If \( \lambda_1 \) and \( \lambda_2 \) are the longest wavelengths of the Lyman and Paschen series respectively, then \( \lambda_1:\lambda_2 \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 7 : 108
The wavelengths of hydrogen lines follow the Rydberg formula \( \frac{1}{\lambda} = R\left(\frac{1}{n_f^{2}}-\frac{1}{n_i^{2}}\right) \). The longest wavelength of a series is the line with the smallest energy jump, \( n_i = n_f+1 \).
Lyman series (\( n_f=1,\ n_i=2 \))
\( \displaystyle \frac{1}{\lambda_1} = R\left(1-\frac{1}{4}\right) = \frac{3R}{4} \Rightarrow \lambda_1 = \frac{4}{3R} \)
Paschen series (\( n_f=3,\ n_i=4 \))
\( \displaystyle \frac{1}{\lambda_2} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144} \Rightarrow \lambda_2 = \frac{144}{7R} \)
\( \displaystyle \frac{\lambda_1}{\lambda_2} = \frac{4}{3}\times\frac{7}{144} = \frac{7}{108} \)
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:
Lyman series:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{n^2}\right) \qquad n = 2,3,4\ldots \qquad (12.6) \)
Paschen series:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{n^2}\right) \qquad n = 4,5,6\ldots \qquad (12.7) \)”