Question:medium

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

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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.
Updated On: Jan 13, 2026
  • \( 1 : 4 \)
  • \( 1 : 3 \)
  • \( 2 : 3 \)
  • \( 1 : 2 \)
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The Correct Option is A

Solution and Explanation

The Lyman and Paschen series are components of the hydrogen atom's emission spectra. The Lyman series involves transitions ending at the \( n = 1 \) state, while the Paschen series involves transitions ending at the \( n = 3 \) state. The wavelength of emitted light for these transitions is calculated using the Rydberg formula for hydrogen: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] In this formula: - \( \lambda \) represents the wavelength of the emitted radiation. - \( R_H \) is the Rydberg constant. - \( n_1 \) and \( n_2 \) are the principal quantum numbers for the initial and final states, respectively. For the Lyman series limit, the transition is from \( n_2 \to 1 \). The corresponding wavelength is derived from the transition where \( n_2 \to \infty \). Therefore, 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 \( n_2 \to 3 \). The corresponding wavelength is derived from the transition where \( n_2 \to \infty \). Therefore, for Paschen: \[ \frac{1}{\lambda_{\text{Paschen}}} = R_H \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{9} \right) \] The ratio of the wavelengths is the inverse of the ratio of these terms: \[ \frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Paschen}}} = \frac{9}{1} = 9 \] Consequently, the ratio of the wavelength of the Lyman series limit to the Paschen series limit is \( 9:1 \), indicating the Lyman series limit wavelength is nine times greater than the Paschen series limit wavelength.
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