Atoms and Nuclei Chapter-Wise Test 18

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the nuclear density of a nucleus with mass \( 1.66 \times 10^{-27} \, \text{kg} \) and radius \( 1.5 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

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

\( R^3 = (1.5 \times 10^{-15})^3 = 3.375 \times 10^{-45} \, \text{m}^3 \).

Volume = \( \frac{4}{3} \times 3.14 \times 3.375 \times 10^{-45} \approx 1.41 \times 10^{-44} \, \text{m}^3 \).

Density = \( \frac{1.66 \times 10^{-27}}{1.41 \times 10^{-44}} \approx 1.18 \times 10^{17} \, \text{kg/m}^3 \).

1.0 × 10¹⁷ kg/m³
1.1 × 10¹⁷ kg/m³
1.2 × 10¹⁷ kg/m³
1.18 × 10¹⁷ kg/m³
4

A nucleus has a mass defect of \( 0.05 \, \text{u} \). What is its binding energy in MeV? (Given \( 1 \, \text{u} = 931.5 \, \text{MeV/c}^2 \))

Binding energy = \( \Delta M \cdot c^2 \).

\( \Delta M = 0.05 \, \text{u} \).

\( E_b = 0.05 \times 931.5 = 46.575 \, \text{MeV} \approx 46.58 \, \text{MeV} \).

46.58 MeV
93.15 MeV
23.29 MeV
139.73 MeV
1

What is the primary challenge in achieving controlled thermonuclear fusion?

Controlled thermonuclear fusion requires extremely high temperatures (e.g., \( 10^8 \, \text{K} \)), producing a plasma that no container can withstand, making confinement (e.g., using magnetic fields) a major challenge.

Lack of fuel availability
Confinement of high-temperature plasma
Excessive neutron production
Low energy output
2

What is the approximate radius of a nucleus with mass number 64? (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 \( A = 64 \).

\( A^{1/3} = 64^{1/3} = 4 \).

\( R = 1.2 \times 10^{-15} \times 4 = 4.8 \times 10^{-15} \, \text{m} \).

4.8 × 10⁻¹⁵ m
3.6 × 10⁻¹⁵ m
6.0 × 10⁻¹⁵ m
2.4 × 10⁻¹⁵ m
1

A nucleus with mass number 50 has a binding energy of \( 425 \, \text{MeV} \). What is its binding energy per nucleon?

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

\( E_b = 425 \, \text{MeV} \), \( A = 50 \).

\( E_{bn} = \frac{425}{50} = 8.5 \, \text{MeV} \).

8.0 MeV
8.2 MeV
8.5 MeV
8.8 MeV
3

What is the energy equivalent of \( 0.15 \, \text{g} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

\( E = m c^2 \).

\( m = 0.15 \times 10^{-3} \, \text{kg} = 1.5 \times 10^{-4} \, \text{kg} \), \( c^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \).

\( E = 1.5 \times 10^{-4} \times 9 \times 10^{16} = 1.35 \times 10^{13} \, \text{J} \).

1.2 × 10¹³ J
1.3 × 10¹³ J
1.4 × 10¹³ J
1.35 × 10¹³ J
4

Why does nuclear fusion require extremely high temperatures?

High temperatures provide nuclei with enough kinetic energy to overcome the electrostatic repulsion (Coulomb barrier) between positively charged nuclei, enabling fusion via the nuclear force.

To increase nuclear density
To reduce binding energy
To overcome electrostatic repulsion
To enhance nuclear force
3

What is the energy equivalent of \( 0.05 \, \text{g} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

\( E = m c^2 \).

\( m = 0.05 \times 10^{-3} \, \text{kg} = 5 \times 10^{-5} \, \text{kg} \), \( c^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \).

\( E = 5 \times 10^{-5} \times 9 \times 10^{16} = 4.5 \times 10^{12} \, \text{J} \).

4.0 × 10¹² J
4.5 × 10¹² J
5.0 × 10¹² J
5.5 × 10¹² J
2

Why is the nuclear density nearly constant across all nuclei?

The volume of a nucleus is proportional to \( A \) (since \( R \propto A^{1/3} \) and \( V \propto R^3 \)), and the mass is also proportional to \( A \), making the density (mass/volume) independent of \( A \).

Due to varying nuclear force
Because of electron scattering
Volume proportional to mass number
Increase in Coulomb repulsion
3

Why is the binding energy per nucleon a useful measure of nuclear stability?

It represents the average energy needed to remove a nucleon, with higher values indicating a more stable nucleus due to stronger binding against disruptive forces like Coulomb repulsion.

Measures nuclear radius
Indicates nuclear density
Shows Coulomb force strength
Average energy per nucleon
4

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