Thershold frequency, `v_(0)` is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency `1.0xx10^(15)s^(-1)` was allowed to hit a metal surface, an electron having `1.988x10^(-19)J` of kinetic energy was emitted. Calculated the threshold frequency of this metal.
equal to 600nm hits the metal surface.

Explanation

We know that `hv=hv_(0)+KE`
`hv-KE=hv_(0)=(6.626xx10^(-34)Jsxx1xx10^(15)s^(-1))-1998xx10^(-19)J` ltbr. `hv_(0)=6.626xx10^(-19)-1.988xx10^(-19)J`
`hv_(0)=4.638xx10^(-19)J`
`v_(0)=(4.638xx10^(-19)J)/(6.626xx10^(-34)Js)=0.699xx10^(15s^(-1)`
when `lambda=600nm=600xx10^(-19)m`
`v=(c)/(lambda)=(30xx10^(8)ms^(-1))/(6.0xx10^(-7)m)=0.5xx10^(15)s^(-1)`
Thus `vltv_(0)` hence , no electron will be emitted.