Question

In fission, a nucleus of uranium–238, which contains 92 protons, divides into two smaller spheres, each having 46 protons and a radius of `5.9×10^-15 m`. What is the magnitude of the repulsive electric force pushing the two spheres apart?

Answer 1

The force's magnitude is approximately 3.5 kN.

Explanation:

`|F|=k(|q_1q_2|)/r^2`

k is Coulomb's constant, which is roughly `8.99times10^9 N C^-2 m^2`
`q_1=q_2=46(e)` where e is the charge on an electron, which is approximately `=1.602times10^-19C`.
`r_1=r_2=5.9times10^-15m`
The separation between the centres of the nuclei =`r=r_1+r_2=2r_1`

Assuming the nuclei are just touching when fission occurs:

`|F|=k(q_1^2)/r^2=k(46^2e^2)/(2times5.9times10^-15m)^2=(2116ke^2)/(34.81times10^-30 m^2)`

`|F| ~= (2116(8.99times10^9 N C^-2 m^2)(1.602times10^-19C)^2)/(34.81times10^-30 m^2)`

`|F| ~= 3506 N`

The force's magnitude is approximately 3.5 kN.