While undergoing fission, a uranium-235 nucleus with its 92 protons splits into two smaller nuclei each of which contains 46 protons and has a radius of 5.90 x 10^-15m. What is the magnitude of the repulsive force between the two nuclei just after the split?

Use Coulomb's law.

The charge of each nucleus is Q = -46e and the separation is twice the radius.

F = k Q^2/d^2

thankyou :)

To find the magnitude of the repulsive force between the two nuclei just after the split, we need to use the Coulomb's Law equation:

F = k * ((Q1 * Q2) / r^2)

Where:
- F is the magnitude of the repulsive force
- k is the electrostatic constant (9 x 10^9 N*m^2/C^2)
- Q1 and Q2 are the charges of the two nuclei
- r is the distance between the centers of the two nuclei

In this case, both nuclei have the same number of protons, so they have the same charge. The charge of one nucleus can be calculated using the elementary charge, e, which is the charge of one proton: e = 1.6 x 10^-19 C.

Q1 = Q2 = (number of protons) * e = 46 * 1.6 x 10^-19 C

Next, we need to convert the radius given from meters to centimeters since the electrostatic constant is given in SI units.

r = (5.90 x 10^-15 m) * (100 cm/m)

Now we can substitute the values into the equation and calculate the magnitude of the force:

F = (9 x 10^9 N*m^2/C^2) * (((46 * 1.6 x 10^-19 C) * (46 * 1.6 x 10^-19 C)) / ((5.90 x 10^-15 m) * (100 cm/m))^2)

By calculating this equation, you will find the magnitude of the repulsive force between the two nuclei just after the split.