Photoelectric work function of a metal is `1eV`, light of wavelength `lambda = 3000 Å` falls on it. The photoelectrons come out with velocity.

Options

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

Answer

D

Explanation

`(1)/(2)mv^_(max)^(2)=(hc)/(lamda)-phi_(0)`
`(1)/(2)mv_(max)^(2)=(12375eV)/(3000)-1eV`
`=3.125xx1.6xx10^(-19)`
`v_(max)=sqrt((2xx3.125xx1.6xx10^(-19))/(9.1xx10^(-31)))approx10^(6)ms^(-1)`