The smallest wavelength in the spectral lines of Paschen series in th…

The correct answer is Option (3) → \( 8204\,\mathring{\mathrm{A}} \)

The lines of the Paschen series end on the level \( n = 3 \):

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

The smallest wavelength is the series limit, \( n \to \infty \):

    \( \displaystyle \frac{1}{\lambda_{\mathrm{min}}} = \frac{R}{9} \;\Rightarrow\; \lambda_{\mathrm{min}} = \frac{9}{R} = \frac{9}{1.097\times10^{7}}\,\mathrm{m} \)

    \( \displaystyle \lambda_{\mathrm{min}} = 8.204\times10^{-7}\,\mathrm{m} = 8204\,\mathring{\mathrm{A}} \)

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 \( \lambda \) = 364.6 nm. This is the shortest wavelength in the Balmer series.”

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:

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