Atoms and Nuclei Chapter-Wise Test 21

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A nucleus with mass number 36 has a binding energy of \( 288 \, \text{MeV} \). What is its binding energy per nucleon?

\( E_{bn} = \frac{E_b}{A} \).

\( E_b = 288 \, \text{MeV} \), \( A = 36 \).

\( E_{bn} = \frac{288}{36} = 8.0 \, \text{MeV} \).

7.5 MeV
7.8 MeV
8.2 MeV
8.0 MeV
4

Which statement is true about the energy release in nuclear processes?

Nuclear processes (fission and fusion) release energy when less tightly bound nuclei transform into more tightly bound ones, increasing the binding energy per nucleon.

Energy is absorbed in all cases
Only fission releases energy
Only fusion releases energy
Energy is released when binding increases
4

Why is the nuclear force considered saturated in large nuclei?

The nuclear force is short-ranged, affecting only a fixed number of neighboring nucleons, so adding more nucleons in large nuclei does not proportionally increase the binding energy, leading to saturation.

Due to charge dependence
Because of Coulomb force
Limited range of interaction
Increase in nuclear volume
3

A nucleus with mass number 12 has a binding energy of \( 96 \, \text{MeV} \). What is its binding energy per nucleon?

\( E_{bn} = \frac{E_b}{A} \).

\( E_b = 96 \, \text{MeV} \), \( A = 12 \).

\( E_{bn} = \frac{96}{12} = 8.0 \, \text{MeV} \).

7.5 MeV
7.8 MeV
8.2 MeV
8.0 MeV
4

What is the role of high temperature in nuclear fusion?

High temperatures provide nuclei with sufficient kinetic energy to overcome the Coulomb barrier (electrostatic repulsion), allowing them to come close enough for the nuclear force to cause fusion.

Increases nuclear density
Reduces binding energy
Enhances Coulomb force
Overcomes Coulomb barrier
4

What is the nuclear density of a nucleus with mass \( 3.67 \times 10^{-27} \, \text{kg} \) and radius \( 2.0 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

Density = \( \frac{\text{mass}}{\text{volume}} \), Volume = \( \frac{4}{3} \pi R^3 \).

\( R^3 = (2.0 \times 10^{-15})^3 = 8.0 \times 10^{-45} \, \text{m}^3 \).

Volume = \( \frac{4}{3} \times 3.14 \times 8.0 \times 10^{-45} \approx 3.35 \times 10^{-44} \, \text{m}^3 \).

Density = \( \frac{3.67 \times 10^{-27}}{3.35 \times 10^{-44}} \approx 1.10 \times 10^{17} \, \text{kg/m}^3 \).

1.0 × 10¹⁷ kg/m³
1.05 × 10¹⁷ kg/m³
1.10 × 10¹⁷ kg/m³
1.15 × 10¹⁷ kg/m³
3

What is the radius of a nucleus with mass number 16? (Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

The radius of a nucleus is given by \( R = R_0 A^{1/3} \), where \( A = 16 \).

\( A^{1/3} = 16^{1/3} = (2^4)^{1/3} = 2^{4/3} \approx 2.52 \).

\( R = 1.2 \times 10^{-15} \times 2.52 \approx 3.0 \times 10^{-15} \, \text{m} \).

3.0 × 10⁻¹⁵ m
2.4 × 10⁻¹⁵ m
3.6 × 10⁻¹⁵ m
4.8 × 10⁻¹⁵ m
1

A nucleus has a binding energy of \( 127.5 \, \text{MeV} \) and mass number 16. What is its binding energy per nucleon?

\( E_{bn} = \frac{E_b}{A} \).

\( E_b = 127.5 \, \text{MeV} \), \( A = 16 \).

\( E_{bn} = \frac{127.5}{16} \approx 7.97 \, \text{MeV} \approx 8 \, \text{MeV} \).

7.5 MeV
7.8 MeV
8.2 MeV
8.0 MeV
4

A nucleus with mass number 28 has a binding energy of \( 224 \, \text{MeV} \). What is its binding energy per nucleon?

\( E_{bn} = \frac{E_b}{A} \).

\( E_b = 224 \, \text{MeV} \), \( A = 28 \).

\( E_{bn} = \frac{224}{28} = 8.0 \, \text{MeV} \).

8.0 MeV
7.5 MeV
8.5 MeV
9.0 MeV
1

A nucleus with mass number 18 has a binding energy of \( 144 \, \text{MeV} \). What is its binding energy per nucleon?

\( E_{bn} = \frac{E_b}{A} \).

\( E_b = 144 \, \text{MeV} \), \( A = 18 \).

\( E_{bn} = \frac{144}{18} = 8.0 \, \text{MeV} \).

7.5 MeV
7.8 MeV
8.2 MeV
8.0 MeV
4

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