91Ó°ÊÓ

The given data can be listed below,

• The radius of the uniformly charged sphere is,R
• The potential energy of the sphere is,$U=\frac{3Q^2}{20\mathrm{Ï€}{∈}_0\mathrm{R}}$

A force field's presence in an area causes an energy to arise. A positive charge, Q, will cause all other charges to have electric potential energy if it is present because of its electric field.

The total potential energy of the lead nucleus is given by,

$U=\frac{3{\left(zQ\right)}^2}{20\mathrm{π}{∈}_0\mathrm{R}}$

Here, zis the number of protons, Qis the charge on the nucleus, and Ris the radius of the sphere,

Substitute all the values in the above,

role="math" localid="1659181281067" $$\begin{gathered} \mathrm{U}=\frac{3}{5}\frac{{\left(82×1.6×{10}^{-19}\mathrm{C}\right)}^2}{4×3.14\left(8.85×{10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right)7×{10}^{-15}\mathrm{m}} \\ =\frac{41.68×{10}^{-11}\mathrm{J}}{3.14} \\ =1.33×{10}^{-11}\mathrm{J} \end{gathered}$$

So, the energy per proton is given by,

$U_p=\frac{U}{z}$

Substitute all the values in the above,

$$\begin{gathered} {\mathrm{U}}_{\mathrm{p}}=\frac{1.33×{10}^{-10}\mathrm{J}}{82\mathrm{proton}}\left(\frac{6.242×{10}^{18}\mathrm{eV}}{1\mathrm{J}}\right) \\ =10.1\mathrm{MeV}/\mathrm{proton} \end{gathered}$$

Thus, the energy per charge in a lead nucleus is$10.1\mathrm{MeV}/\mathrm{proton}$ .

In exercise 87 neutrons do not repel one another, whereas protons do. The protons' energy levels should increase in comparison to the neutrons because of the repelling energy.so, the proton energies should rise due to repulsion.

Thus, the value obtained at least falls within the proper order of magnitude to explain the discrepancy in Exercise 87.