A photon of frequency v is incident on a metal surface whose threshold frequency is `v_(0)`. The kinetic energy of the emitted photoelectrons will be

Options

A. hv
B. `hv_(0)`
C. `h(v+v_(0))`
D. `h(v-v_(0))`

Answer

D

Explanation

`hv=W_(0)+1/2 mU_("max")^(2) " " :.(hc)/(lambda)=W_(0)+E_("max")`
`:.W_(0)=(hc)/(lambda)-E_("max")`
`(hc)/(lambda)=(6.63xx10^(-34)xx3xx10^(8))/(4xx10^(-7))=(19.89xx10^(-19))/(4)J`
`=(19.89xx10^(-19))/(4xx1.6xx10^(-19))eV=3eV`
`W_(0)=3eV-2eV=1eV`