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)(f_1-f_2)]^((1)/(2))`
B. `v_1^2-v_2^2=(2h)/(m)(f_1-f_2)`
C. `v_1+v_2=[(2h)/(m)(f_1-f_2)]^((1)/(2))`
D. `v_1^2+v_2^2=(2h)/(m)(f_1-f_2)`

Answer

B

Explanation

`hv_1-hv_0=(1)/(2)mv_1^2`
`hv_2-hv_0=(1)/(2)mv_2^2`
`h(v_1-v_2)=(1)/(2)m(v_1^2-v_2^2)` `[v_1=f_1` and `v_2=f_2`]
`v_1^2-v_2^2=(2h)/(m)(f_1-f_2)`