The shortest wavelength in the Balmer series of hydrogen atom would b…

The correct answer is Option (1) → \( 3636\,\mathring{\mathrm{A}} \)

In the Balmer series the electron falls to the level \( n_1 = 2 \). The shortest wavelength comes from the largest jump, from \( n_2 = \infty \) (the series limit), using the Rydberg formula.

    \( \displaystyle \frac{1}{\lambda_{\min}} = R\left(\frac{1}{2^{2}} - \frac{1}{\infty^{2}}\right) = \frac{R}{4} \)

    \( \displaystyle \lambda_{\min} = \frac{4}{R} = \frac{4}{1.1\times10^{7}\,\mathrm{m^{-1}}} = 3.636\times10^{-7}\,\mathrm{m} \)

    \( \displaystyle \lambda_{\min} = 3636\times10^{-10}\,\mathrm{m} = 3636\,\mathring{\mathrm{A}} \)

Note: A similar question was also asked in CUET (UG) 2025 on 03-06-2025 (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 – “… 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. … For n = ∞, one obtains the limit of the series, at \( \lambda \) = 364.6 nm. This is the shortest wavelength in the Balmer series.”