The ratio of maximum frequency and minimum frequency of light emitted in Balmer series of hydrogen spectrum in Bohr's model is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9/5
In Bohr’s model the frequency of an emitted line is proportional to the difference of the \( 1/n^{2} \) terms. The Balmer series contains the transitions that end on n = 2.
\( \displaystyle \nu \propto \left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right), \qquad n = 3, 4, 5, \ldots \)
The maximum frequency is the series limit, \( n \rightarrow \infty \); the minimum frequency is the first line, n = 3.
\( \displaystyle \nu_{\max} \propto \frac{1}{4} - 0 = \frac{1}{4} \)
\( \displaystyle \nu_{\min} \propto \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \)
\( \displaystyle \frac{\nu_{\max}}{\nu_{\min}} = \frac{1/4}{5/36} = \frac{36}{20} = \frac{9}{5} \)
Note: The same question was also asked in CUET (UG) 2026 on 18-05-2026 (Slot 1), and a similar question in CUET (UG) 2025 on 22-05-2025 (Slot 1).