Question

In a hydrogen atom the electron makes a transition from $(n+1{)}^{th}$ level to the $n^{th}$ level. If $n>>1$, the frequency of radiation emitted is proportional to

• A. $\frac{1}{n}$
• B. $\frac{1}{n^3}$
• C. $\frac{1}{n^2}$
• D. $\frac{1}{n^4}$

Answer 1

The correct option is B $\frac{1}{n^3}$

Total energy of electron in $n^{th}$ or bit of hydrogen atom
$E_{n+1}=-\frac{Rhc}{(n+1{)}^2}$
When electron makes a transition from $(n+1{)}^{th}$ level to $n^{th}$ level
Change in energy,
$\Delta E=E_{n+1}-E_n$

$h\nu =Rhc.[\frac{1}{n^2}-\frac{1}{(n+1{)}^2}](\because E=hv)$

$\nu =R.c\left[\frac{(n+1{)}^2-n^2}{n^2(n+1{)}^2}\right]$

$\nu =R.c\left[\frac{1+2n}{n^2(n+1{)}^2}\right]$

For $n>>1$

$\Rightarrow \nu =R.c\left[\frac{2n}{n^2\times n^2}\right]=\frac{2RC}{n^3}\Rightarrow \nu \propto \frac{1}{n^3}$