An electron makes a transition from orbit n = 4 to the orbit n = 2 of…

The correct answer is Option (4) → \( \displaystyle\frac{16}{3R} \)

The wavelength of the radiation emitted in a transition of the hydrogen atom from level \( n_i \) to a lower level \( n_f \) is given by the Rydberg formula:

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right) \)

Here \( n_i = 4 \) and \( n_f = 2 \).

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{16}\right) = R\times\frac{3}{16} \)

    \( \displaystyle \lambda = \frac{16}{3R} \)

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 \( \lambda \) is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc.”