Question

A dye absorbs a photon of wavelength 2 and re-emits the same energy into two photons of wavelength 11 and 12 respectively. The wavelength 2 is related with 14 and 12 as: (A) 2=2+2 (B) 2 = M (C) 2 - 2 (D) 2= hh hth hth (2+2)?

JEE/Engineering Exams

Chemistry

Solution

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A dye absects photon of A wamelery th ( t=frac{h c}{lambda} ) (plancte's equation) hye emits same energy in 2 photom ( lambda_{1} ) and ( lambda_{2} ) ( E_{1}=frac{h c}{lambda_{1}} ) and ( f_{2}=frac{h c}{lambda_{2}} ) Since energy remains conserved ( therefore E=E_{1}+E_{2} ) [ begin{array}{l} frac{h c}{lambda}=frac{h c}{lambda_{1}}+frac{h c}{lambda_{2}} frac{1}{lambda}=frac{1}{lambda_{1}}+frac{1}{lambda_{2}} frac{1}{lambda}=frac{lambda_{1} f_{2}}{lambda_{1} lambda_{2}} alpha=lambda_{1} lambda_{2} end{array} ] ( = ) option ( (B) )