Maximum velocity of the photoelectron emitted by a metal is `1.8 xx 10^(6) ms^(-1)`. Take the value of specific charge of the electron is `1.8 xx 10^(11) C kg^(-1)`. Then the stopping potential in volt is

Options

A. 1
B. 3
C. 9
D. 6

Answer

C

Explanation

As `(1)/(2) mv_("max")^(2) = eV_(S)`
where `v_("max")` is the maximum velocity of the electron and `V_(S)` is the stopping potential
`V_(S) = (1)/(2) (m)/(e ) v_("max")^(2)`
Here, `(e )/(m)= 1.8 xx 10^(11)C kg^(-1), v_("max") = 1.8 xx 10^(6) ms^(-1)`
`: V_(S) = (1)/(2) xx (1)/(1.8 xx 10^(11)) xx (1.8 xx 10^(6))^(2) = 9V`