The wavelength of second line of Balmer series in terms of Rydberg constant R is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{16}{3R} \)
The Balmer series has transitions ending on \( n = 2 \). The first line is \( 3 \rightarrow 2 \), so the second line is \( 4 \rightarrow 2 \).
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{2^{2}} - \frac{1}{4^{2}}\right) = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16} \)
\( \displaystyle \lambda = \frac{16}{3R} \)
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 λ is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc. The value of R is 1.097 × 107 m–1. This equation is also called Balmer formula.”