From the relation $R = R_{0} A^{1/3}$,where $R_{0}$ is a constant and $A$ is the mass number of a nucleus,show that the nuclear matter density is nearly constant (i.e.,independent of $A$).

(N/A) The nuclear radius is given by the relation:
$R = R_{0} A^{1/3}$
where $R_{0}$ is a constant and $A$ is the mass number.
The nuclear matter density $\rho$ is defined as the ratio of the mass of the nucleus to its volume:
$\rho = \frac{\text{Mass of the nucleus}}{\text{Volume of the nucleus}}$
Let $m$ be the average mass of a nucleon. The total mass of the nucleus is approximately $m \times A$.
The volume of the nucleus,assuming it is spherical,is $V = \frac{4}{3} \pi R^{3}$.
Substituting the expression for $R$:
$\rho = \frac{mA}{\frac{4}{3} \pi (R_{0} A^{1/3})^{3}}$
$\rho = \frac{mA}{\frac{4}{3} \pi R_{0}^{3} A}$
Canceling $A$ from the numerator and denominator:
$\rho = \frac{3m}{4 \pi R_{0}^{3}}$
Since $m$,$\pi$,and $R_{0}$ are constants,the nuclear density $\rho$ is independent of the mass number $A$ and is nearly constant.