Question 15: Match List I with List II.
| List I (Spectral Lines of Hydrogen for transitions from) | List II (Wavelengths (nm)) |
|---|---|
| A. \mathrm{n}_{2}=3 to \mathrm{n}_{1}=2 | I. 410.2 |
| B. \mathrm{n}_{2}=4 to \mathrm{n}_{1}=2 | II. 434.1 |
| C. \mathrm{n}_{2}=5 to \mathrm{n}_{1}=2 | III. 656.3 |
| D. \mathrm{n}_{2}=6 to \mathrm{n}_{1}=2 | IV. 486.1 |
Choose the correct answer from the options given below:
(1) A-II, B-I, C-IV, D-III
(2) A-III, B-IV, C-II, D-I
(3) A-IV, B-III, C-I, D-II
(4) A-I, B-II, C-III, D-IV
Answer: Option (2)
Explanation:
All the given transitions end at \mathrm{n}_{1}=2,
so they belong to the Balmer series of the hydrogen spectrum.
The transition from \mathrm{n}_{2}=3 to \mathrm{n}_{1}=2
corresponds to the Hα line with wavelength 656.3 nm,
so A matches with III.
The transition from \mathrm{n}_{2}=4 to \mathrm{n}_{1}=2
corresponds to the Hβ line with wavelength 486.1 nm,
so B matches with IV.
The transition from \mathrm{n}_{2}=5 to \mathrm{n}_{1}=2
corresponds to the Hγ line with wavelength 434.1 nm,
so C matches with II.
The transition from \mathrm{n}_{2}=6 to \mathrm{n}_{1}=2
corresponds to the Hδ line with wavelength 410.2 nm,
so D matches with I.
Therefore, the correct matching is A-III, B-IV, C-II, and D-I.