To find the ratio of the wavelengths of the light absorbed by a hydrogen atom during the transitions $ n=2 \rightarrow n=3 $ and $ n=4 \rightarrow n=6 $, we start by calculating the change in energy ($\Delta E$) for each transition.

The energy of an electron in a hydrogen atom at a given level $n$ is given by:

$ E_n = \frac{-R_H}{n^2} $

where $ R_H $ is the Rydberg constant.

Transition $ n=2 \rightarrow n=3 $:

$ \Delta E_{2 \rightarrow 3} = E_3 - E_2 = \frac{-R_H}{3^2} - \left(\frac{-R_H}{2^2}\right) $

$ = R_H\left(\frac{1}{4} - \frac{1}{9}\right) $

$ = R_H \times \frac{5}{36} $

The wavelength $\lambda_{2 \rightarrow 3}$ is then:

$ \lambda_{2 \rightarrow 3} = \frac{h c}{\Delta E_{2 \rightarrow 3}} = \frac{h c \cdot 36}{R_H \cdot 5} $

Transition $ n=4 \rightarrow n=6 $:

$ \Delta E_{4 \rightarrow 6} = E_6 - E_4 = \frac{-R_H}{36} + \frac{R_H}{16} $

$ = \frac{R_H \times 20}{36 \times 16} $

The wavelength $\lambda_{4 \rightarrow 6}$ is then:

$ \lambda_{4 \rightarrow 6} = \frac{h c}{\Delta E_{4 \rightarrow 6}} = \frac{h c \times 36 \times 16}{R_H \cdot 20} $

Calculating the ratio:

The ratio of the wavelengths $\frac{\lambda_{2 \rightarrow 3}}{\lambda_{4 \rightarrow 6}}$ is:

$ \frac{\lambda_{2 \rightarrow 3}}{\lambda_{4 \rightarrow 6}} = \frac{\frac{h c \cdot 36}{R_H \cdot 5}}{\frac{h c \times 36 \times 16}{R_H \cdot 20}} $

$ = \frac{1}{4} $

Therefore, the ratio of the wavelengths for the given transitions is $\frac{1}{4}$.