In the Lyman series, the ratio of maximum wavelength to minimum wavel…

The correct answer is Option (2) → 4/3

The Lyman series contains the transitions that end on n = 1. By the Rydberg formula, a smaller energy gap gives a longer wavelength.

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^{2}} - \frac{1}{n^{2}}\right), \quad n = 2, 3, 4, \ldots \)

The longest wavelength comes from the smallest jump, n = 2 to n = 1:

    \( \displaystyle \frac{1}{\lambda_{\max}} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \)

The shortest wavelength is the series limit, n = ∞ to n = 1:

    \( \displaystyle \frac{1}{\lambda_{\min}} = R(1 - 0) = R \)

    \( \displaystyle \frac{\lambda_{\max}}{\lambda_{\min}} = \frac{4/(3R)}{1/R} = \frac{4}{3} \)

Shortcut: For a series ending on level n, \( \displaystyle\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{(n+1)^{2}}{2n+1} \); with n = 1 this gives 4/3 (and 9/5 for the Balmer series, n = 2).

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Spectral Series – “… Other series of spectra for hydrogen were subsequently discovered. These are known, after their discoverers, as Lyman, Paschen, Brackett, and Pfund series. These are represented by the formulae:

Lyman series:

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{n^2}\right) \qquad n = 2,3,4\ldots \qquad (12.6) \)”