Maximum velocity of photoelectrons emitted by a metal surface is `1.2xx10^(6)m//s`. Assuming the specific charge of the electrons to be `1.8xx10^(11)C//kg` the value of stopping potential in volt will be:

Options

A. 2
B. 3
C. 4
D. 6

Answer

C

Explanation

For the photoelectrons, maximum velocity `V_(m)=1.2xx10^(6)m//s` and `e/m=1.8xx10^(11)` C/kg
The maximum kinetic energy,
`K_("max")=1/2mv_("max")^(2)=eV_(0)`
`:.V_(0)=(1/2mv_(m)^(2))/(e)=1/2[(v_(m)^(2))/(e//m)]`
`:.V_(0)=1/2xx((1.2xx10^(6))^(2))/(1.8xx10^(11))`
`=(1.2xx1.2)/(1.8xx2)xx10=0.4xx10`
`V_(0)=4V`