Question: Q. 6. Binding energy per nucleon versus mass number curve is as shown.
${ }{Z}^{A} \mathrm{~S},{ }{Z 1}^{A 1} \mathrm{~W},{ }{Z 2}^{A 2} \mathrm{X}$ and ${ }{Z 3}^{A 3} \mathrm{Y}$ are four nuclei indicated on the curve.
Based on the graph :
(a) Arrange $X, W$ and $S$ in the increasing order of stability.
(b) Write the relation between the relevant $A$ and $Z$ values for the following nuclear reaction.
$S \rightarrow X+W$
(c) Explain why binding energy for heavy nuclei is low.
[SQP 2018]
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Solution:
Ans. (a) $S, W, X$ 1 (b) $Z=Z_{1}+Z_{2}$ $A=A_{1}+A_{2}$
(c) Reason for low binding energy :
In heavier nuclei, the Coulombian repulsive effects can increase considerably and can match/ offset the attractive effects of the nuclear forces. This can result in such nuclei being unstable. $\mathbf{1}$
