The ratio of maximum frequency and minimum frequency of light emitted in the Balmer series of hydrogen spectrum is-
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 9/5
In the Balmer series the electron falls to n = 2. The frequency is proportional to \( \frac{1}{2^{2}} - \frac{1}{n^{2}} \), so the maximum frequency is for \( n = \infty \) (series limit) and the minimum frequency is for \( n = 3 \) (the H-alpha line).
\( \displaystyle \nu \propto \frac{1}{2^{2}} - \frac{1}{n^{2}} \)
\( \displaystyle \nu_{\max} \propto \frac{1}{4} - 0 = \frac{1}{4} \)
\( \displaystyle \nu_{\min} \propto \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \)
\( \displaystyle \frac{\nu_{\max}}{\nu_{\min}} = \frac{1/4}{5/36} = \frac{36}{20} = \frac{9}{5} \)
Note: The same question was also asked in CUET (UG) 2026 on 31-05-2026 (Slot 2), and a similar question in CUET (UG) 2025 on 22-05-2025 (Slot 1).
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 \( \lambda \) is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc.”