The shortest wavelength present in the Lyman series of spectral lines…

The correct answer is Option (1) → \( 9.1\times10^{-8}\,\mathrm{m} \)

The wavelengths of the hydrogen spectrum are given by the Rydberg formula. The Lyman series consists of transitions to the ground level \( n_f = 1 \).

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

The wavelength is shortest when the energy of the transition is greatest, that is, for the series limit \( n\to\infty \), where \( 1/n^{2}\to0 \):

    \( \displaystyle \frac{1}{\lambda_{\min}} = R \Rightarrow \lambda_{\min} = \frac{1}{R} = \frac{1}{1.097\times10^{7}}\,\mathrm{m} \)

    \( \lambda_{\min} \approx 9.1\times10^{-8}\,\mathrm{m} \)

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 – “… For \( n = \infty \), one obtains the limit of the series, at \( \lambda \) = 364.6 nm. This is the shortest wavelength in the Balmer 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:

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