JEE Main 2026 — Jan 24, Shift 2 — Chemistry solution

JEE Main2026Jan 24, Shift 2Chemistry

The wavelength of spectral line obtained in the spectrum of $Li^{2 +}$ ion, when the transition takes place between two levels whose sum is 4 and difference is 2, is

A
$2.28 \times 1 0^{- 7}$ cm
B
$2.28 \times 1 0^{- 6}$ cm
C
$1.14 \times 1 0^{- 7}$ cm
D
$1.14 \times 1 0^{- 6}$ cm

Solution

Given $n_{1} + n_{2} = 4$ and $n_{2} - n_{1} = 2$ , solving gives $n_{1} = 1$ , $n_{2} = 3$ . For $Li^{2 +}$ , $Z = 3$ .

Rydberg formula: $\frac{1}{λ} = R_{H} Z^{2} [\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}]$

$\frac{1}{λ} = R_{H} (9) [1 - \frac{1}{9}] = R_{H} \cdot 9 \cdot \frac{8}{9} = 8 R_{H}$

With $R_{H} = 1.1 \times 1 0^{5} cm^{- 1}$ :

$λ = \frac{1}{8 \times 1.1 \times 1 0 ^{5}} = \frac{1}{8.8 \times 1 0 ^{5}} cm \approx 1.14 \times 1 0^{- 6} cm$ .

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