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. `150 km s^(-1)`
B. `200 km s^(-1)`
C. `500 km s^(-1)`
D. `250 km s^(-1)`

Answer

C

Explanation

Here , stopping potential m `V_s = 0.71 V`
For electron , `e/m = 1.76 xx10^(11) C" kg "^(-1)`
The maximum kinetic energy of the photoelectron is
`K_(max)=1/2mv_(max)^2=eV_S :. v_(max)^2=2e/mV_s`
`v_(max) =sqrt(2 e/mV_s)=sqrt(2xx1.76 xx10^(11) xx 0.71)`
`= 5 xx 10^(5) ms^(-1) = 500 xx 10^(3) ms^(-1) = 500 km s ^(-1)`