In the study of a photoelectric effect the graph between the stopping potential V and frequency v of the incident radiation on two different metals P and Q is shown in fig.
(i) which one of two metals have higher threshold frequency
(ii) Determine the work function of the metal which has greater value
(iii) Find the maximum kinetic energy of electron emitted by light of frequency `8xx10^(14) Hz` for this metal.

Explanation

(i) Threshold frequency for P, `v_(oP)=3xx10^(14)Hz`
for Q is `v_(oQ)=6xx10^(14)Hz so v_(oQ)gtv_(oP)`
(ii) Work function of metal `phi_0=hv_0 ,e.i., phipropv_0`.
As `v_(oQ)gtv_(oP)`, so work function of metal Q is greather than that of metal P.
`phi_Q=hv_(oQ)=((6.63xx10^(-34))xx(6xx10^(14)))/(1.6xx10^(-19))eV`
`=2.48eV`
(iii) Max KE, `K_(max)=hv-phi_Q`
`(6.63xx10^(-34)xx8xx10^(14))/(1.6xx10^(-19))-2.48`
`=3.315-2.48=0.835eV`