The wavelength of the first member of the Balmer series in the hydrogen spectrum is \(6563\,\mathring{\mathrm{A}}\). The wavelength of first member of Lyman series will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1215\,\mathring{\mathrm{A}}\)
The wavelengths of hydrogen lines follow the Rydberg formula \( \frac{1}{\lambda} = R\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right) \). The first Balmer line is \( n = 3 \to 2 \); the first Lyman line is \( n = 2 \to 1 \).
\( \displaystyle \frac{1}{\lambda_B} = R\left(\frac{1}{2^{2}} - \frac{1}{3^{2}}\right) = \frac{5R}{36} \)
\( \displaystyle \frac{1}{\lambda_L} = R\left(\frac{1}{1^{2}} - \frac{1}{2^{2}}\right) = \frac{3R}{4} \)
\( \displaystyle \frac{\lambda_L}{\lambda_B} = \frac{5R/36}{3R/4} = \frac{5}{27} \)
\( \displaystyle \lambda_L = \frac{5}{27} \times 6563\,\mathring{\mathrm{A}} \approx 1215\,\mathring{\mathrm{A}} \)
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 – “… 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.”
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:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{n^2}\right) \qquad n = 2,3,4\ldots \qquad (12.6) \)”