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 wavelengths ${\lambda }_1: {\lambda }_2$ emitted in the two cases is

• A

  $\frac{7}{5}$
• B

  $\frac{27}{20}$
• C

  $\frac{27}{5}$
• Correct answer
  D

  $\frac{20}{7}$

Solution

According to Rydberg formula

$\frac{1}{\lambda }=R\left[\frac{1}{n_f^2}-\frac{1}{n_i^2}\right]$

In first case, n_f = 3, n_i = 4

$\therefore \frac{1}{{\lambda }_1}=R\left[\frac{1}{3^2}-\frac{1}{4^2}\right]=R\left[\frac{1}{9}-\frac{1}{16}\right]=\frac{7}{144}R\dots \left(\text{i}\right)$

In second case, n_f = 2, n_i = 3

$\therefore \frac{1}{{\lambda }_2}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=R\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5}{36}R\dots \left(\text{i}\text{i}\right)$

Divide (ii) by (i) we get

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{5}{36}\times \frac{144}{7}=\frac{20}{7}$

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