91Ó°ÊÓ

The given data can be listed below as,

• The expression of electric potential energy of a system of two particles is,$U_{el}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{r}$
• The distance between two protons is, $r=9×{10}^{-9}\text{m}$.
• The distance between a proton and an electron is,$r\text{'}=r=9×{10}^{-9}\text{m}$ .

Whenever two charged particles lie at a specific distance in a particular electric field, there would be energy required to move the charged particles. The required energy refers to electric potential energy.

$9×{10}^{-9}m$The relation ofelectric potential energy of two protonsis expressed as,

$$\begin{gathered} U_1=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_1}{r} \\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{{\left(q_1\right)}^2}{r} \end{gathered}$$

Here, U₁ is the electric potential energy of two protons separated by a distance of , represents the charge on a proton whose value is $1.6×{10}^{-19}\text{C}$ and is the vacuum permittivity whose value is $8.85×{10}^{-12}\text{F/m}$.

Substitute all the known values in the above equation.

$$\begin{gathered} U_1=\left[\frac{1}{4\mathrm{π}\left(8.85×{10}^{-12}\text{F/m}\right)}\frac{{\left(1.6×{10}^{-19}\text{C}\right)}^2}{\left(9×{10}^{-9}\text{m}\right)}\right] \\ ≈\left(2.56×{10}^{-20}{\text{C}}^{\text{2}}\text{/F}\right)\left(\frac{1\text{J}}{1{\text{C}}^{\text{2}}\text{/F}}\right) \\ ≈2.56×{10}^{-20}\text{J} \end{gathered}$$

Thus, the electric potential energy of the two protons is $2.56×{10}^{-20}\text{J}$.

The relation ofelectric potential energy of a proton and an electron separated by the same distanceis expressed as,

$U_2=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{r\text{'}}$

Here, is the electric potential energy of a proton and an electron separated by the same distance, represents the charge on a proton whose value is $1.6×{10}^{-19}\text{C}$, represents the charge on an electron whose value is $-1.6×{10}^{-19}\text{C}$ and is the separation distance between a proton and an electron.

Substitute all the known values in the above equation.

$$\begin{gathered} U_2=\left[\frac{1}{4\mathrm{π}\left(8.85×{10}^{-12}\text{F/m}\right)}\frac{\left(1.6×{10}^{-19}\text{C}\right)\left(-1.6×{10}^{-19}\text{C}\right)}{\left(9×{10}^{-9}\text{m}\right)}\right] \\ ≈\left(-2.56×{10}^{-20}{\text{C}}^{\text{2}}\text{/F}\right)\left(\frac{1\text{J}}{1{\text{C}}^{\text{2}}\text{/F}}\right) \\ ≈-2.56×{10}^{-20}\text{J} \end{gathered}$$

Thus, the electric potential energy of a proton and an electron is$-2.56×{10}^{-20}\text{J}$ .