Match the Lists I and II where transition lines and their corresponding wavelengths for Lyman series in hydrogen spectrum are given.
| List-I | List-II | ||
| Transition Line | Wavelength in terms of the Rydberg constant | ||
| (A) | n = 2 to 1 | (I) | \( \displaystyle\frac{9}{8R} \) |
| (B) | n = 3 to 1 | (II) | \( \displaystyle\frac{1}{R} \) |
| (C) | n = 4 to 1 | (III) | \( \displaystyle\frac{4}{3R} \) |
| (D) | n = infinity to 1 | (IV) | \( \displaystyle\frac{16}{15R} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
For the Lyman series the electron ends in n = 1, and the Rydberg formula gives
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^{2}} - \frac{1}{n^{2}}\right) \)
(A) n = 2 to 1 — this matches (III) \( 4/3R \).
\( \displaystyle \frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \)
(B) n = 3 to 1 — this matches (I) \( 9/8R \).
\( \displaystyle \frac{1}{\lambda} = R\left(1 - \frac{1}{9}\right) = \frac{8R}{9} \)
(C) n = 4 to 1 — this matches (IV) \( 16/15R \).
\( \displaystyle \frac{1}{\lambda} = R\left(1 - \frac{1}{16}\right) = \frac{15R}{16} \)
(D) n = infinity to 1 — this matches (II) \( 1/R \). This is the series limit, the shortest wavelength of the series.
\( \displaystyle \frac{1}{\lambda} = R\left(1 - 0\right) = R \)
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:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{n^2}\right) \qquad n = 2,3,4\ldots \qquad (12.6) \)”