Question

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. $\frac{27}{5}$
• D. $\frac{20}{7}$

Answer 1

Answer: (D)

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
$∴ \frac{1}{\left(\lambda \right)_{1}} = R \left[\frac{1}{3^{2}} - \frac{1}{4^{2}}\right] = R ⁡ \left[\frac{1}{9} - \frac{1}{1 6}\right] = \frac{7}{1 4 4} R ⁡ … \left(\text{i}\right)$
In second case, n_f = 2, n_i = 3
$∴ \frac{1}{\left(\lambda \right)_{2}} = R \left[\frac{1}{2^{2}} - \frac{1}{3^{2}}\right] = R ⁡ \left[\frac{1}{4} - \frac{1}{9}\right] = \frac{5}{3 6} R ⁡ … \left(\text{i} \text{i}\right)$
Divide (ii) by (i) we get
$ \frac{\lambda _{1}}{\lambda _{2}} = \frac{5}{3 6} \times \frac{1 4 4}{7} = \frac{2 0}{7}$