If the minimum wavelength corresponding to the Paschen series of hydrogen spectra is 820 nm then that corresponding to the Balmer series would be.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 364.4 nm
By the Rydberg formula, the minimum wavelength of a series (its series limit) is for a transition from \( n_2 = \infty \) to the lower level \( n_1 \) of the series.
\( \displaystyle \frac{1}{\lambda_{\min}} = R\left(\frac{1}{n_1^2} - \frac{1}{\infty}\right) = \frac{R}{n_1^2} \;\Rightarrow\; \lambda_{\min} = \frac{n_1^2}{R} \)
For the Paschen series \( n_1 = 3 \) and for the Balmer series \( n_1 = 2 \):
\( \displaystyle \frac{\lambda_{\mathrm{Balmer}}}{\lambda_{\mathrm{Paschen}}} = \frac{2^2}{3^2} = \frac{4}{9} \)
\( \displaystyle \lambda_{\mathrm{Balmer}} = \frac{4}{9} \times 820\,\mathrm{nm} = 364.4\,\mathrm{nm} \)
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 – “… For n = ∞, one obtains the limit of the series, at λ = 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:
Paschen series:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{n^2}\right) \qquad n = 4,5,6\ldots \qquad (12.7) \)”