In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:
a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from 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 $\stackrel{\circ }{A}$ separation?

Given,
                 $$\begin{gathered} \mathrm{Charge} \mathrm{on} \mathrm{an} \mathrm{electron}, {\mathrm{q}}_1 = -1.6 \times {10}^{-19}\mathrm{C} \\ \mathrm{Charge} \mathrm{on} \mathrm{a} \mathrm{proton}, {\mathrm{q}}_2 = 1.6 \times {10}^{-19} \mathrm{C} \\ \mathrm{Distance} \mathrm{between} \mathrm{electron} \mathrm{and} \mathrm{proton}, \mathrm{r} = 0.53 \mathrm{\AA } = 0.53 \times {10}^{-10}\mathrm{m} \end{gathered}$$
 
a) Potential energy  = $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1 {\mathrm{q}}_2}{\mathrm{r}}$
                                 $$\begin{gathered} = \frac{\left(9 \times {10}^9\right) \left(-1.6 \times {10}^{-19}\right) \left(1.6 \times {10}^{-19}\right)}{\left(0.53 \times {10}^{-10}\right)}\mathrm{J} \\ = -43.47 \times {10}^{-19} \mathrm{J} \end{gathered}$$

$= -\frac{43.47 \times {10}^{-19}}{1.6 \times {10}^{-19}}\mathrm{eV} = -27.17 \mathrm{eV}$

Potential energy is zero at infinite separation.
Hence, the potential energy of the system is (–27.17–0) or 27.17 eV if zero of potential energy is taken at infinite separation.

b) Kinetic Energy of electron is given by

$\frac{1}{2}\mathrm{Potential} \mathrm{Energy} =\frac{27.2}{2}=13.6 \mathrm{eV}$

$\therefore$ Total energy of electron = -27.2 + 13.6
                                      = - 13.6 eV

Amount of work required to free the electron = Increase in energy of electron
= 0 – (–13.6)
= 13.6 eV

c) If the zero of Potential Energy is taken at 1.06 Å (1.06 x 10^–10 m) separation, then the potential energy of system 
      $$\begin{gathered} \mathrm{P}.\mathrm{E} = 9 \times {10}^9\left[\frac{\left(1.6 \times {10}^{-19}\right) \times \left(1.6 \times {10}^{-19}\right)}{1.06 \times {10}^{-10}}\right] \\ = 21.74 \times {10}^{-19}\mathrm{J} \\ = - \frac{21.74 \times {10}^{-19}}{1.6 \times {10}^{-19}} \\ = -13.585 \mathrm{eV} \end{gathered}$$

The amount of work done to free the electron in this case is,

$$\begin{gathered} = -27.17 - (-13.585) \\ = - 13.585 \mathrm{eV} \end{gathered}$$