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A gas absorbs a photon of `310nm` and emits three photons. If energy of emitted photons is in the raio `1:2:1` then select correct statement(s) :A. The wave length of emitted photons is in the ratio `1: 2:1`B. The wave length of emitted photons is in the ratio `2:1:2`C. The wave length of emitted photons is in the ratio `1240 nm`.D. The wavelength of least energatic photon is `1310 nm`. |
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Answer» Correct Answer - B::C `E_(1) = (12400)/(lambda_(1)) :. (E_(1))/(E_(2)) = (lambda_(1))/(lambda_(2)) = (1)/(2) rArr lambda_(1) = 2lambda_(2)` `E_(2) = (12400)/(lambda_(2))` `E = (12400)/(3100) = 2 ((12400)/(lambda_(1))) + (12400)/(lambda_(2))` `:. (12400)/(3100) = 2((12400)/(lambda_(1))) + ((12400)/(lambda_((1)/(2))))` `(12400)/(3100) = 4 ((12400)/(lambda_((1)/(2))))` `lambda_(1) = 4 xx 3100 Å` `= 12400Å` `= 1240 nm` |
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