The nuclear density in general, is independent of mass number $A$.
The radius of nucleus is directly proportional to the mass number $A$ of the nucleus.
The binding energy of a nucleus is inversely proportional to its mass defect.
Energy is observed when heavy nuclei undergo. transmutation into light nuclei.
Solution
We know that, the radius $(R)$ of nucleus is given as
$R=R_0 A^{1 / 3}$ or $R \propto A^{1 / 3}$
Where,
$\begin{aligned} R_0 & =\text { constant } \\ A & =\text { mass number }\end{aligned}$
$\therefore$ Nuclear density, $\rho=\frac{\text { mass }}{\text { volume }}=\frac{m A}{\frac{4}{3} \pi R^3}$ (here, $m$ is mass of each nucleon)
$\rho=\frac{3 m}{4 \pi R_0^3}$
Hence, it is clear from above formula, nuclear density is independent of mass number $A$.
The binding energy, $E=\Delta m c^2$
i. e $E \propto \Delta m$
Where, $\Delta m=$ mass defect
Energy is released when heavy nuclei undergo transmutation into light nuclei.