JEE Physics 2023 Question with Solution

Given: $^{238}_{92} U = 238.05060 \, u$, $^{234}_{90} Th = 234.04360 \, u$, $^{4}_{2} He = 4.00260 \, u$, and $1u = 931.5 \, MeV/c^2$.

Find: The energy released $Q$ during alpha decay.

For nuclear decay, the released energy is obtained from the mass defect:

$$Q = (\Delta m)_{amu} \times 931.5 \, MeV$$

The mass defect is:

$$\Delta m = m_U - m_{Th} - m_{He}$$

Substituting the given values:

$$\Delta m = 238.05060 \, u - 234.04360 \, u - 4.00260 \, u = 0.0044 \, u$$

Now,

$$Q = 0.0044 \times 931.5 \, MeV = 4.0986 \, MeV$$

Therefore, the energy released during the alpha decay of $^{238}_{92} U$ is $4.0986 \, MeV$.

Given: Parent and product nuclear masses are provided.

Find: The decay energy $Q$.

A quick method is to directly subtract the total product mass from the parent mass, then multiply by $931.5$:

$$Q = \left(238.05060 - 234.04360 - 4.00260\right) \times 931.5$$ $$Q = 0.0044 \times 931.5 = 4.0986 \, MeV$$

This works because $1u$ of mass defect corresponds to $931.5 \, MeV$ of energy. Hence, the required answer is $4.0986$.