Atoms and Nuclei Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The binding energy of a nucleus is \( 240 \, \text{MeV} \) and its mass number is 30. What is the binding energy per nucleon?

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

\( E_b = 240 \, \text{MeV} \), \( A = 30 \).

\( E_{bn} = \frac{240}{30} = 8 \, \text{MeV} \).

7.5 MeV
7.8 MeV
8.0 MeV
8.5 MeV
3

What is the radius of a nucleus with mass number 32? (Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

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

\( A = 32 \), \( A^{1/3} = 32^{1/3} \approx 3.17 \).

\( R = 1.2 \times 10^{-15} \times 3.17 \approx 3.8 \times 10^{-15} \, \text{m} \).

3.6 × 10⁻¹⁵ m
4.0 × 10⁻¹⁵ m
4.2 × 10⁻¹⁵ m
3.8 × 10⁻¹⁵ m
4

A nucleus with mass number 70 has a binding energy of \( 595 \, \text{MeV} \). What is its binding energy per nucleon?

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

\( E_b = 595 \, \text{MeV} \), \( A = 70 \).

\( E_{bn} = \frac{595}{70} \approx 8.5 \, \text{MeV} \).

8.0 MeV
8.5 MeV
9.0 MeV
7.5 MeV
2

What is the energy equivalent of a neutron with mass \( 1.6749 \times 10^{-27} \, \text{kg} \) in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

\( E = m c^2 \).

\( m = 1.6749 \times 10^{-27} \, \text{kg} \), \( c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \).

\( E = 1.6749 \times 10^{-27} \times 9 \times 10^{16} \approx 1.507 \times 10^{-10} \, \text{J} \).

1.507 × 10⁻¹⁰ J
1.674 × 10⁻¹⁰ J
1.800 × 10⁻¹⁰ J
2.000 × 10⁻¹⁰ J
1

What happens to the nuclear force when nucleons are separated by more than a few femtometres?

The nuclear force is short-ranged and drops rapidly to zero beyond a few femtometres, leaving only weaker forces like Coulomb repulsion or gravity to act between nucleons.

Becomes stronger
Drops to zero
Becomes repulsive
Remains constant
2

What role does the binding energy play in nuclear stability?

Higher binding energy indicates a more tightly bound nucleus, making it more stable as it requires more energy to break apart, reflecting the strength of the nuclear force.

Reduces nuclear radius
Increases Coulomb force
Indicates tighter binding
Decreases nuclear density
3

Why does the nuclear force not distinguish between protons and neutrons?

The nuclear force is charge-independent, acting equally between neutron-neutron, proton-neutron, and proton-proton pairs, as it is a fundamental property unrelated to electric charge.

It is charge-independent
It depends on mass difference
It is weakened by Coulomb force
It varies with nuclear size
1

What is the mass defect of a nucleus with binding energy \( 216 \, \text{MeV} \)? (Given \( 1 \, \text{u} = 931.5 \, \text{MeV/c}^2 \))

\( \Delta M = \frac{E_b}{c^2} \).

\( \Delta M = \frac{216}{931.5} \approx 0.232 \, \text{u} \approx 0.23 \, \text{u} \).

0.20 u
0.22 u
0.24 u
0.23 u
4

A nucleus with mass number 40 has a binding energy of \( 320 \, \text{MeV} \). What is its binding energy per nucleon?

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

\( E_b = 320 \, \text{MeV} \), \( A = 40 \).

\( E_{bn} = \frac{320}{40} = 8.0 \, \text{MeV} \).

8.0 MeV
7.5 MeV
8.5 MeV
9.0 MeV
1

What is the significance of the mass-energy equivalence in nuclear physics?

Einstein’s \( E = m c^2 \) explains how mass defect converts into binding energy, unifying mass and energy conservation and enabling the understanding of energy release in nuclear reactions.

Explains binding energy
Determines nuclear radius
Measures atomic number
Increases nuclear density
1

What is the mass defect of a nucleus with binding energy \( 167.67 \, \text{MeV} \)? (Given \( 1 \, \text{u} = 931.5 \, \text{MeV/c}^2 \))

\( \Delta M = \frac{E_b}{c^2} \).

\( \Delta M = \frac{167.67}{931.5} \approx 0.18 \, \text{u} \).

0.16 u
0.17 u
0.18 u
0.19 u
3

How much energy is released when \( 5 \, \text{g} \) of matter is converted into energy? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

\( E = m c^2 \).

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

\( E = 5 \times 10^{-3} \times 9 \times 10^{16} = 4.5 \times 10^{14} \, \text{J} \).

4.0 × 10¹⁴ J
4.5 × 10¹⁴ J
5.0 × 10¹⁴ J
5.5 × 10¹⁴ J
2

What is the volume of a nucleus with radius \( 3.0 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

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

\( R^3 = (3.0 \times 10^{-15})^3 = 2.7 \times 10^{-44} \, \text{m}^3 \).

Volume = \( \frac{4}{3} \times 3.14 \times 2.7 \times 10^{-44} \approx 1.13 \times 10^{-43} \, \text{m}^3 \).

1.0 × 10⁻⁴³ m³
1.13 × 10⁻⁴³ m³
1.5 × 10⁻⁴³ m³
2.0 × 10⁻⁴³ m³
2

Which characteristic of the nuclear force explains its dominance over the Coulomb force in the nucleus?

The nuclear force is much stronger than the Coulomb force at short distances (within the nucleus), allowing it to overcome proton repulsion and bind nucleons together.

Long-range nature
Greater strength
Charge dependence
Weak interaction
2

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