Question

In a hydrogen atom, the electron and proton and bound at a distance of about 0.53 A. 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? Taking the zero of the potential energy at infinite separation of the electron from proton

Answer 1

Total energy $=-273.2eV+13.6eV=-13.6eV$

The work required $=13.6eV$

The distance between electron-proton of a hydrogen atom, $d=0.53A$

Charge on an electron, $q_1=-1.6\times {10}^{-19}C$

Charge on a proton, $q_2=+1.6\times {10}^{-19}C$

Now,

Potential energy of the system, $p_e$ = Potential energy at infinity separation - Potential energy at distance, d

$\Rightarrow$ $p_e=0-\frac{q_1{q}_2}{4\pi {ϵ}_{0}d}$ (where${ϵ}_0$ is the permittivity of free space and $\frac{1}{4\pi {ϵ}_0}=9\times {10}^{9}N{m}^2{C}^{-2}$)

$\Rightarrow$ $p_e=0-\frac{(1.6\times {10}^{-19}{)}^2\times 9\times {10}^9}{0.53\times {10}^{10}}=-43.7\times {10}^{-19}J=-27.2eV$

Since, kinetic energy in the orbit is half the magnitude of potential energy obtained

$K.E=\frac{|-27.2|}{2}=13.6eV$

Minimum work required = $27.2eV-13.6eV=13.6eV$