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.