The unit of length convenient on the nuclear scale is a fermi: $1 \; f = 10^{-15} \; m$. Nuclear sizes obey roughly the following empirical relation: $r = r_{0} A^{1/3}$,where $r$ is the radius of the nucleus,$A$ is its mass number,and $r_{0}$ is a constant equal to about $1.2 \; f$. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of a sodium nucleus.

(N/A) The radius of a nucleus $r$ is given by the relation: $r = r_{0} A^{1/3}$ $(i)$,where $r_{0} = 1.2 \; f = 1.2 \times 10^{-15} \; m$.
The volume of the nucleus $V$ is given by: $V = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (r_{0} A^{1/3})^{3} = \frac{4}{3} \pi r_{0}^{3} A$.
The mass of a nucleus $M$ is approximately equal to its mass number $A$ in atomic mass units (amu): $M = A \times 1.66 \times 10^{-27} \; kg$.
The density of the nucleus $\rho$ is given by: $\rho = \frac{\text{Mass of nucleus}}{\text{Volume of nucleus}} = \frac{A \times 1.66 \times 10^{-27}}{\frac{4}{3} \pi r_{0}^{3} A} = \frac{3 \times 1.66 \times 10^{-27}}{4 \pi r_{0}^{3}} \; kg/m^{3}$.
This relation shows that the nuclear mass density depends only on the constant $r_{0}$ and is independent of $A$. Hence,the nuclear mass densities of all nuclei are nearly the same.
For a sodium nucleus,the density is: $\rho_{\text{sodium}} = \frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14 \times (1.2 \times 10^{-15})^{3}} \approx 2.29 \times 10^{17} \; kg/m^{3}$.