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

According to Einstein's photoelectric equation
`hg=hg_(0)+1/2 mv^(2) [f=n " or " v]`
`:.1/2mv^(2)=hf-hf_(0)=h(f-f_(0))`
`:.v^(2)=(2h)/(m)(f-f_(0))`
`:.v_(1)^(2)=(2h)/(m)(f_(1)-f_(0))` and `v_(2)^(2)=(2h)/(m)(f_(2)-f_(0))`
`:.v_(1)^(2)-v_(2)^(2)=(2h)/(m)(f_(1)-f_(0)-f_(2)+f_(0))`
`=(2h)/(m)(f_(1)-f_(2))`