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

Explanation

`KE = hv - hv_(0)` we have
`hv_(1) hv_(0)=2( hv_(2)-hv_(0))`
`v_(0) - 2v_(2) - v_(1)`
`= 2(2.0 xx 10^(16)s^(-1)) - (3.2 xx 10^(16)s^(-1))`
`= 0.8 xx 10^(16)s^(-1))`