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

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 Å separation?

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 $d$

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

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

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

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

$=-13.6 \mathrm{eV}$

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