The smallest wavelength in the spectral lines of the Paschen series in the hydrogen atom is: (R is the Rydberg's constant in \( \mathring{\mathrm{A}}^{-1} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \frac{9}{R}\,\mathring{\mathrm{A}} \)
In the Paschen series the electron ends in the level \( n_1 = 3 \). The smallest wavelength belongs to the largest energy jump, which is from \( n_2 = \infty \) to \( n_1 = 3 \) (the series limit).
\( \displaystyle \frac{1}{\lambda_{\min}} = R\left(\frac{1}{3^{2}} - \frac{1}{\infty^{2}}\right) = \frac{R}{9} \)
\( \displaystyle \lambda_{\min} = \frac{9}{R} \)
With \( R \) in \( \mathring{\mathrm{A}}^{-1} \), the wavelength is \( 9/R \) in \( \mathring{\mathrm{A}} \).
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 λ = 364.6 nm. This is the shortest wavelength in the Balmer series.”
Spectral Series – “… These are represented by the formulae: … Paschen series:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{n^2}\right) \qquad n = 4,5,6\ldots \qquad (12.7) \)”