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MHT CET 2024 (9 May, Shift 1) · Hydrogen Spectrum & Spectral Series

Hydrogen Spectrum & Spectral Series question from MHT CET 2024 (9 May, Shift 1)

Q18MHT CET 2024 · 9 MAY · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series

In hydrogen atom, ratio of the shortest wavelength in the Balmer series to that in the Paschen series is

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Answer: (c)

Shortest wavelength = series limit: $\lambda = \frac{n_1^2}{R}$ .
Balmer: $\frac{4}{R}$ ; Paschen: $\frac{9}{R}$ .
Ratio $= 4 : 9$ .
So the answer is (c) $4 : 9$ .

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