Light of wavelength `3000Å` is incident on a metal surface whose work function is 1 eV. The maximum velocity of emitted photoelectron will be

Options

A. `10 ms^(-1)`
B. `10^(3) ms^(-1)`
C. `10^(4) ms^(-1)`
D. `10^(6) ms^(-1)`

Answer

D

Explanation

`(1)/(2)mv^(2)=(hc)/(lambda)-phi_(0)`
`=(6.63xx10^(-34)xx3xx10^(8))/((3xx10^(-7))xx1.6xx10^(-19))-1`
`=4.14-1=3.14 eV`
or `v sqrt((2xx3.14xx1.6xx10^(-19))/(99.1xx10^(-31)))=10^(6) ms^(-1)`