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

`1/2 mv_(1)^(2)=hf_(1)-phi_(1)" "( :. E_(1)=hf_(1),E_(2)=hf_(2))`
`1/2mv_(2)^(2)=hf_(2)-phi_(0)`
`rArr h(f_(1)-f_(2))=1/2m(v_(1)^(2)-v_(2)^(2))`
`rArrv_(1)^(2)-v_(2)^(2)=(2h)/m(f_(1)-f_(2))`