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

Correct option is A

​The Paschen series in the hydrogen spectrum corresponds to the transitions of electrons from higher energy levels (n ≥ 4) to n = 3. The longest wavelength corresponds to the transition from n=4 to n=3, and the shortest wavelength corresponds to the transition from n=∞ to n=3.

The formula for wavelength is: $$\frac{1}{\lambda_{\text{longest}}} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$$

$$\text{Here, } p = 3 \text{ and } n = 4 \text{ for the longest wavelength in the Paschen series. Using the formula:} \\\frac{1}{\lambda_{\text{longest}}} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$$

$$\frac{1}{\lambda_{\text{longest}}} = R_H \left( \frac{1}{9} - \frac{1}{16} \right) = R_H \left( \frac{7}{144} \right)$$

$$\lambda_{\text{longest}} = \frac{144}{7 R_H}$$

For the shortest wavelength, the transition is from n=∞ to n=3. Using the formula:

$$\frac{1}{\lambda_{\text{shortest}}} = R_H \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{9} \right)$$

$$\lambda_{\text{shortest}} = \frac{9}{R_H}$$

$$\frac{\lambda_{\text{longest}}}{\lambda_{\text{shortest}}} = \frac{\frac{144}{7 R_H}}{\frac{9}{R_H}} = \frac{144}{7} \times \frac{R_H}{9} = \frac{16}{7}$$​​​​​​​​​