In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:

Solution:

The distance between electron-proton of a hydrogen atom, $d=0.53 \mathrm{~A}$

Charge on an electron, $q_{1}=-1.6 \times 10^{-19} \mathrm{C}$

Charge on a proton, $q_{2}=+1.6 \times 10^{-19} \mathrm{C}$

(a) Potential at infinity is zero.

Potential energy of the system, = Potential energy at infinity − Potential energy at distance d

$=0-\frac{q_{1} q_{2}}{4 \pi \in_{0} d}$

where,

$\epsilon_{0}$ is the permittivity of free space

$\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$

$\therefore$ Potential energy $=0-\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{0.53 \times 10^{-10}}=-43.47 \times 10^{-19} \mathrm{~J}$

$\because 1.6 \times 10^{-19} \mathrm{~J}=1 \mathrm{eV}$

$\therefore$ Potential energy $=-43.7 \times 10^{-19}=\frac{-43.7 \times 10^{-19}}{1.6 \times 10^{-19}}=-27.2 \mathrm{eV}$

Therefore, the potential energy of the system is −27.2 eV.

(b) Kinetic energy is half of the magnitude of potential energy.

Kinetic energy $=\frac{1}{2} \times(-27.2)=13.6 \mathrm{eV}$

Total energy = 13.6 − 27.2 = 13.6 eV

Therefore, the minimum work required to free the electron is 13.6 eV.

(c) When zero of potential energy is taken, $d_{1}=1.06 \mathrm{~A}$

$\therefore$ Potential energy of the system $=$ Potential energy at $d_{1}-$ Potential energy at $d$

$=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0} d_{1}}-27.2 \mathrm{eV}$

$=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{1.06 \times 10^{-10}}-27.2 \mathrm{eV}$

$=21.73 \times 10^{-19} \mathrm{~J}-27.2 \mathrm{eV}$

$=13.58 \mathrm{eV}-27.2 \mathrm{eV}$

$=-13.6 \mathrm{eV}$