Question

Consider a gold nucleus to be a sphere of radius 6.9 fermi in which protons and neutrons are distributed. Find the force of repulsion between two protons situated at largest separation. Why do these protons not fly apart under this repulsion?

Answer 1

Given, radius of the sphere, R = 6.9 fermi
So, the largest separation between two protons = 2R = 13.8 fermi
Charge on a proton, q = $1.6\times {10}^{-19}\mathrm{C}$
By Coulomb's Law, force of repulsion,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1{q}_2}{r^2}$
$\Rightarrow F=9\times {10}^9\times \frac{{\left(1.6\times {10}^{-19}\right)}^2}{{\left(2R\right)}^2}=1.2\mathrm{N}$
Inside the nucleus, another short-range attractive force (nuclear force) acts on the protons. That's why these protons do not fly apart due to the Coulombian repulsion.