Question

Sol. [Key Concept: Ratio of wavelengths absorbed in H-atom transitions (Rydberg: 1/λ ∝ ΔE)] $\Delta E = \frac{hc}{\lambda} = E_{\text{final}\text{~}} - E_{\text{initial}\text{~}}\ \left( E_{n} = \frac{- R_{H}}{n^{2}} \right)$

$$\begin{matrix} & \begin{matrix} \Delta E_{2 \rightarrow 3} = \frac{hc}{\lambda_{2 \rightarrow 3}} = E_{3} - E_{2} = \frac{- R_{H}}{3^{2}} - \left( \frac{- R_{H}}{2^{2}} \right) \\ = R_{H}\left( \frac{1}{4} - \frac{1}{9} \right) = R_{H} \times \frac{5}{36} \end{matrix} \\ & \ \\ & \begin{matrix} & \ \therefore\lambda_{2 \rightarrow 3} = \frac{hc \cdot 36}{R_{H} \cdot 5} \\ & \Delta E_{4 \rightarrow 6} = E_{6} - E_{4} = \frac{- R_{H}}{36} + \frac{R_{H}}{16} = \frac{R_{H} \times 20}{36 \times 16} \\ & \frac{hc}{\lambda_{4 \rightarrow 6}} = \frac{R_{H} \times 20}{36 \times 16} \\ & \lambda_{4 \rightarrow 6} = \frac{hc \times 36 \times 16}{R_{H} \cdot 20} \\ & \frac{\lambda_{2 \rightarrow 3}}{\lambda_{4 \rightarrow 6}} = \frac{hc \cdot 36}{R_{H} \cdot 5} \\ & \frac{hc \times 36 \times 16}{R_{H} \cdot 20} \\ & \ = \frac{1}{4} \end{matrix} \end{matrix}$$