If the wave length of 1st line of Balmer series of hydrogen is 6561 Å, the wavelength of the 2nd line of series will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4860 Å
The Balmer series is the set of lines in which the electron falls to \( n = 2 \). The first line is the transition \( 3\to2 \) and the second line is \( 4\to2 \). By the Rydberg formula \( \frac{1}{\lambda} = R\left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right) \):
\( \displaystyle \frac{1}{\lambda_1} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36} \)
\( \displaystyle \frac{1}{\lambda_2} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16} \)
Dividing, the unknown R cancels:
\( \displaystyle \frac{\lambda_2}{\lambda_1} = \frac{5/36}{3/16} = \frac{5\times16}{36\times3} = \frac{20}{27} \)
\( \displaystyle \lambda_2 = 6561\,\mathrm{\mathring{A}}\times\frac{20}{27} = 4860\,\mathrm{\mathring{A}} \)
Shortcut: \( 6561 = 27\times243 \), so \( \frac{6561}{27} = 243 \) and \( \lambda_2 = 243\times20 = 4860\,\mathrm{\mathring{A}} \) with no decimals. The second line must also be shorter than the first, which removes Options (1) and (4).
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
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right) \qquad (12.5) \)”