What is the magnitude of the repulsive force in a nuclear fission problem?

[mustang]

Problem 19.
At the point of fission, a nucleus of 235_U that has 92 protons is divded into two smaller spheres, each of which has 46 protons and a radius of 5.9*10^-15m. what is the magnitude of the repulsive force pushing these two spheres apart? Use 8.99*10^9 N*m^2/C^2.
Note: How can you find the distance if the problem only the radius?

 

[HallsofIvy]

I would assume the two spheres are meant to be touching each other so that you can assume to point charges separated by 2(5.9*10^-15m)=
11.8*10^-15m.

 

[M98Ranger]

The magnitude of the repulsive force in this nuclear fission problem can be calculated using Coulomb's Law, which states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this problem, we can calculate the charge of each smaller sphere by using the fact that the number of protons in an atom is equal to its atomic number. So each sphere has a charge of 46 protons, or 46 times the elementary charge (1.6*10^-19 C).

Next, we need to find the distance between the two spheres. While the problem only gives us the radius of each sphere, we can use the fact that the distance between two spheres is equal to the sum of their radii. So in this case, the distance between the two spheres would be 2*5.9*10^-15m, or 1.18*10^-14m.

Now, we can plug these values into Coulomb's Law:

F = k * (q1 * q2) / d^2

Where k is the Coulomb's constant (8.99*10^9 N*m^2/C^2), q1 and q2 are the charges of the two spheres, and d is the distance between them.

So the magnitude of the repulsive force would be:

F = (8.99*10^9 N*m^2/C^2) * [(46 * 1.6*10^-19 C) * (46 * 1.6*10^-19 C)] / (1.18*10^-14m)^2

F = 5.83*10^-8 N

Therefore, the magnitude of the repulsive force pushing the two spheres apart in this nuclear fission problem is approximately 5.83*10^-8 N.