For a hydrogen atom, the ratio of the largest wavelength of Lyman… | JEE Main 2025 (7 Apr, Shift 1)

Q11JEE Main 2025 · 7 APR · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is

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Answer: (a)

Largest wavelength = first line.
Lyman ( $2 \to 1$ ): $\frac{1}{\lambda_L} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$ , so $\lambda_L = \frac{4}{3R}$ .
Balmer ( $3 \to 2$ ): $\frac{1}{\lambda_B} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$ , so $\lambda_B = \frac{36}{5R}$ .
$\frac{\lambda_L}{\lambda_B} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27}$ .
So the answer is (a) $5 : 27$ .

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