The kinetic energy of the electron in an orbit of radius r in hydrogen atom is proportional to ( e = electronic charge)

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TestHub · Accepted Answer

The electrostatic force between the electron and the nucleus provides the necessary centripetal force for the circular orbit.

$\frac{m v ^{2}}{r} = \frac{1}{4 π ϵ _{0}} \frac{e ^{2}}{r ^{2}}$

$m v^{2} = \frac{1}{4 π ϵ _{0}} \frac{e ^{2}}{r}$

Kinetic energy, $K = \frac{1}{2} m v^{2} = \frac{1}{4 π ϵ _{0}} \frac{e ^{2}}{2 r}$

In CGS units, $\frac{1}{4 π ϵ _{0}} = 1$ , so $K = \frac{e ^{2}}{2 r}$ .

Answer: $\frac{e ^{2}}{2 r}$

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