Question

At the point of fission, a nucleus of ^235 U that has 92 protons is divided into two smaller spheres, each of which has 46 protons and a radius of 5.90 X 10 ^-15m. What is the magnitude of the repulsive force pushing these two spheres apart

Answer 1

Answer : $F = 3.5\times10^{3}\ N$

Explanation :

Given that

Radius of sphere $r = 5.90\times 10^{-15}\ m$

The distance between the centers of the two spheres is

$r = 2\times 5.90\times 10^{-15}\ m$

The charge of the sphere $q = 46\times1.6\times10^{-19} C$

The magnitude of the repulsive force between the charges pushing them a part is

Using coulomb law

$F = \dfrac {kq_{1}q_{2}}{r^{2}}$

$F = \dfrac{9\times10^{9}\times (46\times1.6\times10^{-19})^{2}C^{2}}{2\times(5.90\times10^{-15})^{2}\ m^{2}}$

$F = 3501.3\ N$

$F = 3.5\times10^{3}\ N$

Hence, this is the required solution.