Two identical photocathode receive light of frequencies `f_(1)` and `f_(2)`. If the maximum velocities of the photoelectrons (of mass m) coming out are respectively `v_(1)` and `v_(2)` then:

Options

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

Answer

A

Explanation

`hf = hf_0 + 1/2 mv^2`
Hence , `v_1^2 = (2hf_1)/(m) - (2hf_0)/(m) " and " v_2^2 = (2hf_2)/(m)-(2hf_0)/(m)`
`therefore v_1^(2)-v_(2)^(2) =(2h)/(m)[f_1-f_2]`