if e/m of electron is `1.76xx10^(11)C(kg)^(-1)` andn stopping potential is 0.71 V, then the maximum velocity of the photoelectron is

Options

A. `150kms^(-1)`
B. `200kms^(-1)`
C. `500kms^(-1)`
D. `250kms^(-1)`

Answer

C

Explanation

We can write, `(1)/(2)mv^(2)=eV`
`implies v=sqrt((2eV)/(m))=sqrt(2Vxx(e)/(m))`
or `v=sqrt(2xx0.71xx1.76xx10^(11))`
`=5xx10^(5)ms^(-1)=500kms^(-1)`