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 $\lambda_{1}: \lambda_{2}$ emitted in the two cases is

• A

  $7 / 5$
• B

  $27 / 20$
• Correct answer
  C

  $27 / 5$
• D

  $20 / 7$

Solution

The wave number $(\bar{v})$ of the radiation $$ \begin{array}{l} =\frac{1}{\lambda} \\ =R_{\infty}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \end{array} $$ Now for case (I) $n_{1}=3, n_{2}=2$ $$ \begin{array}{l} \frac{1}{\lambda_{1}}=R_{\infty}\left[\frac{1}{9} \frac{-1}{4}\right], R_{\infty}=\text { Rydberg constant } \\ \frac{1}{\lambda_{1}}=R_{\infty}\left[\frac{4-9}{36}\right]=\frac{-5 R_{\infty}}{36} \\ \Rightarrow \quad \lambda_{1}=\frac{-36}{5 R_{\infty}} \\ \frac{1}{\lambda_{2}}=R_{\infty}\left[\frac{1}{4}-\frac{1}{1}\right]=\frac{-3 R_{\infty}}{4} \Rightarrow \lambda_{2}=\frac{-4}{3 R_{\infty}} \\ \Rightarrow \frac{\lambda_{1}}{\lambda_{2}}=\frac{-36}{5 R_{\infty}} \times \frac{3 R_{\infty}}{-4} \Rightarrow \frac{\lambda_{1}}{\lambda_{2}}=\frac{27}{5} \end{array} $$

Hence, the correct answer is option: 'C'.