Question

The threshold frequency `v_(0)` for a metal is `6xx10^(14) s^(-1)`. Calculate the kinetic energy of an electron emitted when radiation of frequency `v=1.1xx10^(15) s^(-1)` hits the metal.

Answer 1

`K.E.=1/2 m_(e)V^(2)=h(v-v_(0))`
`:. K.E.=(6.626xx10^(-34))(1.1xx10^(15)-6xx10^(14))`
`:. K.E.=(6.626xx10^(-34))(5xx10^(14))`
`=3.313xx10^(-19) J`