Question

A gas absorbs a photon of 355 nm and emit at two wavelengths. If one of the emission is at 680
A. 325 nm
B. 743 nm
C. 518 nm
D. 1035 nm

Answer 1

Correct Answer - B
As energies are additive, `E = E_(1) + E_(2)`
or `(hc)/(lamda) = (hc)/(lamda_(1)) + (hc)/(lamda_(2)) or (1)/(lamda) = (1)/(lamda_(1)) + (1)/(lamda_(2))`
But `lamda = 355 nm, lamda_(1) = 680nm,`
`:. (1)/(355) = (1)/(680) + (1)/(lamda_(2))`
or `(1)/(lamda_(2)) = (1)/(355) - (1)/(680) = (325)/(355 xx 680)`
or `lamda_(2) = (355 xx 680)/(325) = 745nm`