Question

In a hydrogen like atom electron make transition from an energy level with quantum number $n$ to another with quantum number $(n-1)$. If $n>>1$, the frequency of radiation emitted is proportional to :

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

Answer 1

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

According to Bohr's theory of hydrogen atom, frequency of radiation emitted,
$\Delta E=hv$

$v=\frac{\Delta E}{h}=k[\frac{1}{(n-1{)}^2}-\frac{1}{n^2}]$
where, $K$ is a constant.

$=\frac{k(n^2-(n-1{)}^2)}{n^2(n-1{)}^2}$
$=\frac{k(2n-1)}{n^2(n-1{)}^2}$

As, $n>>1,v=\frac{k\left(2n\right)}{n^4}$
$v\propto \frac{1}{n^3}$

Hence, option $D$ is correct.