How is the radius of the nucleus estimated? Write its relation to the radius and atomic mass number.

(N/A) At Rutherford's suggestion,Geiger and Marsden performed their experiment on the scattering of $\alpha$-particles from thin gold foils. Their experiments revealed that the actual size of the nucleus of gold must be less than $4.0 \times 10^{-14} \ m$.
By performing scattering experiments in which fast electrons,instead of $\alpha$-particles,are projectiles that bombard targets made up of various elements,the size of the nuclei of various elements has been accurately measured,leading to the following formula:
$A$ nucleus of mass number $A$ has a radius $R = R_{0} A^{1/3}$,where $R_{0} = 1.2 \times 10^{-15} \ m = 1.2 \ fm$ and $1 \ fm = 10^{-15} \ m$.
The value of this constant is in the order of the range of the nuclear force. The volume of the nucleus is:
$V = \frac{4}{3} \pi R^{3} = \frac{4}{3} \pi (R_{0} A^{1/3})^{3} = \frac{4}{3} \pi R_{0}^{3} A$.
Therefore,$V \propto A$,which implies that the volume is directly proportional to the mass number. The density of the nucleus is:
$\rho = \frac{M}{V} = \frac{m A}{\frac{4}{3} \pi R_{0}^{3} A} = \frac{3m}{4 \pi R_{0}^{3}}$.
Hence,the density of the nucleus does not depend on the mass number $A$. Calculating the density:
$\rho = \frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14 \times (1.2 \times 10^{-15})^{3}} \approx 2.3 \times 10^{17} \ kg \ m^{-3}$.
This density is approximately $2.3 \times 10^{14}$ times that of water,indicating that the nucleus is extremely dense due to the large amount of empty space in an atom.