In a hydrogen atom the electron makes a transition from $(n+1)^{\text {th }}$ level to the $n^{\text {th }}$ level. If $n \gg 1$, the frequency of radiation emitted is proportional to :
Correct Option: , 2
(2)
Total energy of electron in $n^{\text {th }}$ orbit of hydrogen atom
$E_{n}=-\frac{R h c}{n^{2}}$
Total energy of electron in $(n+1)^{\text {th }}$ level of hydrogen atom
$E_{n+1}=-\frac{R h c}{(n+1)^{2}}$
When electron makes a transition from $(n+1)^{\text {th }}$ level to $n^{\text {th }}$
level
Change in energy,
$\Delta E=E_{n+1}-E_{n}$
$h v=R h c \cdot\left[\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right]$ $(\because E=h v)$
$\mathrm{v}=R \cdot c\left[\frac{(n+1)^{2}-n^{2}}{n^{2}(n+1)^{2}}\right]$
$\mathrm{v}=R \cdot c\left[\frac{1+2 n}{n^{2}(n+1)^{2}}\right]$
For $n>>1$
$\Rightarrow v=R \cdot c\left[\frac{2 n}{n^{2} \times n^{2}}\right]=\frac{2 R C}{n^{3}} \Rightarrow v \propto \frac{1}{n^{3}}$