In the atomic hydrogen line spectrum, the wavelength of the first line of the Lyman series is
(Given: Rydberg constant \( R = 1.097 \times 10^{7}\,\mathrm{m^{-1}} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 122\,\mathrm{nm} \)
The Lyman series is the set of lines for transitions to \( n = 1 \), and its first line is the transition \( n = 2 \to 1 \). The Rydberg formula is
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right) \)
\( \displaystyle \frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \)
\( \displaystyle \lambda = \frac{4}{3R} = \frac{4}{3 \times 1.097\times10^{7}\,\mathrm{m^{-1}}} = 1.215\times10^{-7}\,\mathrm{m} \approx 122\,\mathrm{nm} \)
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:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{n^2}\right) \qquad n = 2,3,4\ldots \qquad (12.6) \)”