A dye absorbs a photon of wavelength `lamda` and re-emits the same energy into two photns of wavelength `lamda_(1)` and `lamda_(2)` respectively. The wavelength `lamda` is related to `lamda_(1)` and `lamda_(2)` as :

The energy of a photon of wavelength lamda is given by

Momentum of a photon of wavelength lamda is

Energy of a photon = E , kinetic energy of a proton is also the same as energy of the photon. Corresponding wavelength of the photon and de Broglie wavelength of the proton are lamda_(1) and lamda_(2) , respectively. What is the relation between E and the ratio of lamda_(1) and lamda_(2) ?

K_(1) and K_(2) are the maximum kinetic energies of the photo electrons emitted when light of wave lengths lamda_(1) , and lamda_(2) , respectively are incident on a metallic surface. If lamda_(1)=3lamda_(2) then