# 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 $n>>1$, the frequency of radiation emitted is proportional to :

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

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

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

4. $\frac{1}{n}$

Correct Option: , 2

Solution:

In hydrogen atom,

$E_{n}=\frac{-E_{0}}{n^{2}}$

Where $\mathrm{E}_{0}$ is Ionisation Energy of $\mathrm{H}$.

$\rightarrow$ For transition from $(\mathrm{n}+1)$ to $\mathrm{n}$, the energy of emitted radiation is equal to the difference in energies of levels.

$\Delta \mathrm{E}=\mathrm{E}_{\mathrm{n}+1}-\mathrm{E}_{\mathrm{n}}$

$\Delta \mathrm{E}=\mathrm{E}_{0}\left(\frac{1}{\mathrm{n}^{2}}-\frac{1}{(\mathrm{n}+1)^{2}}\right)$

$\Delta \mathrm{E}=\mathrm{h} v=\mathrm{E}_{0}\left(\frac{(\mathrm{n}+1)^{2}-\mathrm{n}^{2}}{\mathrm{n}^{2}(\mathrm{n}+1)^{2}}\right)$

$\mathrm{h} v=\mathrm{E}_{0}\left[\frac{2 \mathrm{n}+1}{\mathrm{n}^{4}\left(1+\frac{1}{\mathrm{n}}\right)^{2}}\right]$

$\mathrm{h} v=\mathrm{E}_{0}\left[\frac{\mathrm{n}\left(2+\frac{1}{\mathrm{n}}\right)}{\mathrm{n}^{4}\left(1+\frac{1}{\mathrm{n}}\right)^{2}}\right]$

Since $n \gg \gg 1$

Hence, $\frac{1}{\mathrm{n}} \simeq 0$

$h v=E_{0}\left[\frac{2}{n^{3}}\right]$

$v \alpha \frac{1}{\mathrm{n}^{3}}$