Question

At the point of fission, a nucleus of 235U that has 92 protons is divided into two smaller spheres, each of which has 46 protons and a radius of 5.83 ✕ 10-15 m. What is the magnitude of the repulsive force pushing these two spheres apart?

Answer 1

Answer:

3581.93928018 N

Explanation:

Number of protons = 46

r = Radius = $2\times 5.83\times 10^{-15}=1.166\times 10^{-14}\ m$

q = Charge = 46e

e = Charge of electron = $1.6\times 10^{-19}\ C$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

From Coulomb's law

$F=k\dfrac{q_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times (46\times 1.6\times 10^{-19})^2}{(1.166\times 10^{-14})^2}\\\Rightarrow F=3581.93928018\ N$

The magnitude of the repulsive force pushing these two spheres apart is 3581.93928018 N