Question

9.
The threshold frequencies of two photosensitive surfaces are v1 and v2,
respectively. What is the ratio of the velocities of the photoelectrons
emitted from these surfaces when light of frequency v is incident on them
and photoemission occurs ?​

Answer 1

The ratio of the velocities of the photo electrons emitted from the two surface is √((v-v1)/(v-v2)).

Using Einstein's photoelectric emission equation -

K.E = hv - work function of the surface

For the surface 1 -

1/2 mv'² = hv - hv1 [v1 is the threshold frequency of surface 1]

=> v'² = 2h(v-v1)/m

For the surface 2 -

1/2 mv''² = hv - hv2 [v2 is the threshold frequency of surface2]

=> v''² = 2h(v-v2)/m

Ratio of the velocity of the photoelectrons emitted -

v'/v'' = √(2h(v-v1)/m) / √(2h(v-v2)/m)

=> v'/v'' = √((v-v1)/(v-v2))