Two identical photo-cathodes receive light of frequencies `v_1` and `v_2`. If the velocities of the photoelectrons (of mass m) coming out are `v_1` and `v_2` respectively, then

Options

A. `v_1- v_2 = [((2h)/(m))(v_1-v_2))]^(1//2)`
B. `v_1^2 - v_2^2=(2h)/(m) (v_1 - v_2)`
C. `v_1- v_2 = [((2h)/(m))(v_1-v_2))]^(1//2)`
D. `v_1^2 - v_2^2=(2h)/(m) (v_1 - v_2)`

Answer

B

Explanation

`(1)/(2) mv_1^2=hv_1 - W`
`(1)/(2) mv_2^2 = hv_2 -W`
From these two equation, we can see that
`V_1^2 -v_2^2= (2h)/(m) (v_1 - v_2)`