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

`F = 3501.34 N`

Step-by-step explanation:

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

`F = (kq_1q_2)/(r^2)`

now we know that

`q_1 = q_2 = 46 * (1.6 * 10^(-19))`

`q_1 = q_2 = 7.36 * 10^(-18) C`

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

`r = 5.90 * 10^(-15) + 5.90 * 10^(-15)`

`r = 1.18 * 10^(-14) m`

now from above formula we have

`F = ((9 * 10^9)(7.36 * 10^(-18))^2)/((1.18 * 10^(-14))^2)`

`F = 3501.34 N`