Question

Consider a gold nucleus to be a sphere of radius $6\cdot 9$ 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

According to the question,

maximum separation between protons $=2r=2\times 6.9$

$=13.8ferni$

$d=13.8\times {10}^{-15}m$

Charge on each proton $=1.6\times {10}^{-19}$

$\begin{array}{c}F=\frac{9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{{\left(1.38\times {10}^{14}\right)}^2}\\ =\frac{9\times 2.56}{1.38\times 1.38\times {10}^{9-38+28}}\\ =12.09\times {10}^{-1}N\\ =1.209N\end{array}$