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The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes $n = 2 \rightarrow n = 3$ and $n = 4 \rightarrow n = 6$ transitions, respectively, is:
A
$\frac{1}{36}$
B
$\frac{1}{16}$
C
$\frac{1}{9}$
D
$\frac{1}{4}$
Explanation
Energy absorbed is $\Delta E_1 = R_H \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5}{36} R_H$ and $\Delta E_2 = R_H \left(\frac{1}{16} - \frac{1}{36}\right) = \frac{5}{144} R_H$. Since $\lambda \propto 1/\Delta E$, $\lambda_1 / \lambda_2 = \Delta E_2 / \Delta E_1 = \frac{5/144}{5/36} = \frac{36}{144} = \frac{1}{4}$.
Detailed Solution
For hydrogen-like transitions, the energy difference between energy levels $n_1$ and $n_2$ is $\Delta E = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. For $n = 2 \rightarrow 3$: $\Delta E_1 = R_H \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5}{36} R_H$. For $n = 4 \rightarrow 6$: $\Delta E_2 = R_H \left(\frac{1}{16} - \frac{1}{36}\right) = \frac{5}{144} R_H$. Since wavelength $\lambda = \frac{hc}{\Delta E}$, the ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} = \frac{5/144}{5/36} = \frac{36}{144} = \frac{1}{4}$.