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JEE Main 2024 (6 Apr, Shift 1) · Hydrogen Spectrum & Spectral Series

Hydrogen Spectrum & Spectral Series question from JEE Main 2024 (6 Apr, Shift 1)

Q20JEE Main 2024 · 6 APR · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is:

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Answer: (d)

Shortest wavelength = series limit: $\lambda = \frac{n_1^2}{R}$ .
Balmer: $\frac{4}{R}$ ; Lyman: $\frac{1}{R}$ .
Ratio $= 4 : 1$ .
So the answer is (d) $4 : 1$ .

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