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)` or `(v-v_(0))= (K. E)/(h)`
Given `K.E_(2) = 2 K. E_(1)`
`v_(2) - v_(0) = (K. E_(2) )/(h) --------1` & `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`
or `v_2 - v_0 = 2v_1` or `v_(0) = 2v_1 = v_2`
`= 2 (2.0 xx 10^(16) ) - (3.2 xx 10^(16) ) rArr 8.0 xx 10^(15)` Hz