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 :
Correct Option: , 2
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, $\frac{1}{\mathrm{n}} \simeq 0$
$h v=E_{0}\left[\frac{2}{n^{3}}\right]$
$v \alpha \frac{1}{\mathrm{n}^{3}}$
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