When the electron in hydrogen atom jumps from the 2nd orbit to the 1st orbit, the wavelength of the radiation emitted is \( \lambda \). When the electron jumps from the 3rd orbit to 1st orbit, the wavelength of emitted radiation is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{27}{32}\lambda \)
For hydrogen the wavelength of the emitted line is given by the Rydberg formula, so only the ratio of the two brackets is needed.
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right) \)
\( \displaystyle 2 \rightarrow 1: \quad \frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \)
\( \displaystyle 3 \rightarrow 1: \quad \frac{1}{\lambda'} = R\left(1 - \frac{1}{9}\right) = \frac{8R}{9} \)
\( \displaystyle \frac{\lambda'}{\lambda} = \frac{3/4}{8/9} = \frac{27}{32} \;\Rightarrow\; \lambda' = \frac{27}{32}\lambda \)
The jump from the 3rd orbit releases more energy, so its wavelength is shorter than \( \lambda \), as expected.
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 – “… Other series of spectra for hydrogen were subsequently discovered. These are known, after their discoverers, as Lyman, Paschen, Brackett, and Pfund series. These are represented by the formulae:
Lyman series:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{n^2}\right) \qquad n = 2,3,4\ldots \qquad (12.6) \)”