1. In the given nuclear reaction, the element X is

$_{11}^{22} N a \to X + e^{+} + v$ ⇒ (NEET 2022 Phase 1)

A. $_{11}^{23} N a$

B. $_{10}^{23} N e$

C. $_{10}^{22} N e$

D. $_{12}^{22} M g$

The correct answer is option (C)

The nuclear reaction is given as

$_{11}^{22} N a \to_{Z}^{A} X +_{+ 1} e^{0} + v$

From conservation of atomic number

$11 = Z + 1 \Rightarrow Z = 10 \Rightarrow N e$

From conservation of mass number

$22 = A + 0 \Rightarrow A = 22$

$∴$ $_{Z}^{A} X =_{10}^{22} N e$

2. A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding per nucleon of unfragmented nuclei is 7.6MeV while that of fragments is 8.5MeV. The total gain in the Binding Energy in the process is : ⇒ (NEET 2021)

A. 216MeV

B. 0.9MeV

C. 9.4MeV

D. 804MeV

The correct answer is option (A)

Given binding energy per nucleon of X, Y & Z are

7.6MeV, 8.5MeV & 8.5MeV respectively,

Gain in binding energy is :

Q = Binding Energy of products $-$ Binding energy of reactants

= (120 $\times$ 8.5 $\times$ 2) $-$ (240 $\times$ 7.6) MeV

= 216 MeV

3. When a uranium isotope $_{92}^{235} U$ is bombarded with a neutron, it generates $_{36}^{89} K r$ three neutrons and : ⇒ (NEET 2020 Phase 1)

A. $_{40}^{91} Z r$

B. $_{36}^{101} K r$

C. $_{36}^{103} K r$

D. $_{56}^{144} B a$

The Correct Answer is Option (D)

$_{92}^{235} U$ + $_{0}^{1} n$ = $_{36}^{89} U$ + $_{0}^{1} n$ + $_{Z}^{A} X$

Balancing atomic number on both side, we get

92 + 0 = 36 + Z

So, Z = 56

Balancing mass number on both side, we get

235 + 1 = 89 + 3 + A

So, A = 144

It gives $_{56}^{144} B a$

4. The energy equivalent of 0.5 g of a substance is : ⇒ ( NEET 2020 Phase 1)

A. $4.5 \times 10^{13} J$

B. $1.5 \times 10^{13} J$

C. $0.5 \times 10^{13} J$

D. $4.5 \times 10^{16} J$

The Correct Answer is Option (A)

$E = m c^{2} = 0.5 \times 10^{- 3} \times {(3 \times 10^{8})}^{2} = 4.5 \times 10^{13} J$

5. The binding energy per nucleon of and nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction

$_{3}^{7} L i +_{1}^{1} H \to_{2}^{4} H e +_{2}^{4} H e + Q$

the value of energy Q released is ⇒ (AIPMT 2014)

A. 19.6 MeV

B. $-$ 2.4 MeV

C. 8.4 MeV

D. 17.3 MeV

The Correct Answer is Option (D)

Given:

Binding energy per nucleon of ₃Li⁷ and ₂He⁴ nuclei are 5.60 MeV and 7.06 MeV

Energy released = 7.06 × 8 – 5.60 × 7

= 17.3 MeV

6. How does the Binding Energy per nucleon vary with the increase in the number of nucleons ? ⇒ ( NEET 2013 (Karnataka))

A. Decrease continuously with mass number.

B. First decreases and then increases with increase in mass number.

C. First increases and then decreases with increase in mass number.

D. increases continuously with mass number.

The Correct Answer is Option (C)

Currently no explanation available

7. A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 u. The energy liberated per u is (given 1 u = 931 MeV) ⇒ (NEET 2013)

A. 6.675 MeV

B. 13.35 MeV

C. 2.67 MeV

D. 26.7 MeV

The Correct Answer is Option (A)

Mass defect $Δ$ m = 0.02866 a.m.u.

Energy = 0.02866 × 931 = 26.7 MeV

As ₁H² + ₁H² $\to$ ₂He⁴

Energy liberated per a.m.u = $\frac{13.35}{2}$ MeV = 6.675 MeV

8. The power obtained in a reactor using U²³⁵ disintegration is 1000 kW. The mass decay of U²³⁵ per hour is ⇒ (AIPMT 2011 Prelims)

A. 10 microgram

B. 20 microgram

C. 40 microgram

D. 1 microgram

The Correct Answer is Option (C)

From Einstein relation, E = mc²

Rearranging, m = $\frac{E}{c^{2}}$

We see that mass decay per second :

$\frac{d m}{d t}$ = $\frac{1}{c^{2}}$ × $\frac{d E}{d t}$ = $\frac{1000 \times 10^{3}}{{(3 \times 10^{8})}^{2}}$

