The wavelength of the first line of Lyman series of hydrogen is 121.6…

The correct answer is Option (2) → \( \sim 97.3\,\mathrm{nm} \)

For hydrogen the wavelengths follow the Rydberg formula; for the Lyman series the lower level is \( n_{1} = 1 \), and the first, second and third lines have \( n_{2} = 2, 3, 4 \).

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^{2}} - \frac{1}{n^{2}}\right) \)

    \( \displaystyle \text{first line } (n = 2): \ \frac{1}{\lambda_{1}} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \)

    \( \displaystyle \text{third line } (n = 4): \ \frac{1}{\lambda_{3}} = R\left(1 - \frac{1}{16}\right) = \frac{15R}{16} \)

Taking the ratio, R cancels:

    \( \displaystyle \frac{\lambda_{3}}{\lambda_{1}} = \frac{3/4}{15/16} = \frac{3}{4}\times\frac{16}{15} = \frac{4}{5} \)

    \( \displaystyle \lambda_{3} = \frac{4}{5}\times121.6\,\mathrm{nm} = 97.28\,\mathrm{nm} \approx 97.3\,\mathrm{nm} \)

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:

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