A gas absorbs photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at

Options

A. 1035 nm
B. 325 nm
C. 743 nm
D. 518 nm

Answer

C

Explanation

We know that, `E=hv=hc//lambda," "E=E_(1)+E_(2)or (hc)/(lambda)=(hc)/(lambda_(1))+(hc)/(lambda_(2))`
`rArr" "(1)/(lambda)=(1)/(lambda_(1))+(1)/(lambda_(2)) rArr (1)/(355)=(1)/(680)+(1)/(lambda_(2)) therefore" "lambda=(355xx680)/(680-355)=742.796nm~~743nm`