Now mass decay of U²³⁵ per hour,

= $\frac{d m}{d t}$ $\times$ 60 $\times$ 60

= $\frac{1000 \times 10^{3}}{{(3 \times 10^{8})}^{2}}$ $\times$ 3600

= 4 × 10^–8 kg = 40 microgram

9. A radioactive nucleus of mass M emits a photon of frequency $v$ and the nucleus recoils. The recoil energy will be ⇒ ( AIPMT 2011 Prelims)

A. Mc² $-$ h $v$

B. h² $υ$ ²/2Mc²

C. Zero

D. h $v$

The Correct Answer is Option (B)

Momentum Mu = $\frac{E}{c} = \frac{h v}{c}$

Recoil energy :

$\frac{1}{2} M u^{2}$ = $\frac{1}{2} \frac{M^{2} u^{2}}{M}$

= $\frac{1}{2 M} {(\frac{h v}{c})}^{2}$

= $\frac{h^{2} ν^{2}}{2 M c}$

10. Fusion reaction takes place at high temperature because ⇒ (AIPMT 2011 Prelims)

A. nuclei break up at high temperature

B. atoms get ionised at high temperature

C. kinetic energy is high enough to overcome the coulomb repulsion between nuclei

D. molecules break up at high temperature

The Correct Answer is Option (C)

Extremely high temperature needed for fusion make kinetic energy large enough to overcome coulomb repulsion between nuclei.

11. The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is ⇒ (AIPMT 2010 Mains)

A. 23.6 MeV

B. 2.2 MeV

C. 28.0 MeV

D. 30.2 MeV

The Correct Answer is Option (A)

₁H² + ₁H² $\to$ ₂He⁴ + $Δ$ E

The binding energy per nucleon of a deuteron = 1.1 MeV

$∴$ Total binding energy = 2 × 1.1 = 2.2 MeV

The binding energy per nucleon of a helium nuclei = 7 MeV

$∴$ Total binding energy = 4 × 7 = 28 MeV

Hence, energy released

$Δ$ E = (28 – 2 × 2.2) = 23.6 MeV

12. The mass of a $_{3}^{7} L i$ Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of $_{3}^{7} L i$ nucleus is nearly⇒ (AIPMT 2010 Prelims)

A. 46 MeV

B. 5.6 MeV

C. 3.9 MeV

D. 23 MeV

The Correct Answer is Option (B)

For $_{3}^{7} L i$ nucleus,

Mass defect, $Δ$ M = 0.042 u

$∴$ 1 u = 931.5 MeV/c²

$∴$ $Δ$ M = 0.042 × 931.5 MeV/c² = 39.1 MeV/c²

Number of nucleons in $_{3}^{7} L i$ is 7.

Binding energy per nucleon,

E_bn = $\frac{39.1}{7}$ = 5.6 MeV

13. If M(A; Z), M_p and M_n denote the masses of the nucleus $_{Z}^{A} X,$ proton and neutron respectively in units of u (1 u = 931.5 MeV/c²) and BE represents its bonding energy in MeV, then ⇒ (AIPMT 2008)

A. M(A, Z) = ZM_p + (A $-$ Z)M_n $-$ BE

B. M(A, Z) = ZM_p + (A $-$ Z)M_n + BE/c²

C. M(A, Z) = ZM_p + (A $-$ Z)M_n $-$ BE/c²

D. M(A, Z) = ZM_p + (A $-$ Z)M_n + BE

The Correct Answer is Option (C)

Mass defect = ZM_p + (A –Z)M_n – M(A, Z)

$\Rightarrow$ $\frac{B . E}{c^{2}}$ = ZM_p + (A –Z)M_n – M(A, Z)

$∴$ M(A, Z) = ZM_p + (A $-$ Z)M_n $-$ $\frac{B . E}{c^{2}}$

14. A nucleus $_{Z}^{A} X$ has mass represented by M(A, Z). If M_p and M_n denote the mass of proton and neutron respectively and B.E. the binding energy in MeV, then ⇒ (AIPMT 2007 )

A. B.E. = [ZM_p + (A $-$ Z)M_n $-$ M(A, Z)]c²

B. B.E. = [ZM_p + AM_p $-$ M(A, Z)]c²

C. B.E. = M(A, Z) $-$ ZM_p $-$ (A $-$ Z)M_n

D. B.E. = [M(A, Z) $-$ ZM_p $-$ (A $-$ Z)M_n]c²

The Correct Answer is Option (A)

The difference in mass of a nucleus and its constituents, $Δ$ M, is called the mass defect and is given by

$Δ$ M = [ZM_p + (A $-$ Z)M_n] - M(A, Z)

binding energy = $Δ$ Mc²

= [ZM_p + (A $-$ Z)M_n $-$ M(A, Z)]c²

⇒ Nuclei

Topic 2 : Mass-Energy and Nuclear Reaction