Maximum velocity of the photoelectron emitted by a metal is `1.8 xx 10^(6) ms^(-1)`. Take the value of specific charge of the electron is `1.8 xx 10^(11) C kg^(-1)`. Then the stopping potential in volt is

Options

A. 1
B. 8
C. 9
D. 6

Answer

C

Explanation

Given `v=1.8xx10^(6)m//s, (e)/(m)=1.8xx10^(11) C//kg`
We have `eV_(0)=(1)/(2) mv^(2)`
`V_(0)(e)/(m)=(v^(2))/(2)`
`rArr V_(0)xx1.8xx10^(11)=(1.8xx1.8xx(10^(6))^(2))/(1)`
`V_(0)=9V`.