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 has to be less than $4.0 \times 10^{-14} \mathrm{~m}$.

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 have been accurately measured and got the following formula.

A nucleus of mass number $\mathrm{A}$ has a radius $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ where $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}=1.2 \mathrm{fm}$ where $1 \mathrm{fm}=10^{-15} \mathrm{~m}$

The value of this constant is in the order of the range of the nuclear force and the value of $\mathrm{R}_{0}$ depends on the type of projectile particle.

Volume of nucleus,

$\mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}$

$=\frac{4}{3} \pi\left(\mathrm{R}_{0} \mathrm{~A}^{1 / 3}\right)^{3}$

$=\frac{4}{3} \pi \mathrm{R}_{0}^{3} \mathrm{~A}$

$\therefore \mathrm{V} \propto \mathrm{A}$ $\left[\because \frac{4}{3} \pi R_{0}^{3}\right.$ is constant $]$

and density of nucleus

$\rho=\frac{\mathrm{M}}{\mathrm{V}}=\frac{\mathrm{mA}}{\frac{4}{3} \pi \mathrm{R}_{0}^{3} \mathrm{~A}}=\frac{3 \mathrm{~m}}{4 \pi \mathrm{R}_{0}^{3}}$

Hence, density of nucleus does not depend on the mass number $A$. The density of substance of nucleus,

$\rho=\frac{3 \mathrm{~m}}{4 \pi \mathrm{R}_{0}^{3}}=\frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14 \times\left(1.2 \times 10^{-15}\right)^{3}}$

$\therefore \rho=0.22945 \times 10^{18}$

$\therefore \rho \approx 2.3 \times 10^{17} \mathrm{~kg} \mathrm{~m}^{-3}$

Hence, density of nucleus is $2.3 \times 10^{14}$ times that of density of water $\left(10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\right)$.

The density of the nucleus is very large because the atoms has a large amount of empty space.