An expression for the frequency of revolution of the electron in the…

The correct answer is Option (1) → \( \displaystyle\frac{me^4}{4\varepsilon_0^2 h^3}\times\frac{1}{n^3} \)

The frequency of revolution is the speed divided by the circumference of the orbit, \( f = v/(2\pi r) \). The speed and the radius follow from Bohr's quantisation condition and the Coulomb force providing the centripetal force.

    \( \displaystyle m v r = \frac{n h}{2\pi} \)

    \( \displaystyle \frac{m v^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2} \;\Rightarrow\; m v^2 r = \frac{e^2}{4\pi\varepsilon_0} \)

Dividing the second relation by the first:

    \( \displaystyle v = \frac{e^2}{2\varepsilon_0 n h} \)

Putting this \( v \) back into \( m v r = nh/2\pi \):

    \( \displaystyle r = \frac{\varepsilon_0 n^2 h^2}{\pi m e^2} \)

So the frequency of revolution is

    \( \displaystyle f = \frac{v}{2\pi r} = \frac{e^2}{2\varepsilon_0 n h}\times\frac{\pi m e^2}{2\pi\,\varepsilon_0 n^2 h^2} = \frac{m e^4}{4\varepsilon_0^2 h^3}\times\frac{1}{n^3} \)

Shortcut: Check the units: \( m e^4/(\varepsilon_0^2 h^3) \) has the unit \( \mathrm{s^{-1}} \). Options (3) and (4) contain an extra \( c \), and Option (2) has \( h^2 \) instead of \( h^3 \), so their units are not those of frequency.