Question: 46. The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes \mathrm{n}=2 \rightarrow \mathrm{n}=3 and \mathrm{n}=4 \rightarrow \mathrm{n}=6 transitions, respectively, is
(1) \frac{1}{36}
(2) \frac{1}{16}
(3) \frac{1}{9}
(4) \frac{1}{4}
Answer: Option (4)
Explanation:
For absorption in a hydrogen atom, the electron moves from a lower energy level to a higher energy level.
The energy difference between two energy levels of a hydrogen atom is given by
\Delta E = 13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\text{ eV}The wavelength of absorbed light is related to energy by
\Delta E = \frac{hc}{\lambda}Hence, wavelength is inversely proportional to energy difference
\lambda \propto \frac{1}{\Delta E}For transition 2 \rightarrow 3, energy absorbed is
\Delta E_1 = 13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \Delta E_1 = 13.6\left(\frac{1}{4}-\frac{1}{9}\right) = 13.6\times\frac{5}{36}For transition 4 \rightarrow 6, energy absorbed is
\Delta E_2 = 13.6\left(\frac{1}{4^2}-\frac{1}{6^2}\right) \Delta E_2 = 13.6\left(\frac{1}{16}-\frac{1}{36}\right) = 13.6\times\frac{5}{144}The ratio of wavelengths is
\frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} \frac{\lambda_1}{\lambda_2} = \frac{\frac{5}{144}}{\frac{5}{36}} = \frac{36}{144} = \frac{1}{4}Therefore, the correct answer is Option (4).