. Read the following statements: (C) The volume energy bv = a3A
(A) Volume of the nucleus is directly proportional to the (D) Decrease in the binding energy is proportional to
mass number. surface area.
(B) Volume of the nucleus is independent of mass number. (E) While estimating the surface energy, it is assumed that
(C) Density of the nucleus is directly proportional to the each nucleon interacts with 12 nucleons, (a1, a2 and a3
mass number.
are constants)
(D) Density of nucleus is directly proportional to the cube
root of the mass number. Choose the most appropriate answer from the options
(E) Density of the nucleus is independent of the mass given below:
number. (a) (C), (D) only (b) (B), (C), (E) only
Choose the correct option from the following options. (c) (A), (B), (C), (D) only (d) (B), (C) only
 [29 July, 2022 (Shift-II)]
6. The mass of proton, neutron and helium nucleus are
(a) (A) and (D) only. (b) (A) and (E) only.
respectively 1.0073 u, 1.0087 u and 4.0015 u. The binding
(c) (B) and (E) only. (d) (A) and (C) only.
energy of helium nucleus is: [01 Feb, 2023 (Shift-I)]
3. Mass numbers of two nuclei are in the ratio of 4 : 3. Their (a) 14.2 MeV (b) 28.4 MeV
nuclear densities will be in the ratio of
 [26 July, 2022 (Shift-II)] (c) 56.8 MeV (d) 7.1 MeV
1 7. 238
92 A→ 234
90
4
B+ D+Q
2
 3 3
(a) 4 : 3 (b)   In the given nuclear reaction, the approximate amount of
4
energy released will be:
1
 4 3 [Given, mass
= 238
of 92 A 238.05079 × 931.5 MeV/c 2 , mass
(c) 1 : 1 (d)  
3 234
4. The radius R of a nucleus of mass number A can be o=
f 90 B 234.04363 × 931.5 MeV/c 2 , mass of
estimated by the formula R = (1.3 × 10–15)A1/3 m. It follows =D 4.00260 × 931.5 MeV/c ] [13 April, 2023 (Shift-I)]
4
2
2

that the mass density of a nucleus is of the order of:

(a) 3.82 MeV
(Mprot = Mneut.  1.67 × 10–27 kg)
 [3 Sep, 2020 (Shift-II)] (b) 5.9 MeV
(a) 10 kg m
3 –3
(b) 10 kg m
17 –3
(c) 2.12 MeV
(c) 10 kg m
10 –3
(d) 10 kg m
24 –3
(d) 4.25 MeV

