Atoms and Nuclei Chapter-Wise Test 20

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the energy equivalent of \( 0.001 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

\( E = m c^2 \).

\( m = 0.001 \, \text{kg} \), \( c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \).

\( E = 0.001 \times 9 \times 10^{16} = 9 \times 10^{13} \, \text{J} \).

9 × 10¹³ J
8 × 10¹³ J
10 × 10¹³ J
7 × 10¹³ J
1

A nucleus has a mass defect of \( 0.136 \, \text{u} \). What is its binding energy in MeV? (Given \( 1 \, \text{u} = 931.5 \, \text{MeV/c}^2 \))

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

\( \Delta M = 0.136 \, \text{u} \).

Energy = \( 0.136 \times 931.5 \approx 126.7 \, \text{MeV} \).

125.0 MeV
126.7 MeV
130.0 MeV
132.5 MeV
2

Which process cannot produce elements beyond the peak of the binding energy curve?

Fusion produces heavier nuclei with higher binding energy per nucleon up to the peak (around A = 56), but beyond this, further fusion does not release energy, limiting element synthesis.

Nuclear fission
Chemical reactions
Nuclear fusion
Mass-energy conversion
3

What is the volume of a nucleus with radius \( 2.7 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

Volume = \( \frac{4}{3} \pi R^3 \).

\( R^3 = (2.7 \times 10^{-15})^3 = 1.9683 \times 10^{-44} \, \text{m}^3 \).

Volume = \( \frac{4}{3} \times 3.14 \times 1.9683 \times 10^{-44} \approx 8.25 \times 10^{-44} \, \text{m}^3 \).

7.5 × 10⁻⁴⁴ m³
8.25 × 10⁻⁴⁴ m³
9.0 × 10⁻⁴⁴ m³
10.0 × 10⁻⁴⁴ m³
2

What is the nuclear density of a nucleus with mass \( 5.01 \times 10^{-27} \, \text{kg} \) and radius \( 2.1 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

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

\( R^3 = (2.1 \times 10^{-15})^3 = 9.261 \times 10^{-45} \, \text{m}^3 \).

Volume = \( \frac{4}{3} \times 3.14 \times 9.261 \times 10^{-45} \approx 3.88 \times 10^{-44} \, \text{m}^3 \).

Density = \( \frac{5.01 \times 10^{-27}}{3.88 \times 10^{-44}} \approx 1.29 \times 10^{17} \, \text{kg/m}^3 \).

1.29 × 10¹⁷ kg/m³
2.31 × 10¹⁷ kg/m³
1.15 × 10¹⁷ kg/m³
1.50 × 10¹⁷ kg/m³
1

A nucleus with mass number 216 has a radius of \( 7.2 \times 10^{-15} \, \text{m} \). What is the value of \( R_0 \)?

\( R = R_0 A^{1/3} \).

\( A = 216 \), \( A^{1/3} = 6 \).

\( R_0 = \frac{R}{A^{1/3}} = \frac{7.2 \times 10^{-15}}{6} = 1.2 \times 10^{-15} \, \text{m} \).

1.0 × 10⁻¹⁵ m
1.1 × 10⁻¹⁵ m
1.3 × 10⁻¹⁵ m
1.2 × 10⁻¹⁵ m
4

What is the primary outcome of nuclear fission in terms of energy?

Nuclear fission converts mass defect into energy, initially as kinetic energy of fragments and neutrons, which is eventually transferred as heat, due to the increase in binding energy per nucleon.

Increase in nuclear radius
Mass converted to energy
Decrease in nuclear density
Reduction in nucleon count
2

Which of the following best describes the nuclear force?

The nuclear force is a strong, short-range attractive force that binds protons and neutrons in the nucleus, overcoming the Coulomb repulsion between protons. It is much stronger than the Coulomb force and does not depend on the electric charge of the nucleons.

Strong, short-range, and charge-independent
Weak, long-range, and charge-dependent
Strong, long-range, and charge-dependent
Weak, short-range, and charge-independent
1

What is the binding energy of a nucleus if its mass defect is \( 0.15 \, \text{u} \)? (Given \( 1 \, \text{u} = 931.5 \, \text{MeV/c}^2 \))

\( E_b = \Delta M \cdot c^2 \).

\( \Delta M = 0.15 \, \text{u} \).

\( E_b = 0.15 \times 931.5 = 139.725 \, \text{MeV} \approx 139.73 \, \text{MeV} \).

130.0 MeV
135.0 MeV
139.73 MeV
145.0 MeV
3

A nucleus with mass number 100 has a binding energy of \( 850 \, \text{MeV} \). What is its binding energy per nucleon?

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

\( E_b = 850 \, \text{MeV} \), \( A = 100 \).

\( E_{bn} = \frac{850}{100} = 8.5 \, \text{MeV} \).

8.0 MeV
8.5 MeV
9.0 MeV
7.5 MeV
2

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