Excited hydrogen atom emits light in the ultraviolet region at 2.47 × 1 0 15 Hz . With this frequency, the energy of a single photon is: ( h = 6.63 × 1 0 − 34 Js )

E = hv = 6.63 × 1 0 − 34 × 2.47 × 1 0 15 = 1.640 × 1 0 − 18 J

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ChemistryAtomic StructureSpectrumEasy2 minPYQ_2014

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Excited hydrogen atom emits light in the ultraviolet region at $2.47 \times 1 0^{15} Hz$ . With this frequency, the energy of a single photon is: $(h = 6.63 \times 1 0^{- 34} Js)$

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Answer:

Solution:

$E = hv$ $= 6.63 \times 1 0^{- 34} \times 2.47 \times 1 0^{15} = 1.640 \times 1 0^{- 18} J$

Stream:JEESubject:ChemistryTopic:Atomic StructureSubtopic:Spectrum

⏱ 2mℹ️ Source: PYQ_2014

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