1 JEE PYQs Physics

 8. The Q-value of a nuclear reaction and kinetic energy of 14. How many alpha and beta particles are emitted when
the projectile particle, Kp are related as: Uranium 92U238 decays to lead 82Pb206?
 [28 June, 2022 (Shift-I)]
[26 June, 2022 (Shift-I)]
(a) Q = Kp (b) (Kp + Q) < 0
(a) 3 alpha particles and 5 beta particles
(c) Q < Kp (d) (Kp + Q) > 0
(b) 6 alpha particles and 4 beta particles
9. Given the masses of various atomic particles mp = 1.0072
(c) 4 alpha particles and 5 beta particles
u, mn = 1.0087 u, me = 0.000548 u, mν = 0, md = 2.0141 u,
where p ≡ proton, n ≡ neutron, e ≡ electron, –v≡ antineutrino (d) 8 alpha particles and 6 beta particles
and d ≡ deuteron. Which of the following process is 15. A nucleus of mass M emits g–ray photon of frequency
allowed by momentum and energy conservation? ‘hv’. The loss of internal energy by the nucleus is: [Take
 [6 Sep, 2022 (Shift-II)] ‘c’ as the speed of electromagnetic wave]
(a) n + n → deuterium atom  [20 July, 2021 (Shift-I)]
(electron bound to the nucleus) hv 
(a) hv 1 + 2 
(b) 0
(b) n + p → d + γ  2 Mc 
(c) p → n + e+ + –v
 hv 
(d) e+ + e– → γ (c) hv 1 − 2 
(d) hv
 2 Mc 
10. Find the binding energy per nucleon for 120 50
Sn. Mass of
proton mP = 1.00783 u, mass of neutron mn = 1.00867 u Integer Type Questions
and mass of tin nucleus mSn = 119.902199 u.
(Take 1 u = 931 MeV) [4 Sep, 2020 (Shift-II)] 16. A nucleus disintegrates into two smaller parts, which have
(a) 9.0 MeV (b) 8.5 MeV their velocities in the ratio 3 : 2. The ratio of their nuclear
1
(c) 8.0 MeV (d) 7.5 MeV  x 3
sizes will be   . The value of ‘x’ is _______.
11. Consider the nuclear fission Ne20 → 2He4 + C12.Given 3
that the binding energy / nucleon of Ne20, He4 and C12 are,  [25 Jan, 2023 (Shift-II)]
respectively, 8.03 MeV, 7.07 MeV, and 7.86 MeV, identify
the correct statement: [10 Jan, 2019 (Shift-II)] 17. Assume that protons and neutrons have equal masses.
Mass of a nucleon is 1.6 × 10–7 kg and radius of nucleus
(a) energy of 12.4 MeV will be supplied
is 1.5 × 10–15 A1/3 m. The approximate ratio of the nuclear
(b) 8.3 MeV energy will be released density and water density is n × 1013. The value of n is
(c) energy of 9.72 MeV will be released _________ [24 Jan, 2023 (Shift-I)]
(d) energy of 11.9 MeV has to be supplied 18. Nucleus A having Z = 17 and equal number of protons and
12. Consider the following radioactive decay process neutrons has 1.2Me V binding energy per nucleon.
− Another nucleus B of Z = 12 has total 26 nucleons and
218 α β γ α
84 A  → A1  → A2  → A3  → 1.8Me V binding energy per nucleons.
β +
γ The difference of binding energy of B and A will be _____
A4  → A5  → A6
Me V. [1 Feb, 2023 (Shift-II)]
The mass number and the atomic number A6 are given by:
19. A nucleus with mass number 242 and binding energy per
 [24 Jan, 2023 (Shift-I)]
nucleon as 7.6 MeV breaks into fragment each with mass
(a) 210 and 82 (b) 210 and 84 number 121. If each fragment nucleus has binding energy
(c) 210 and 80 (d) 211 and 80 per nucleon as 8.1 MeV, the total gain in binding energy
13. In the following nuclear reaction, is ________ MeV [8 April, 2023 (Shift-I)]
α β− α γ
D → D1 → D2 → D3 → D4 20. The minimum kinetic energy needed by an alpha particle
to cause the nuclear reaction 167N + 24He → 11H + 198O in a
Mass number of D is 182 and atomic number is 74. Mass
laboratory frame is n (in MeV). Assume that 167N is at rest
number and atomic number of D4 respectively will be ____
in the laboratory frame. The masses of 167N, 24He, 11H and 198O
 [29 June, 2022 (Shift-II)] can be taken to be 16.006 u, 4.003 u, 1.008 u and 19.003
(a) 174 and 71 (b) 174 and 69 u, respectively, where 1 u = 930 MeVc–2. The value of n
is________. [JEE Adv, 2022]
(c) 172 and 69 (d) 172 and 71

2 JEE PYQs Physics

21. From the given data, the amount of energy required to 23. A common example of alpha decay is
27
break the nucleus of aluminium 13 Al is ________ x ×  [12 April, 2023 (Shift-I)]
238 234
10–3 J. [25 July, 2021 (Shift-II)] 92 U 
→ 90 Th + 2 He 4 + Q
Mass of neutron = 1.00866u Given :
Mass of proton = 1.00726 u 238
92 U = 238.05060u

