The shortest wavelength in the Balmer series of a hydrogen atom is (Given: \( R = 1.1 \times 10^{7}\,\mathrm{m^{-1}} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3637 Å
In the Balmer series the electron ends in the level \( n_1 = 2 \). The shortest wavelength comes from the largest jump, from \( n_2 = \infty \) to \( n_1 = 2 \) (the series limit), given by the Rydberg formula.
\( \displaystyle \frac{1}{\lambda_{\min}} = R\left(\frac{1}{2^{2}} - \frac{1}{\infty^{2}}\right) = \frac{R}{4} \)
\( \displaystyle \lambda_{\min} = \frac{4}{R} = \frac{4}{1.1\times10^{7}\,\mathrm{m^{-1}}} = 3.636\times10^{-7}\,\mathrm{m} \)
\( \displaystyle \lambda_{\min} \approx 3636\,\mathring{\mathrm{A}} \approx 3637\,\mathring{\mathrm{A}} \)
Note: A similar question was also asked in CUET (UG) 2025 on 15-05-2025 (Slot 2).
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 – “… For n = ∞, one obtains the limit of the series, at λ = 364.6 nm. This is the shortest wavelength in the Balmer series.”