In nuclear fission, a nucleus splits roughly in half.

(a) What is the potential 4.00 10-14 m from a fragment that has 58 protons in it?

(b) What is the potential energy in MeV of a similarly charged fragment at this distance?

To calculate the potential at a certain distance from a charged object, you can use Coulomb's Law. Coulomb's Law states that the electric potential V between two point charges is given by the equation:

V = k * (Q1 * Q2) / r

where V is the potential, k is the electrostatic constant (9.0 x 10^9 N.m^2/C^2), Q1 and Q2 are the charges of the objects (in this case, the charges of the nucleus and the fragment), and r is the distance between them.

Let's solve part (a) first. The nucleus has 58 protons, which means it has a charge of +58e, where e is the elementary charge (1.6 x 10^-19 C). The fragment is similarly charged, so its charge is -58e. We can plug in these values and the given distance (4.00 x 10^-14 m) into the equation:

V = (9.0 x 10^9 N.m^2/C^2) * ((+58e) * (-58e)) / (4.00 x 10^-14 m)

Now you can calculate the value of V.

To answer part (b), we need to convert the potential energy from Joules to MeV (mega electron volts) because MeV is a commonly used unit in nuclear physics. We can use the conversion factor that 1 MeV is equal to 1.6 x 10^-13 J.

Once you have the value of V in Joules from part (a), you can convert it to MeV using the conversion factor:

Potential Energy (in MeV) = Potential Energy (in J) / (1.6 x 10^-13 J/MeV)

Now you can calculate the potential energy in MeV using the value of V found in part (a).

Remember to perform the appropriate unit conversions and use the given values to obtain the final answers.

jbm