When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9 volts . If `e//m` for the electrons is `1.8 xx 10^(11) C kg^(-1)` the maximum velocity of the ejected electrons is

Options

A. `6 xx 10^(5) ms^(-1)`
B. `8 xx 10^(5) ms^(-1)`
C. `1.8 xx 10^(6) ms^(-1)`
D. `1.8 xx 10^(5) ms^(-1)`

Answer

C

Explanation

`(1)/(2) mv_(max)^(2) = eV_(0) rArr v_(max) = sqrt(2 ((e )/(m))) V_(0)`
`= sqrt( 2 xx 1.8 xx 10^(11) xx 9) = 1.8 xx 10^(6) m//s`.