InterviewSolution
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InterviewSolution
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The electron in hydrogen atom is in the first Bohr orbit `(n = 1)`. The ratio of transition energies, `E(n-1 rarr n =3)` to `E(n=1 rarr n=2)`, is-A. 32/27B. 16/27C. 32/9D. `8//9` |
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Answer» Correct Answer - A `DeltaE=13.6 (1/n_(1)^(2)-1/n_(2)^(2))eV//"atom" , (DeltaE_(1 rarr 3))/(DeltaE_(1 rarr 2))=(1/1^(2)-1/3^(2))/(1/1^(2)-1/2^(2))=.32/27` |
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