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_1^2-v_2^2=(2h)/m(upsilon_1-upsilon_2)`
B. `v_1+v_2=[(2h)/m(upsilon_1+upsilon_2)]^(1//2)`
C. `v_1^2+v_2^2=(2h)/m(upsilon_1+upsilon_2)`
D. `v_1-v_2=[(2h)/m(upsilon_1-upsilon_2)]^(1//2)`

Answer

A

Explanation

According to Einstein.s equation,
Kinetic energy of emitted electron `= hupsilon` - (work function `phi_0` )
`:. 1/2 mv_(1)^(2) =hupsilon_(1) -phi_(0) " "...(i) and 1/2 mv_2^2 =hupsilon_(2) -phi_(0) " "...(ii)`
Subtracting (ii) from (i) we get
`1/2 m(v_1^2 -v_2^2)=h (upsilon_(1) -upsilon_2):. v_1^2-v_(2)^(2)=(2h)/(m) (upsilon_(1)-upsilon_(2))`