The ratio of maximum wavelength to minimum wavelength in Balmer series is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9/5
The Balmer series contains the transitions that end on the level n = 2.
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right), \quad n = 3, 4, 5, \ldots \)
The longest wavelength comes from the smallest energy gap, n = 3 to n = 2. The shortest wavelength is the series limit, n → ∞.
\( \displaystyle \frac{1}{\lambda_{\mathrm{max}}} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36} \)
\( \displaystyle \frac{1}{\lambda_{\mathrm{min}}} = R\left(\frac{1}{4} - 0\right) = \frac{R}{4} \)
\( \displaystyle \frac{\lambda_{\mathrm{max}}}{\lambda_{\mathrm{min}}} = \frac{36/5R}{4/R} = \frac{36}{20} = \frac{9}{5} \)
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 – “… This series is called Balmer series (Fig. 12.6). The line with the longest wavelength, 656.3 nm in the red is called Hα; the next line with wavelength 486.1 nm in the blue-green is called Hβ, the third line 434.1 nm in the violet is called Hγ; and so on. … 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.”