Question 126: Atoms & Nuclei

Question

The binding energy for the following nuclear reactions are expressed in MeV .
$$ \begin{aligned} & { }_2 \mathrm{He}^3+{ }_0 \mathrm{n}^1 ightarrow{ }_2 \mathrm{He}^4+20 \mathrm{MeV} \ & { }_2 \mathrm{He}^4+{ }_0 \mathrm{n}^1 ightarrow{ }_2 \mathrm{He}^5-0.9 \mathrm{MeV} \end{aligned} $$
If $\mathrm{X}_3, \mathrm{X}_4, \mathrm{X}_5$ denote the stability of ${ }_2 \mathrm{He}^3,{ }_2 \mathrm{He}^4$ and ${ }_2 \mathrm{He}^5$, respectively, then the correct order is :

Accepted Answer

A nucleus is more stable if it has a higher binding energy i.e., is in a lower energy state.
In exothermic reaction energy is released, $\Delta Q>0$. The products are more stable than the reactants.
In endothermic reaction energy is absorbed, $\Delta Q\mathrm{X}_3$.
In ${ }_2^4 \mathrm{He}+{ }_0^1 \mathrm{n} \rightarrow{ }_2^5 \mathrm{He}-0.9 \mathrm{MeV}$ the absorbed energy is $0.9 \mathrm{MeV}(\mathrm{Q}=-0.9 \mathrm{MeV})$.
Since energy must be supplied to this reaction, the product ${ }_2^5 \mathrm{He}$ is less stable than the reactant ${ }_2^4 \mathrm{He}$.
Therefore, $\mathrm{X}_4>\mathrm{X}_5$.
From reaction 1 , the jump in stability from $\mathrm{X}_3$ to $\mathrm{X}_4$ is very large ( 20 MeV ).
From reaction 2, the drop in stability from $\mathrm{X}_4$ to $\mathrm{X}_5$ is very small ( 0.9 MeV ).
Because $\mathrm{X}_5$ is only slighty less stable than $\mathrm{X}_4$, but $\mathrm{X}_3$ is significantly less stable than $\mathrm{X}_4$, we can conclude that $\mathrm{X}_5$ is still more stable than $\mathrm{X}_3$.
Therefore, the order of stability is $X_4>X_5>X_3$.
Hence, the correct option is (A).

Answer

$X_4>X_5>X_3$

Answer

$X_4 < X_5 < X_3$

Answer

$X_4 > X_5 < X_3$

Answer

$X_4=X_5=X_3$

Atoms & Nuclei

Choose the correct answer: