Two identical photocathode receive light of frequency `f_(1)` and `f_(2)` if the velocity of photoelectrons (of mass m) coming out are `v_(1)` and `v_(2)` respectively 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)[(2h)/(m)(f_(1)- f_(2))]^(1//2)`

Answer

A

Explanation

`hf_(1)=phi_(0)+1/2 mv_(1)^(2), hf_(2)=phi_(0)+1/2 mv_(2)^(2)implies v_(1)^(2)-v_(2)^(2)=(2h)/m(f_(1)-f_(2))`