When a certain metal was irradiated with light of frequency `3.2 xx 10^(16)s^(-1)` the photoelectrons emitted had twice the KE as did photoelectrons emitted when the same metal was irradiated with light of frequency`2.0 xx 10^(16)s^(-1)` . Calculate the thereshold frequency of the metal.

Explanation

Applying photoelectric equation,
`K.E. = hv - hv_(0) rArr (v-v_(0)) = (K.E)/(h)`
Given `K.E_(2) = 2K.E_(1)`
`v_(2) - v_(0) = (K.E_(2))/(h)` ......1
And `v_(1) - v_(0) = (K.E_(1))/(h)` ........2
Dividing equation (1) with (2),
`(v_(2)-v_(0))/(v_(1)-v_(0)) = (K.E_(2))/(K.E_(1)) = (2K.E_(1))/(K.E_(1)) = 2`
`rArr v_(2) -v_(0) = 2v_(1) - 2v_(0) rArr v_(0) = 2v_(1) - v_(2)`
`rArr v_(0) = 2(2.0 xx 10^(16)) - (3.2 xx 10^(16)) = 8.0 xx 10^(15)Hz`