The binding energy for the following nuclear | vedclass

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(A) The stability of a nucleus is directly related to its binding energy per nucleon. However, in these reactions, we can compare the binding energy $

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$X_4 > X_3 > X_5$

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(A) The stability of a nucleus is directly related to its binding energy per nucleon. However, in these reactions, we can compare the binding energy $(BE)$ directly to determine relative stability.From the first reaction: ${ }_2 He ^3 + { }_0 n ^1 \rightarrow { }_2 He ^4 + 20 \ MeV$. The energy released $(Q = 20 \ MeV)$ implies that $BE({ }_2 He ^4) - BE({ }_2 He ^3) = 20 \ MeV$. Thus, $BE({ }_2 He ^4) > BE({ }_2 He ^3)$.From the second reaction: ${ }_2 He ^4 + { }_0 n ^1 \rightarrow { }_2 He ^5 - 0.9 \ MeV$. The energy absorbed $(Q = -0.9 \ MeV)$ implies that $BE({ }_2 He ^5) - BE({ }_2 He ^4) = -0.9 \ MeV$. Thus, $BE({ }_2 He ^4) > BE({ }_2 He ^5)$.Comparing the two, we find $BE({ }_2 He ^4) > BE({ }_2 He ^3)$ and $BE({ }_2 He ^4) > BE({ }_2 He ^5)$.Since ${ }_2 He ^4$ is a very stable alpha particle (magic number $Z=2, N=2$), it has the highest stability. Comparing ${ }_2 He ^3$ and ${ }_2 He ^5$, ${ }_2 He ^3$ is more stable than ${ }_2 He ^5$ because ${ }_2 He ^5$ is highly unstable and decays rapidly.Therefore, the order of stability is $X_4 > X_3 > X_5$.

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