When frequencies `f_(1)` and `f_(2)` are incident on two indentical photo sensitive surfaes ,maximum velocities of photo-electrons of mass m are `v_(1)` and `v_(2)` hence……

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

From Eiestien.s photoelectric equation,
`(1)/(2)mv_(max)^(2)=hf-phi`
`therefore hf=(1)/(2)mv_(max)^(2)+phi`
`therefore hf_(1)=(1)/(2)mv_(1)^(2)+phi [because v_(max)=v_(1)]`
and `hf_(2)=(1)/(2)mv_(2)^(2)+phi[because v_(max)=v_(2)]`
`therefore h(f_(1)-f_(2))=(1)/(2)m(v_(1)^(2)-v_(2)^(2))`
`therefore (2h)/(m)(f_(1)-f_(2))=v_(1)^(2)therefore v_(1)^(2)-v_(2)^(2)=(2h)/(m)=(2h)/(m)(f_(1)-f_(2))`