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)-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

Using Einstein photoelectric equation, `E=W_(0)+K_("max")`
`hf_(1)=W_(0)+(1)/(2)mv_(1)^(2) ` . . . (i)
`hf_(2)=W_(0)+(1)/(2)mv_(2)^(2)` . . (ii)
`implies h(f_(1)-f_(2))=(1)/(2)m(v_(1)^(2)-v_(2)^(2))`
`implies (v_(1)^(2)-v_(2)^(2))=(2h)/(m)(f_(1)-f_(2))`