Light of frequency `7.21xx10^(14)Hz` is incident on a metal surface. Electrons with a maximum speed of `6.0xx10^(5)ms^(-1)` are ejected from the surface. What is the threshold frequency for photoemission of electrons? `h=6.63xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg`.

Explanation

Frequency of the incident photon, v=488 nm =`488xx10^(-9)` m
Maximum speed of the electrons , `v=6.0xx10^5` m/s
Planck's constant , `h=6.626xx10^(-34) Js`
Mass of an electron , m`=9.1xx10^(-31) kg`
For threshold frequency `v_0`, the relation for kinetic energy is written as
`1/2 mv^2=h(v-v_0)`
`v_0=v-(mv^2)/(2h)`
`=7.21xx10^14 -((9.1xx10^(-31))xx(6xx10^5)^2)/(2xx(6.626xx10^(-34))`
`=7.21xx10^14-2.472xx10^14`
`=4.738xx10^14 Hz`
Therefore, the threshold frequency for the photoemission of electrons is `4.738xx10^14 Hz`