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)(v_(1)-v_(2))`
B. `v_(1)+v_(2)[=(2h)/(m)(v_(1)+v_(2))]^(1//2)`
C. `v_(1)^(2)+v_(2)^(2)=(2h)/(m)(v_(1)+v_(2))`
D. `v_(1)-v_(2)[=(2h)/(m)(v_(1)+v_(2))]^(1//2)`

Answer

A

Explanation

(a) : According to Einstein's equation, kineticenergy of emitted electron=hv-(work function `phi_(0))`
`:.(1)/(2)mv_(1)^(2)=hv_(1)-phi and (1)/(2)mv_(2)^(2)=hv_(2)-phi`
`:.(1)/(2)m(v_(1)^(2)-v_(2)^(2))=(v_(1)-v_(2))`
or `v_(1)^(2)-v_(2)^(2)=(2h)/(m)(v_(1)-v_(2))`