The ratio of shortest wavelength of lines in Balmer and Paschen series is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4 : 9
In a hydrogen series ending at level \( n_f \), the shortest wavelength is the series limit, where the electron falls from \( n_i = \infty \). By the Rydberg formula:
\( \displaystyle \frac{1}{\lambda_{\min}} = R\left(\frac{1}{n_f^{2}} - \frac{1}{\infty^{2}}\right) = \frac{R}{n_f^{2}} \)
Balmer series: \( n_f = 2 \Rightarrow \lambda_B = \frac{4}{R} \). Paschen series: \( n_f = 3 \Rightarrow \lambda_P = \frac{9}{R} \).
\( \displaystyle \frac{\lambda_B}{\lambda_P} = \frac{4}{9} \)
Shortcut: The shortest wavelength of a series is proportional to \( n_f^{2} \), so the ratio is simply \( 2^{2} : 3^{2} = 4 : 9 \).
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 – “… Balmer found a simple empirical formula for the observed wavelengths
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right) \qquad (12.5) \)
where λ is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc.”
Spectral Series – “… These are known, after their discoverers, as Lyman, Paschen, Brackett, and Pfund series. These are represented by the formulae:
Paschen series:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{n^2}\right) \qquad n = 4,5,6\ldots \qquad (12.7) \)”