JEE Physics 2024 Question with Solution

Given:

• Reaction 1:

$$^6_3\text{Li} + ^1_0\text{n} \rightarrow ^4_2\text{He} + ^3_1\text{H}$$

• Reaction 2:

$$^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n}$$

• Atomic masses:
• $M(^6_3\text{Li}) = 6.01690 \, \text{amu}$
• $M(^3_1\text{H}) = 3.01605 \, \text{amu}$
• $M(^4_2\text{He}) = 4.00388 \, \text{amu}$
• $M(^1_0\text{n}) = 1.008665 \, \text{amu}$
• $M(^2_1\text{H}) = 2.01471 \, \text{amu}$

Find: The total energy released during the explosion.

Use

$$\Delta E = \Delta m \times 931.5 \, \text{MeV/amu}$$

For the first reaction:

$$\text{Reactant mass} = 6.01690 + 1.008665 = 7.025565 \, \text{amu}$$ $$\text{Product mass} = 4.00388 + 3.01605 = 7.01993 \, \text{amu}$$ $$\Delta m_1 = 7.025565 - 7.01993 = 0.005635 \, \text{amu}$$ $$\Delta E_1 = 0.005635 \times 931.5 = 5.25 \, \text{MeV} \text{ (approximately)}$$

For the second reaction:

$$\text{Reactant mass} = 2.01471 + 3.01605 = 5.03076 \, \text{amu}$$ $$\text{Product mass} = 4.00388 + 1.008665 = 5.012545 \, \text{amu}$$ $$\Delta m_2 = 5.03076 - 5.012545 = 0.018215 \, \text{amu}$$ $$\Delta E_2 = 0.018215 \times 931.5 = 16.97 \, \text{MeV}$$

Therefore,

$$\Delta E_{\text{total}} = \Delta E_1 + \Delta E_2 = 5.25 + 16.97 = 22.22 \, \text{MeV}$$

Therefore, the energy released during the explosion is $22.22 \, \text{MeV}$. The correct option is D.

Given: The two reactions occur successively, and the neutron as well as tritium cancel when the reactions are added.

Find: The net energy released.

Combine the reactions to get the total reaction:

$$^6_3\text{Li} + ^2_1\text{H} \rightarrow 2\,^4_2\text{He}$$

Now calculate the Q-value directly:

$$Q = \left[M(^6_3\text{Li}) + M(^2_1\text{H}) - 2M(^4_2\text{He})\right] \times 931.5 \, \text{MeV}$$ $$Q = \left[6.01690 + 2.01471 - 2 \times 4.00388\right] \times 931.5 \, \text{MeV}$$ $$Q = \left[8.03161 - 8.00776\right] \times 931.5 \, \text{MeV}$$ $$Q = 0.02385 \times 931.5 = 22.22 \, \text{MeV}$$

So, the total energy released is $22.22 \, \text{MeV}$, hence the correct option is D.