In a hydrogen atom the electron makes a transition from

Question:

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 :

  1. (1) $\frac{1}{n}$

  2. (2) $\frac{1}{n^{3}}$

  3. (3) $\frac{1}{n^{2}}$

  4. (4) $\frac{1}{n^{4}}$


Correct Option: , 2

Solution:

(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 $n>>1$

$\Rightarrow v=R \cdot c\left[\frac{2 n}{n^{2} \times n^{2}}\right]=\frac{2 R C}{n^{3}} \Rightarrow v \propto \frac{1}{n^{3}}$

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