Consider a ${}^{\mathbf{238}}\mathbf{\cup }$nucleus to be made up of an alpha particle (${}^{\mathbf{4}}\mathbf{He}$) and a residual nucleus (${}^{\mathbf{234}}\mathbf{Th}$). Plot the electrostatic potential energy U(r), where r is the distance between these particles. Cover the approximate range10 fm < r < 100 fm and compare your plot with that of Fig. 42-10.

A nuclear reaction is given:${}^{238}\mathrm{U}\to {}^4\mathrm{He}+{}^{234}\mathrm{Th}$

Electric potential energy is the energy that is needed to move a charge against an electric field. The energy required to move the electron from any surface is less hard in comparison to the energy required to move it from a strong electric field. Thus, stronger field gives larger energy.

Formula:

The electrostatic potential energy of a system of charges,$\cup \left(r\right)=\frac{k{q}_1{q}_2}{r}\dots ........\left(1\right)$

Equation (1) gives the electrostatic potential energy between two uniformly charged spherical charges (in this case $q_1=2e$and$q_1=90e$ ) with rbeing the distance between their centers. Assuming the “uniformly charged spheres” condition is met in this instance, we write the equation in such a way that we can make use of$K=1/4{\mathrm{\pi \epsilon }}_0$and the electron-volt unit as follows:

$$\begin{gathered} U\left(r\right)=\frac{\left(9\times {10}^{9}N.m^2/C^2\right)\left(2\times 1.6\times {10}^{-19}C\right)\left(90e\right)}{r} \\ =\frac{2.59\times {10}^{-7}}{r}eV \end{gathered}$$

with runderstood to be in meters. It is convenient to write this for rin femtometers, in this case U = (259/r)MeV. This is plotted below.

Hence, the graph for electrostatic potential energy versus the distance between the particles is plotted.