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

Here, `e/m=1.8xx10^(11)C//kg`
and `K_(max)=1/2mv_(max)^(2)=eV_(0)`
`:. V_(0)=1/2(mv_(max)^(2))/e`
or `V_(0)=(v_(max)^(2))/(2(e//m))=((1.2xx10^(6))^(2))/(3.6xx10^(11))=4`volt