Question

In a hydrogen atom, the electron and proton are bound at a distance of about $0.53\stackrel{˚}{}A$:

(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?

Answer 1

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

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

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

(a) Potential at infinity is zero.

The 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,

${\in }_0$ is the permittivity of free space.

Substituting the values in the above equation we get,

P.E.= -43.7 $\times$ 10${}^{-19}$ J

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

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

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

Total energy $=13.6-27.2=13.6eV$

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

(c) When zero of potential energy is taken, $d_1=1.06\mathring{A}$.

$\therefore$ Potential energy of the system $=$ Potential energy at $d_1$ − Potential energy at d

$=\frac{q_1{q}_2}{4\pi {\in }_0{d}_1}-27.2eV$

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

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

$=13.58eV-27.2eV$

$=-13.6eV$