Maximum velocity of the photoelectrons emitted by a metal surface `1.2xx10^(6)ms^(-1)`. Assuming the specific charge of the electron to be `1.8xx10^(11)C kg^(-1)`, the value of the stopping potential in volt will be

Options

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

Answer

C

Explanation

Specific charge of electron,
`e//m=1.8xx10^(11)C kg^(-1)`
Maximum kinetic energy of photoelectrons
`=(1)/(2)mv_(max)^(2)=eV_(s)`
where `V_(s)` is the stopping potential.
`therefore (1)/(2)mv_(max)^(2)=eV_(s)`
`rArr V_(s)=(mv_(max)^(2))/(2e)=((1.2xx10^(6))^(2))/(2xx1.8xx10^(11))=4V`.