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If a hydrogen atom emit a photon of energy `12.1 eV` , its orbital angular momentum changes by `Delta L. then`Delta L` equalsA. `(h)/(2pi)`B. `(3h)/(2pi)`C. `(h)/(pi)`D. `(2h)/(pi)` |
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Answer» Correct Answer - C `DeltaE = E_(3) - E_(1) = 12.1eV` Electron jumps from 3rd orbit to 1st orbit `mvr = (3h)/(2pi) - (h)/(2pi)` |
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