Question

In a semiclassical model of the hydrogen atom, the electron

orbits the proton at a distance of 0.053 nm.

a. What is the electric potential of the proton at the position of the electron?

b. What is the electron’s potential energy?

Short answer

The potential energy of the electric of the photon is $27V$

The potential energy of the electric electron is$-4.3\times {10}^{-18}J$

Step-by-step solution

Definition of the potential energy of the electric of the photon 

The formula of the equation of electric potential energy of the photon

$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r} \\ \frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N.m^2 \\ q=1.6\times {10}^{19}C \\ r=0.053\times {10}^{-9}m \\ V=9\times {10}^{9}N.m^2\frac{1.6\times {10}^{19}C}{0.053\times {10}^{-9}m}=27V \end{gathered}$$

Definition of the electric potential  energy of the electron  

The formula of the equation of electric potential energy of the photon

$U=QV$

where,

$$\begin{gathered} Q=-1.6\times {10}^{-19}C \\ V=27V \\ U=\left(-1.6\times {10}^{-19}C\right)\times \left(27V\right) \end{gathered}$$

Thus, the Electric Potential of the proton at the position is$-4.3\times {10}^{-18}J$