The maximum kinetic energy of photoelectrons ejected from a metal, when it is irradiated with radiation of frequency `2xx10^(14)s^(-1)` is `6.63xx10^(-20)`J. the threshold frequency of the metal is:

Options

A. `2 xx 10^(14) s^(-1)`
B. `3 xx 10^(14) s^(-1)`
C. `2 xx 10^(-14) s^(-1)`
D. `1 xx 10^(14) s^(-1)`

Answer

D

Explanation

`1/2 mv^2 = h (v - v_0)`
`6.63 xx 10^(-20) kg m^2 s^(-2)`
`= 6.63 xx 10^(-34) kg m^2 s^(-1) (2 xx 10^(14) - v_0)`
`(6.63 xx 10^(20))/(6.63 xx 10^(-34)) = 2 xx 10^(14) - v_0`
`1 xx 10^(14) - 2 xx 10^(14) = -v_0 "or " v_0 = 1 xx 10^(14) s^(-1)`.