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

3581.93928018 N

Step-by-step explanation:

Number of protons = 46

r = Radius =
`2* 5.83* 10^(-15)=1.166* 10^(-14)\ m`

q = Charge = 46e

e = Charge of electron =
`1.6* 10^(-19)\ C`

k = Coulomb constant =
`8.99* 10^(9)\ Nm^2/C^2`

From Coulomb's law

`F=k(q_1q_2)/(r^2)\\\Rightarrow F=(8.99* 10^(9)* (46* 1.6* 10^(-19))^2)/((1.166* 10^(-14))^2)\\\Rightarrow F=3581.93928018\ N`

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