JEE Main 2026 — Apr 4, Shift 1 — Chemistry solution

JEE Main2026Apr 4, Shift 1Chemistry

What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

Solution

By the Rydberg formula $\overset{ν}{ˉ} = R_{H} Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ with $Z = 1$ .

First line of Balmer ( $n_{1} = 2 \to n_{2} = 3$ ): $\overset{ν}{ˉ}_{1} = R_{H} (\frac{1}{4} - \frac{1}{9}) = \frac{5 R _{H}}{36}$ .

First line of Brackett ( $n_{1} = 4 \to n_{2} = 5$ ): $\overset{ν}{ˉ}_{2} = R_{H} (\frac{1}{16} - \frac{1}{25}) = \frac{9 R _{H}}{400}$ .

$\frac{ν ˉ _{1}}{ν ˉ _{2}} = \frac{5/36}{9/400} = \frac{5 \times 400}{36 \times 9} = \frac{2000}{324} = \frac{500}{81} .$

Expressed as $5 : 0.81$ (dividing both sides of $500 : 81$ by $100$ ).

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