A photon of 300 nm is absorbed by a gas and then, it re-emits two photons and attains the same initial energy level. One re-emitted photon has wavelength 500 nm. Calculate the wavelength of other photon re-emitted out

Explanation

`E=E_(1)+E_(2),(hC)/(lambda)=(hC)/(lambda_(1))+(hC)/(lambda_(2)),(1)/(lambda)=(1)/(lambda_(1))+(1)/(lambda_(2)),(1)/(300)=(1)/(700)+(1)/(lambda_(2))`
`7lambda_(2)=3lambda_(2)+2100 (or) lambda_(2)=525nm`.
The wave length of second photon is 525 nm