The expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1), for a large value of n become

$\frac{m e^4}{4 \varepsilon_0^2 h^3} \times \frac{1}{n^3}$

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

The total energy of electron in n^th orbit

$E_n=\frac{-m e^4 z^2}{8 \varepsilon_0^2 h^2 n^2}$

$\Delta E=\frac{m e^4 z^2}{8 \varepsilon_0^2 h^2}\left[\frac{1}{n^2}-\frac{1}{(n-1)^2}\right]$

$\Delta E=\frac{m e^4 z^2}{8 \varepsilon_0^2 h^2}\left[\frac{(n-1)^2-n^2}{n^2(n-1)^2}\right]$

$\Delta E=\frac{m e^4 z^2}{8 \varepsilon_0^2 h^2}\left[\frac{2 n-1}{n^2(n-1)^2}\right]$

$\Delta E=\frac{m e^4 z^2}{8 \varepsilon_0^2 h^2}\left[\frac{2 n\left(1-\frac{1}{2 n}\right)}{h^4\left(1-\frac{1}{h}\right)^2}\right]$

$\Delta E=\frac{m e^4 z^2}{4 \varepsilon_0 h^2 n^3}$

$v=\frac{m e^4 z^2}{4 \varepsilon_0 h^2 n^3}$