What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum lines?
(Take \( h=6.63\times10^{-34}\,\mathrm{Js} \); \( c=3\times10^{8}\,\mathrm{ms^{-1}} \); \( e=1.6\times10^{-19}\,\mathrm{C} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 122 nm
The Lyman series contains the lines emitted when the electron of a hydrogen atom jumps to the ground state \( n = 1 \). The least energetic photon comes from the smallest jump, \( n = 2 \rightarrow n = 1 \), and has the longest wavelength of the series.
By the Bohr model, \( E_n = -13.6/n^{2}\,\mathrm{eV} \):
\( \displaystyle \Delta E = 13.6\left(\frac{1}{1^{2}} - \frac{1}{2^{2}}\right)\mathrm{eV} = 13.6 \times \frac{3}{4}\,\mathrm{eV} = 10.2\,\mathrm{eV} \)
\( \displaystyle \Delta E = 10.2 \times 1.6\times10^{-19}\,\mathrm{J} = 1.632\times10^{-18}\,\mathrm{J} \)
\( \displaystyle \lambda = \frac{hc}{\Delta E} = \frac{6.63\times10^{-34} \times 3\times10^{8}}{1.632\times10^{-18}}\,\mathrm{m} \)
\( \displaystyle \lambda = 1.22\times10^{-7}\,\mathrm{m} \approx 122\,\mathrm{nm} \)
Shortcut: With \( hc \approx 1240\,\mathrm{eV\,nm} \): \( \lambda = 1240/10.2 \approx 122\,\mathrm{nm} \).
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:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{n^2}\right) \qquad n = 2,3,4\ldots \qquad (12.6) \)
… The Lyman series is in the ultraviolet, and the Paschen and Brackett series are in the infrared region.”