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Calculate the two highest wavelength of the radiation emitted when hydrogen atoms make transition from higher state to `n = 2` |
Answer» Correct Answer - `6561 Å, 4863 Å` (Approx) `lambda_(3 to 2)=12400/(DeltaE_(3 to 2))=12400/1.89=6561 Å, lambda_(4 to 2)=12400/(DeltaE_(4 to 2))=12400/2.55=4863 Å` |
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