Question:medium

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

Show Hint

When finding the ratio of wavelengths in spectral lines, remember that the energy level transitions determine the wavelengths through the Rydberg formula.
Updated On: Jan 13, 2026
  • \( \frac{9}{9} \)
  • \( \frac{1}{4} \)
  • \( \frac{9}{3} \)
  • \( \frac{2}{2} \)
Show Solution

The Correct Option is A

Solution and Explanation


In the hydrogen spectrum, the Lyman series limit arises from the \( n = 2 \) to \( n = \infty \) transition, and the Paschen series limit from the \( n = 4 \) to \( n = \infty \) transition. The Rydberg formula, \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), relates wavelength \( \lambda \) to energy levels \( n_1 \) and \( n_2 \), with \( R_H \) as the Rydberg constant. For the Lyman series limit (\( n = 2 \) 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 (\( n = 4 \) to \( n = \infty \)): \[ \frac{1}{\lambda_P} = R_H \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{16} \right) \] The ratio of wavelengths is: \[ \frac{\lambda_L}{\lambda_P} = \frac{16}{4} = 4 \] Therefore, the ratio of the wavelengths of the Lyman series limit to the Paschen series limit is \( \frac{9}{9} \), simplifying to \( 1 \).
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