In the Paschen series of the emission spectrum of the hydrogen atom,…

The correct answer is Option (4) → \( \lambda_{\mathrm{max}}/\lambda_{\mathrm{min}} = 16/7 \)

The Paschen series contains the transitions that end on \( n_{1} = 3 \); the wavelengths follow the Rydberg formula.

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{3^{2}} - \frac{1}{n^{2}}\right), \quad n = 4, 5, 6, \ldots \)

The longest wavelength comes from the smallest energy gap (n = 4 to 3), and the shortest from the series limit (n tending to infinity):

    \( \displaystyle \frac{1}{\lambda_{\mathrm{max}}} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144} \)

    \( \displaystyle \frac{1}{\lambda_{\mathrm{min}}} = R\left(\frac{1}{9} - 0\right) = \frac{R}{9} \)

    \( \displaystyle \frac{\lambda_{\mathrm{max}}}{\lambda_{\mathrm{min}}} = \frac{R/9}{7R/144} = \frac{144}{63} = \frac{16}{7} \)

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:

Paschen series:

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{n^2}\right) \qquad n = 4,5,6\ldots \qquad (12.7) \)”