Question
A dye absorbs a photon of wavelength $$\lambda $$ and reemits
the same energy into two photons of
wavelength $${\lambda _1}$$ and $${\lambda _2}$$ respectively. The
wavelength $$\lambda $$ is related with $${\lambda _1}$$ and $${\lambda _2}$$ as:
A.
$$\lambda = \frac{{{\lambda _1} + {\lambda _2}}}{{{\lambda _1}{\lambda _2}}}$$
B.
$$\lambda = \frac{{{\lambda _1}{\lambda _2}}}{{{\lambda _1} + {\lambda _2}}}$$
C.
$$\lambda = \frac{{\lambda _1^2\lambda _2^2}}{{{\lambda _1} + {\lambda _2}}}$$
D.
$$\lambda = \frac{{{\lambda _1}{\lambda _2}}}{{{{\left( {{\lambda _1} + {\lambda _2}} \right)}^2}}}$$
Answer :
$$\lambda = \frac{{{\lambda _1}{\lambda _2}}}{{{\lambda _1} + {\lambda _2}}}$$
Solution :
$$\eqalign{
& E = {E_1} + {E_2};\,\,\frac{{hc}}{\lambda } = \frac{{hc}}{{{\lambda _1}}} + \frac{{hc}}{{{\lambda _2}}} \cr
& \Rightarrow \frac{{hc}}{\lambda } = hc\left( {\frac{{{\lambda _2} + {\lambda _1}}}{{{\lambda _1}{\lambda _2}}}} \right);\,\lambda = \frac{{{\lambda _1}{\lambda _2}}}{{{\lambda _{`1}} + {\lambda _2}}} \cr} $$