Question

What is the ratio of the wavelength of the Lyman series limit to that of the Paschen series limit in the hydrogen spectrum?

• A. \( \frac{9}{9} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{9}{3} \)
• D. \( \frac{2}{2} \)

Hint:

When finding the ratio of wavelengths in spectral lines, remember that the energy level transitions determine the wavelengths through the Rydberg formula.

Answer 1

The Correct Option is A

In the hydrogen spectrum, the Lyman series limit corresponds to the transition from \( n = 2 \) to \( n = \infty \), and the Paschen series limit corresponds to the transition from \( n = 4 \) to \( n = \infty \). The wavelength of the spectral lines can be derived from the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Where \( R_H \) is the Rydberg constant, and \( n_1 \) and \( n_2 \) are the initial and final energy levels, respectively. For the Lyman series limit (transition from \( n = 2 \) to \( n = \infty \)): \[ \frac{1}{\lambda_L} = R_H \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{4} \right) \] For the Paschen series limit (transition from \( n = 4 \) to \( n = \infty \)): \[ \frac{1}{\lambda_P} = R_H \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{16} \right) \] Now, the ratio of the wavelengths is: \[ \frac{\lambda_L}{\lambda_P} = \frac{16}{4} = 4 \] Thus, the ratio of the wavelengths of the Lyman series limit to the Paschen series limit is \( \frac{9}{9} \), which simplifies to \( 1 \).