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
✓
Answer
According to Rydberg formula
$\frac{1}{\lambda}=R\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right]$
In first case, $n_{f}=3, n_{i}=4$
$\therefore \quad \frac{1}{\lambda_{1}}=R\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]=R\left[\frac{1}{9}-\frac{1}{16}\right]=\frac{7}{144} R$ .... $(i)$
In second case, $n_{f}=2, n_{i}=3$
$\therefore \quad \frac{1}{\lambda_{2}}=R\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=R\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5}{36} R$ .... $(ii)$
Divide $(ii)$ by $(i)$, we get
$\frac{\lambda_{1}}{\lambda_{2}}=\frac{5}{36} \times \frac{144}{7}=\frac{20}{7}$
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