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Question
Q278) The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H = 2.18 × 10–18 J atom–1 and h = 6.625 × 10–34 Js)



SOLUTION

Here E1 is ionization energy of H = 2.18 × 10–18 J atom–1
Now energy difference emitted as radiation givenby E=hν

Answer:(c)