Match the LIST-I with LIST-II (Symbols have their usual meaning)
| LIST-I | LIST-II | ||
| (Wavelength) | (Transition of electron in Hydrogen atom) | ||
| A. | \( \displaystyle\frac{4}{3R} \) | I. | \( n = 3 \) to \( 1 \) |
| B. | \( \displaystyle\frac{9}{8R} \) | II. | \( n = 4 \) to \( 2 \) |
| C. | \( \displaystyle\frac{16}{3R} \) | III. | \( n = 3 \) to \( 2 \) |
| D. | \( \displaystyle\frac{36}{5R} \) | IV. | \( n = 2 \) to \( 1 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-I, C-II, D-III
The wavelengths of hydrogen lines follow the Rydberg formula, with \( n_f \) the lower and \( n_i \) the upper level.
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right) \)
(A) \( 4/3R \) — this matches (IV) \( n = 2 \) to \( 1 \), the first line of the Lyman series.
\( \displaystyle \frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \Rightarrow \lambda = \frac{4}{3R} \)
(B) \( 9/8R \) — this matches (I) \( n = 3 \) to \( 1 \), the second Lyman line.
\( \displaystyle \frac{1}{\lambda} = R\left(1 - \frac{1}{9}\right) = \frac{8R}{9} \Rightarrow \lambda = \frac{9}{8R} \)
(C) \( 16/3R \) — this matches (II) \( n = 4 \) to \( 2 \), the second line of the Balmer series.
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16} \Rightarrow \lambda = \frac{16}{3R} \)
(D) \( 36/5R \) — this matches (III) \( n = 3 \) to \( 2 \), the first Balmer line.
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36} \Rightarrow \lambda = \frac{36}{5R} \)