Question

In nuclear fission, a nucleus of uranium-238, containing 92 protons, can divide into two smaller spheres, each having 46 protons and a radius of 5.90 ✕ 10-15 m. What is the magnitude of the repulsive electric force pushing the two spheres apart?

Answer 1

Answer:

$F = 3501.34 N$

Explanation:

As we know that electrostatic repulsion force between two protons is given as

$F = \frac{kq_1q_2}{r^2}$

now we know that

$q_1 = q_2 = 46 \times (1.6 \times 10^{-19})$

$q_1 = q_2 = 7.36 \times 10^{-18} C$

now the distance between the two atoms is same as the distance between two center

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

$r = 1.18 \times 10^{-14} m$

now from above formula we have

$F = \frac{(9 \times 10^9)(7.36 \times 10^{-18})^2}{(1.18 \times 10^{-14})^2}$

$F = 3501.34 N$