The ratio maximum wavelength to minimum wavelength in Lyman series is:

The correct answer is Option (1) → \( \displaystyle\frac{4}{3} \)

The Lyman series contains the lines of transitions ending on \( n = 1 \):

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

The longest wavelength comes from the smallest energy gap, \( n = 2 \to 1 \):

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

The shortest wavelength (series limit) comes from \( n = \infty \to 1 \):

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

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

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 – “… These are known, after their discoverers, as Lyman, Paschen, Brackett, and Pfund series. These are represented by the formulae:

Lyman series:

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