Mass of Aluminium nucleus = 27.18846 u
234
(Assume 1 u corresponds to x J of energy) 90 Th = 234.04360u

(Round off to the nearest integer)
4
2 He = 4.00260u
, and
22. A nucleus disintegrates into two nuclear parts, in such MeV
a way that ratio of their nuclear sizes is 1 : 21/3. Their 1u = 931.5 2
c
respective speed have a ratio of n : 1. The value of n is
The energy released (Q) during the alpha decay of
 [11 April, 2023 (Shift-II)] 238
92 U is __________ MeV

3 JEE PYQs Physics

 ANSWER KEY
1. (c) 2. (b) 3. (c) 4. (b) 5. (a) 6. (b) 7. (d) 8. (d) 9. (b) 10. (b)
11. (c) 12. (c) 13. (a) 14. (d) 15. (a) 16. [2] 17. [11] 18. [6] 19. [121] 20. [2.32]
21. [27] 22. [2] 23. [4]

EXPLANATIONS

Am A is incorrect
1. (c) Q Nuclear density = Coulomb contribution to the binding energy is
4 3
πR −a Z ( Z − 1)
3 bc = 2 ⇒ B is incorrect
Where A is the mass number and m is mass of one A1/3
nucleon. Volume energy ∝ A ⇒ C is correct
1
As we consider only surface energy contribution then

Q R = R0 A 3 option is correct. ⇒ D is correct.
In case of surface energy, only 3 interactions

⇒ R3 = R30A contribute to surface energy. ⇒ E is incorrect.
Am

⇒ Nuclear density = 6. (b) B.E of Helium = (2mP + 2mN – mHe) × 931.5 MeV
4 3
πR0 A = 28.4 MeV
3
7. (d) Q = (mA – mB – mD) × c2
3m   = (238.05079 – 234.04363 – 4.00260)
Nuclear density =
4πR03
  = (0.00456 c2).u

\ Nuclear density is independent of A MeV  MeV 
0.00456c 2 ⋅ 931.5 2  1u = 931 2 
2. (b) Radius, R = R0 A1/3 ⇒ R ∝ A1/3 c   c 

V ∝ R3 ⇒ 4.25 MeV

∴V∝A 8. (d) A + p → B + b
Q = KB + Kb – KA
And mass ∝ A
⇒ Q + KP = KB + Kb
So, density is independent of A.
⇒ Q + KP > 0
3. (c) Radius of nucleus, R = R0 A1/3
9. (b) ∆m should be positive
mass mA mA

=δ = = (mp + mn) > md
Volume 4 πR3 4 πR3 A ⇒ only (2) is possible
0
3 3

⇒ Density is independent of mass number 10. (b) B.E. = (∆m)C2

Mass
∆m = (50 mP + 70 mn) – (mSn)
4. (b) Density of Nucleus =
Volume = (50 × 1.00783 + 70 × 1.00867) – (119.902199)

B.E. = ∆m × 931 MeV
1.67 × 10 –27 × A
= ≈ 1017
⇒ B.E. = 1020.5631 MeV
4 ( )3
π 1.3 × 10 –45 × A B.E. 1020.5631
3
⇒= = 8.5MeV = 8.5 MeV
Nucl. 120
5. (a) Q Mass number A ∝ r3 ⇒ r ∝ A1/3.
11. (c) Energy = (Binding Energy)Reactant – (Binding Energy)Product
r2 1
\ Surface energy per nucleon ∝ ∝ 1/3
A A = 20 × 8.03 – (8 × 7.07 + 12 × 7.86)

4 JEE PYQs Physics

 = 160.6 – (56.56 + 94.32) = 160.6 – 150.88 = 9.72 m1 m2
MeV
⇒ =
4 3 4 3
πr1 πr2
Hence, 9.72 MeV released. No option is correct. 3 3
Hence, this question is bonus. 3
 r1  m1
12. (c) 218 A 
α 214
→ 82 β−
A1  214
→ 83 γ
A2  214
→ 83 A3 ⇒  =

