The wavelength of emitted wave due to energy transition in hydrogen atom from n = 3 to n = 2 energy level will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6563 Å
The wavelength emitted in a transition of the hydrogen atom is given by the Rydberg formula; the jump from \( n=3 \) to \( n=2 \) is the first line (H-alpha) of the Balmer series.
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{2^{2}} - \frac{1}{3^{2}}\right) = R\times\frac{5}{36} \)
With \( R = 1.097\times10^{7}\,\mathrm{m^{-1}} \):
\( \displaystyle \lambda = \frac{36}{5R} = \frac{36}{5\times1.097\times10^{7}}\,\mathrm{m} = 6.563\times10^{-7}\,\mathrm{m} \)
\( \lambda = 6563\,\mathrm{\mathring{A}} \)
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 – “… This series is called Balmer series (Fig. 12.6). The line with the longest wavelength, 656.3 nm in the red is called Hα; … 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 λ is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc. The value of R is 1.097 × 107 m–1.”