The relation below correctly represents the frequency of light emitted when an electron in a hydrogen atom jumps from a higher \( (n_i) \) energy level to \( (n_f) \) a lower energy level: \( \nu = Rc\left(\displaystyle\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right) \)
Predict the values of \( n_i, n_f \), respectively for the Paschen series, which will give the highest wavelength in the series
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4, 3
In the Paschen series the electron jumps down to the third level, so \( n_f = 3 \). The wavelength is largest when the frequency, and so the energy of the photon, is smallest. That happens for the smallest energy step, which is to the third level from the next level above it, \( n_i = 4 \).
\( \displaystyle \nu = Rc\left(\frac{1}{3^{2}} - \frac{1}{4^{2}}\right) = Rc\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7Rc}{144} \)
So \( n_i = 4 \) and \( n_f = 3 \).
Why the other options are incorrect
Option (1) 3, 2 — Incorrect. \( n_i = 3,\ n_f = 2 \) is the longest-wavelength line of the Balmer series, not the Paschen series.
Option (3) 5, 4 — Incorrect. \( n_i = 5,\ n_f = 4 \) is the longest-wavelength line of the Brackett series, whose final level is 4.
Option (4) \( \infty \), 3 — Incorrect. \( n_i = \infty,\ n_f = 3 \) is the series limit of the Paschen series, which gives the shortest wavelength, not the highest.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), but it is still in the CUET 2026 syllabus, so do not skip it.
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:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{n^2}\right) \qquad n = 4,5,6\ldots \qquad (12.7) \)”