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Chemistry - Atomic Structure Question with Solution | TestHub

ChemistryAtomic StructureSpectrumEasy2 minQB

ChemistryEasysingle choice

The wavelength of the radiation emitted when an electron of a hydrogen atom falls from infinity to the stationary state $n = 2$ is $(R = 1.097 \times 1 0^{7} m^{- 1})$ :-

Options:

Answer:

Solution:

$1/ λ = R (1/4 - 0) = R /4$ , so $λ = 4/ R = 4/ (1.097 \times 1 0^{7}) = 3.65 \times 1 0^{- 7} m = 365$ nm.

Answer: (B).

Stream:JEESubject:ChemistryTopic:Atomic StructureSubtopic:Spectrum

⏱ 2mℹ️ Source: QB

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