Question

From the relation $R = R_0A^{\frac{1}{3}}$, 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).

Answer 1

We have the expression for nuclear radius as:
$R = R_0A^{\frac{1}{3}}$
Where,
$R_0$ = Constant.
A = Mass number of the nucleus
Nuclear matter density,$ρ =\frac{ Mass \space of\space the\space Nucleus}{Volume \space of \space the \space Nucleus}$
Let m be the average mass of the nucleus.
Hence, mass of the nucleus = mA
$ρ = \frac{mA}{\frac{4}{3}\pi R^3}$ = $\frac{3mA}{4\pi (R_oA\frac{1}{3})^3}$ = $\frac{3mA}{4πR_{o}^{3}A}$ = $\frac{3m}{4\pi R_{o}^{3}}$
Hence, the nuclear matter density is independent of A. It is nearly constant.