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A certain dye absorbs `4530A^(@)` and fluoresence at `5080A^(@)` these being wavelength of maximum absorption that under given condition `47%` of the absorbed energy is emitted. Calculate the ratio of the no of quanta emitted to the number absorbed. |
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Answer» Correct Answer - `0.527` E of light absorbed in one photon `=(hc)/(lambda_("absorbed"))` Let `n_(1)` photons are absorbed , therefore, Total energy absorbed `=(n_(1)hc)/(lambda_("absorbed"))` Now, E of light re-emitted out in one photon `=(hc)/(lambda_("emitted"))` Let `n_(2)` photons are re-emitted then Total energy re-emitted out `=n_(2)xx(hc)/(lambda_(emitted))` As given `E_("absorbed")xx(47)/(100)=E_("re-emitted out")` `(hc)/(lambda_("absorbed"))xxn_(1)xx(47)/(100)=n_(2)xx(hc)/(lambda_("emitted"))` `therefore (n_(1))/(n_(2))=(47)/(100)xx(lambda_("emitted"))/(lambda_("absorbed"))=(47)/(100)xx(5080)/(4530)` `therefore (n_(1))/(n_(2))=0.527` |
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