Question

What is the ratio of the wavelength of the Lyman series limit to the Paschen series limit?

• A. \( 1 : 4 \)
• B. \( 1 : 3 \)
• C. \( 2 : 3 \)
• D. \( 1 : 2 \)

Hint:

To find the wavelength ratio of two series limits in the hydrogen spectrum, use the Rydberg formula and compare the limits where \( n_2 \to \infty \) for both series.

Answer 1

The Correct Option is A

The Lyman and Paschen series are part of the hydrogen atom's emission spectra. The Lyman series corresponds to transitions where the final state is \( n = 1 \), while the Paschen series corresponds to transitions where the final state is \( n = 3 \). The wavelength of light emitted during these transitions can be found using the Rydberg formula for hydrogen: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where: - \( \lambda \) is the wavelength of the emitted radiation, - \( R_H \) is the Rydberg constant, - \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states, respectively. For the Lyman series limit, the transition is from \( n_2 \to 1 \), and the wavelength corresponds to the transition where \( n_2 \to \infty \). Thus, for Lyman: \[ \frac{1}{\lambda_{\text{Lyman}}} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H \] For the Paschen series limit, the transition is from \( n_2 \to 3 \), and the wavelength corresponds to the transition where \( n_2 \to \infty \). Thus, for Paschen: \[ \frac{1}{\lambda_{\text{Paschen}}} = R_H \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{9} \right) \] Now, the ratio of the wavelengths is the inverse of the ratio of the terms: \[ \frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Paschen}}} = \frac{9}{1} = 9 \] Thus, the ratio of the wavelength of Lyman to Paschen is \( 9:1 \), meaning the wavelength of the Lyman series limit is 9 times greater than that of the Paschen series limit.