The wavelength of photon emitted by a hydrogen atom when an electron makes a transition from \( n = 2 \) to \( n = 1 \) state is
(Take the value of the Rydberg constant, \( R = 1.0973\times10^{7}\,\mathrm{m^{-1}} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 121.5 nm
The wavelengths of the hydrogen spectrum follow the Rydberg formula. A jump from \( n_i = 2 \) to \( n_f = 1 \) gives the first line of the Lyman series.
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right) = R\left(\frac{1}{1} - \frac{1}{4}\right) = \frac{3R}{4} \)
\( \displaystyle \frac{1}{\lambda} = 0.75 \times 1.0973\times10^{7}\,\mathrm{m^{-1}} = 8.230\times10^{6}\,\mathrm{m^{-1}} \)
\( \displaystyle \lambda = \frac{1}{8.230\times10^{6}}\,\mathrm{m} = 1.215\times10^{-7}\,\mathrm{m} = 121.5\,\mathrm{nm} \)
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:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{n^2}\right) \qquad n = 2,3,4\ldots \qquad (12.6) \)”