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Match the List-I (spectral lines of Hydrogen for transitions from) with List-II (Wavelengths) below.
Column I
- A. $n_2=3$ to $n_1=2$
- B. $n_2=4$ to $n_1=2$
- C. $n_2=5$ to $n_1=2$
- D. $n_2=6$ to $n_1=2$
Column II
- I. 410.2 nm
- II. 434.1 nm
- III. 656.3 nm
- IV. 486.1 nm
Correct answer: A → III, B → IV, C → II, D → I
Detailed Solution
$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
Here $n_1 = 2$ is fixed (Balmer series), so as $n_2$ increases, $\frac{1}{\lambda}$ increases, i.e. the wavelength decreases.
$n_2 = 3 \rightarrow 2$: 656.3 nm (longest)
$n_2 = 4 \rightarrow 2$: 486.1 nm
$n_2 = 5 \rightarrow 2$: 434.1 nm
$n_2 = 6 \rightarrow 2$: 410.2 nm (shortest)
Correct match: A-III, B-IV, C-II, D-I.