Question

Photoelectric emission is observed from a metallic surface for frequencies v1 and v2 of the incident light (v1 > v2). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio 1 : n, then the threshold frequency of the metallic surface is

Answer 1

Explanation:

let maximum kinetic energy =Ek

for frequency v1,maximum kinetic energy =Ek1

and for v2 maximum frequency =Ek2

according to the Einstein photoelectric effect

Ek1=hv1-hv0. where v0=threshold frequency.

and Ek2=hv2-hv0

given that, the ratio of the maximum kinetic energy =1:n

so, Ek1/Ek2=h(v1-v0)/h(v2-v0)

1/n=(v1-v0)/(v2-v0)

n(v1-v0)=v2-v0

v0(1-n)=(v2-nv1)

Therefore, v0=(v2-nv1)/(1-n).