Question

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) )

# 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)?

Solution