In the hydrogen atom spectrum, which of the following electronic transitions give rise to radiations in the visible region?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( n = 4 \rightarrow n = 2 \)
In hydrogen, the transitions ending on \( n = 2 \) form the Balmer series, which lies in the visible region. The Lyman series (ending on \( n = 1 \)) is ultraviolet, and the Paschen \( (n = 3) \), Brackett \( (n = 4) \) and Pfund \( (n = 5) \) series are infrared.
For \( n = 4 \rightarrow n = 2 \):
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{2^{2}} - \frac{1}{4^{2}}\right) = \frac{3R}{16} \)
\( \displaystyle \lambda = \frac{16}{3\times1.097\times10^{7}\,\mathrm{m^{-1}}} \approx 486\,\mathrm{nm} \)
This is the blue-green H-beta line, in the visible range (400–700 nm).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), but it is still in the CUET 2026 syllabus, so do not skip it.
NCERT (Pre-rationalised edition, 2022-23):
Spectral Series – “… In 1885, the first such series was observed by a Swedish school teacher Johann Jakob Balmer (1825–1898) in the visible region of the hydrogen spectrum. This series is called Balmer series (Fig. 12.6). … 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.”