Define the term 'work function' of a metal. The threshold frequency of a metal is `f_(0)`. When the light of frequency `2f_(0)` is incident on the metal plane, the maximum velocity of electrons is `v_(1)`. When the frequency of the incident is increased to `5f_(0)`, the maximum velocity of electrons emitted is `v_(2)`. find the ratio of `v_(1)` to `v_(2)`.

Explanation

According to Einstein.s photoelectric equation, we know that
`h(v-v_(0))=(1)/(2)mv_(max)^(2)`
In present problem `v_(0)=f_(0)` and when `v=2f_(0),v_(max)=v_(1)`
`therefore h(2f_(0)-f_(0))=(1)/(2)mv_(1)^(2) or hf_(0)=(1)/(2)mv_(1)^(2)` . . (i)
Again, when `v=5f_(0),v_(max)=v_(2)`. hence we have
`h(5f_(0)-f_(0))=(1)/(2)mv_(2)^(2) or 4hf_(0)=(1)/(2)mv_(2)^(2)`
Dividing (i) by (ii), we get
`(1)/(4)=(v_(1)^(2))/(v_(2)^(2)) or (v_(1))/(v_(2))=sqrt((1)/(4))=(1)/(2)`.