In a hydrogen atom,the electron and proton are bound at a distance of about $0.53 \; \mathring{A}$.
$(a)$ Estimate the potential energy of the system in $eV$,taking the zero of the potential energy at infinite separation of the electron from the proton.
$(b)$ What is the minimum work required to free the electron,given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in $(a)$?
$(c)$ What are the answers to $(a)$ and $(b)$ above if the zero of potential energy is taken at $1.06 \; \mathring{A}$ separation?

(N/A) The potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by distance $r$ is given by $U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$.
For a hydrogen atom,$q_1 = e = 1.6 \times 10^{-19} \; C$ and $q_2 = -e = -1.6 \times 10^{-19} \; C$,with $r = 0.53 \times 10^{-10} \; m$.
$U = \frac{9 \times 10^9 \times (1.6 \times 10^{-19}) \times (-1.6 \times 10^{-19})}{0.53 \times 10^{-10}} \; J = -43.58 \times 10^{-19} \; J$.
Converting to $eV$: $U = \frac{-43.58 \times 10^{-19}}{1.6 \times 10^{-19}} \; eV \approx -27.2 \; eV$.
$(b)$ Kinetic energy $K = -\frac{1}{2} U = -\frac{1}{2} (-27.2) = 13.6 \; eV$.
Total energy $E = K + U = 13.6 - 27.2 = -13.6 \; eV$.
Work required to free the electron is the energy needed to bring the total energy to $0$,which is $13.6 \; eV$.
$(c)$ If zero potential energy is at $r_0 = 1.06 \; \mathring{A}$,the new potential energy $U' = U(r) - U(r_0)$.
$U(r_0) = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{1.06 \times 10^{-10}} \; J = 21.73 \times 10^{-19} \; J = 13.58 \; eV$.
New $U' = -27.2 - 13.58 = -40.78 \; eV$.
New $K$ remains $13.6 \; eV$. New total energy $E' = K + U' = 13.6 - 40.78 = -27.18 \; eV$.
Work required to free the electron is $27.18 \; eV$.