84
r
 2 m2
+
214 α 210 β 210 γ 210

83 A3  → 81 A4  → 80 A5  → 80 A6 1
r  2 3  m1 2 
13. (a) Let us suppose for D4, ⇒ 1 =  
 = 
r2  3   m2 3 
Atomic number = z, Mass number = A
⇒x=2

⸫ A = 182 – 4 – 4 = 174
mass of nuclei
Z = 74 – 2 – 2 + 1 = 71 17. [11] Density of nuclei =
volume of nuclei

14. (d) When a radioactive nucleus decays by emitting an 1.6 × 10−27 A

=ρ = 0.113 × 1018
4
( )
3
α-particle, its mass number decreases by 4 units and π 1.5 × 10 −15
A
its atomic number decreases by 2 units. 3
32
\ Number of α-particles emitted = = 8. 103
ρw =
4
Change in atomic number = 92 – 16 = 76. ρ
Hence = 11.31× 1013
But the required atomic number is 82. This difference ρw
in atomic number is due to emission of b-particles. 18. [6] For nucleus A mass number = 34
There is no change in mass number but its atomic
number increases by 1 unit on the emission of a Total binding energy = 1.2 × 34 = 40.8MeV
b-particle. For nucleus B mass number = 26
\ Number of b – particles emitted = 82 – 76 = 6. Total binding energy= 1.8 × 26MeV
15. (a) Using conservation linear momentum, we can write = 46.8Me V
h hv hv Difference of BE = 46.8 – 40.8 = 6 MeV
Mv = = ⇒v=
λ c Mc 19. [121] Initial binding energy = 242 × 7.6 MeV
v Final binding energy = (121 + 121) × 8.1 MeV
Rest
M M -ray = 242 × 8.1 MeV
JBC JBC Total gain in binding energy = (Final binding energy
Loss of internal energy = Gain in kinetic energy – Initial binding energy)
+ energy of gamma ray = 242 (8.1 – 7.6)
2
1 1  hv   hv  = 242 × 0.5

= Mv 2 + hv= M  + hv= hv 1 +
2  Mc   2 Mc 
2
2
= 121 MeV
16. [2] 20.[2.32]
16
7
N  +  24He  →  11H  +  198O
16.006   4.003     1.008   19.003
v1 3
Given, = 4v0 = 1v1 + 19v2 = 20v2   (For max loss of KE)
v2 2
v0
Using conservation of linear momentum, v2 =

5
m1v1 = m2 v2
E required = (1.008 + 19.003 – 16.006 – 4.003) × 930
m 2 = 1.86
⇒ 1 =

m2 3 1 2 1
4v0 − 20v22 = 1.86 ,
Since, Nuclear mass density remains constant 2 2

5 JEE PYQs Physics

 = [(13.09438 + 14.12124) – 27.18846] u
1 v2
4v02 − 10 0 =
1.86 = [27.21562 – 27.1884]u
2 25
= 0.02716 u
2 8 2

2v02 − v02 = 1.86 , ⇒ v0 = 1.86 = 0.02716x J
5 5
= 27.16x ×10–3
1.86 × 5
v02 = 22. [2]
8
1 2 1.86 × 5 23. [4] Mass defect, ∆m = (∑m)Product – (∑m)reactant
KE = 4= v02
v0 2= = 2.325 MeV Energy released,
2 4
2 2 MeV
21. [27] Mass defect, Q = ∆m . C = (m u − mTh − mµe )C .931.5
C2
∆m = [Zmp + (A – Z)mn – MAl] = 0.0044 × 931.5 MeV = 4.0986 MeV
= [13 × 1.00726 + 14 × 1.00866 – 27.18846]

6 JEE PYQs Physics