Work function of metal A,B and C are 4.5 eV 4.3 eV and 3.8eV respectively .If radiation of 3000 Å is incident on it then …..(`h=6.62xx10^(-34)js,c=3xx10^(8)m//s,1 eV1.6xx10^(-19)C)`

Options

A. photo-electron will be emitted from metal surface A.
B. photo-electron will be emitted from metal B
C. Photo-electron will be emitted from metal C.
D. Photo-electron will be emitted from metal A.B and C

Answer

C

Explanation

If work function of metal is less than energy of incident radiation photo-electron will be emitted.
Energy of incident radiation ,
`E=(hc)/(lambda)J=(hc)/(lambdae)eV`
`=(6.62xx10^(-34)xx3xx10^(8))/(3xx10^(-7)xx1.6xx10^(-19))`
`therefore E=4.14 eV`
This energy is more than work function of metal C but less than function of metal A and B hence answer (C ) is correct.