Discussion Forum

Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelength λ₁: λ₂ emitted in the two cases is

Answer:

Explanation:

The wave number ($\bar{v}$) of the radiation = $\frac{1}{\lambda}$

= $R_{\infty}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Now for case (I) n₁= 3, n₂= 2

$\frac{1}{\lambda_{1}}=R_{\infty}\left[\frac{1}{9}-\frac{1}{4}\right]$

 R_∞ =  Rydberg constant

$\frac{1}{\lambda_{1}}=R_{\infty}\left[\frac{4-9}{36}\right]$

$=\left[\frac{-5R_{\infty}}{36}\right]$

$\lambda_{1}=\left[\frac{-36}{5R_{\infty}}\right]$

$\frac{1}{\lambda_{2}}=R_{\infty}\left[\frac{1}{4}-\frac{1}{1}\right]$

= $\left[\frac{-3R_{\infty}}{4}\right]$

$\lambda_{2}=\left[\frac{-4}{3R_{\infty}}\right]$

$\frac{\lambda_{1}}{\lambda_{2}}=\frac{-36}{5R_{\infty}}\times\frac{3R_{\infty}}{-4}$

$\frac{\lambda_{1}}{\lambda_{2}}$ = $\frac{27}{5}$