If the photons of frequency v are incident on the surfaces of metal A and B of threshold frequencies v/2 and v/3 respectively, then what will be the ratio of the maximum kinetic energy of electrons emitted from A to that of B ?
Answer & explanation
Correct answer: option 2
hv=hv/2 + Ka ⇒ Ka=hv/2
hv=hv/3 + Kb ⇒ Kb=2hv/3
divide the two equations to get 3:4.