Photoelectric emission is observed from a metal surface for frequencies `v_(1)` and `v_(2)` incidents radiations `(v_(1) gt v_(2))`. If the maximum kinetic energy of the two electrons are in the ratio `1:2`, then how is threshold frequency `(v^(@))` experssed.

Answer

`2 v_(1)-v_(2)`

Explanation

`hv_(1)=hv_(0)+(K.E.)_(1) or h(v_(1)-v_(0))=(KE)_(1)`
`hv_(2)=hv_(0)+(K.E.)_(2) or h (v_(2)-v_(0))=(KE)_(2)`
`(h(v_(1)-v_(0)))/(h(v_(2)-v_(0)))=((KE)_(1))/(KE)_(2)=(1)/(2) or 2 (v_(1)-v_(0))=(v_(2)-v_(0))`
`2v_(0)-v_(0)=2v_(1)-v_(2) or v_(0)=2v_(1)-v_(2)`