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(A) We know that the radius $(R)$ of a nucleus is given by the relation: $R = R_0 A^{1/3}$,which implies $R \propto A^{1/3}$.Here,$R_0$ is a constant

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The nuclear density,in general,is independent of mass number $A$.

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(A) We know that the radius $(R)$ of a nucleus is given by the relation: $R = R_0 A^{1/3}$,which implies $R \propto A^{1/3}$.Here,$R_0$ is a constant and $A$ is the mass number.Nuclear density $(ho)$ is defined as the ratio of mass to volume: $ho = \frac{	ext{mass}}{	ext{volume}} = \frac{m A}{\frac{4}{3} \pi R^3}$,where $m$ is the mass of a single nucleon.Substituting $R = R_0 A^{1/3}$,we get: $ho = \frac{m A}{\frac{4}{3} \pi (R_0 A^{1/3})^3} = \frac{3m}{4 \pi R_0^3}$.Since $m$ and $R_0$ are constants,the nuclear density $ho$ is independent of the mass number $A$.Regarding other options: Binding energy $E = \Delta m c^2$,so $E \propto \Delta m$ (directly proportional,not inversely).Energy is released (not just observed) during nuclear fission of heavy nuclei into lighter ones.

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Show that the volume of a nucleus is proportional to its atomic mass number $A$.

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