What is the approximate wavelength of the radiation emitted when the…

The correct answer is Option (2) → 822 nm

A jump that ends on \( n = 3 \) belongs to the Paschen series, and the jump from \( n = \infty \) gives its series limit (shortest wavelength). By the Rydberg formula,

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{3^{2}} - \frac{1}{\infty^{2}}\right) = \frac{R}{9} \)

    \( R = 1.097\times10^{7}\,\mathrm{m^{-1}} \)

    \( \displaystyle \lambda = \frac{9}{R} = \frac{9}{1.097\times10^{7}\,\mathrm{m^{-1}}} = 8.20\times10^{-7}\,\mathrm{m} \approx 820\,\mathrm{nm} \)

This is in the infrared, and the nearest option is 822 nm.

Shortcut: The photon energy is \( 13.6/3^{2} = 1.51\,\mathrm{eV} \), and \( \lambda\,(\mathrm{nm}) = 1240/E\,(\mathrm{eV}) = 1240/1.51 \approx 821\,\mathrm{nm} \).

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:

Paschen series:

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

The Lyman series is in the ultraviolet, and the Paschen and Brackett series are in the infrared region.”