The ratio of the longest wavelength to the shortest wavelength ($$\frac{λ_L}{λ_S}$$​) in Brackett series of hydrogen spectrum is:

Correct option is A

​$$\textbf{Given:} \\\bullet \text{ The \textbf{Brackett series} of the hydrogen spectrum corresponds to transitions where the electron moves to } n = 4. \\\bullet \textbf{Longest wavelength} \text{ corresponds to the transition from } n = 5 \text{ to } n = 4. \\\bullet \textbf{Shortest wavelength} \text{ corresponds to the transition from } n = \infty \text{ to } n = 4.$$

$$\textbf{For the {longest wavelength}}, \text{ the transition is from } n = 5 \text{ to } n = 4. \\[5pt]\text{We use the Rydberg formula for the wavelength:} \\[5pt]\frac{1}{\lambda_{\text{L}}} = R_H \left( \frac{1}{4^2} - \frac{1}{5^2} \right) \\[5pt]\lambda_{\text{L}} = \frac{25 \times 16}{9 R_H} \quad \text{(Equation 1)}$$

​​​$$\textbf{For the {shortest wavelength}}, \text{ the transition is from } n = \infty \text{ to } n = 4. \\[5pt]\frac{1}{\lambda_{\text{S}}} = R_H \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = \frac{1}{16 R_H} \\[5pt]\lambda_{\text{S}} = \frac{16}{R_H} \quad \text{(Equation 2)}$$

$$\frac{\lambda_{\text{L}}}{\lambda_{\text{S}}} = \frac{\left( \frac{25 \times 16}{9 R_H} \right)}{\left( \frac{16}{R_H} \right)} \\[5pt]\frac{\lambda_{\text{L}}}{\lambda_{\text{S}}} = \frac{25}{9}$$​​​