JEE Physics 2026 Question with Solution

Given: The nuclear reactions are

$${}^{3}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{4}_{2}\text{He} + 20 \, \text{MeV}$$

and

$${}^{4}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{5}_{2}\text{He} - 0.9 \, \text{MeV}$$

Find: The correct order of stability $X_3, X_4, X_5$.

Principle: Greater binding energy implies greater nuclear stability. If energy is released during formation, the formed nucleus is more stable. If energy is required, the formed nucleus is less stable.

From the first reaction,

$${}^{3}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{4}_{2}\text{He} + 20 \, \text{MeV}$$

a large amount of energy is released. Therefore, $^{4}_{2}\text{He}$ is more stable than $^{3}_{2}\text{He}$, so

$$X_4 > X_3$$

From the second reaction,

$${}^{4}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{5}_{2}\text{He} - 0.9 \, \text{MeV}$$

the negative energy value means energy is required to form $^{5}_{2}\text{He}$. Hence $^{5}_{2}\text{He}$ is less stable than $^{4}_{2}\text{He}$.

Using the given solution conclusion, the stability order is

$$X_4 > X_5 > X_3$$

Therefore, the correct option is A.

Given: Stability is to be compared using the energy released or absorbed in the nuclear reactions.

Find: Relative stability of $^{3}_{2}\text{He}$, $^{4}_{2}\text{He}$ and $^{5}_{2}\text{He}$.

1. Higher binding energy means the nucleus is more tightly bound and hence more stable.
2. In the reaction forming $^{4}_{2}\text{He}$, energy released is $20 \, \text{MeV}$, which shows strong binding.
3. In the reaction forming $^{5}_{2}\text{He}$, the term $-0.9 \, \text{MeV}$ indicates energy input is needed, so this nucleus is less stable than $^{4}_{2}\text{He}$.
4. Thus $^{4}_{2}\text{He}$ has the greatest stability, and from the extracted solution the remaining order is

$$X_4 > X_5 > X_3$$

Therefore, the correct option is A.