The ratio of maximum frequency and minimum frequency of light emitted in Balmer series of hydrogen spectrum is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 9 : 5
The Balmer series contains the transitions that end on \( n = 2 \). The frequency of a line is proportional to the energy difference of the two levels.
\( \displaystyle \nu = Rc\left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right), \quad n = 3, 4, 5, \ldots \)
The maximum frequency is the series limit, \( n \to \infty \):
\( \displaystyle \nu_{\max} = Rc\left(\frac{1}{4} - 0\right) = \frac{Rc}{4} = \frac{9Rc}{36} \)
The minimum frequency is the first line, \( n = 3 \to 2 \):
\( \displaystyle \nu_{\min} = Rc\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5Rc}{36} \)
\( \displaystyle \frac{\nu_{\max}}{\nu_{\min}} = \frac{9}{5} \)
So the ratio is 9 : 5.
Note: A similar question was also asked in CUET (UG) 2026 on 31-05-2026 (Slot 2) and on 18-05-2026 (Slot 1).