Threshold 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 `1.0 xx 10^(15) s^(-1)` was allowed to hit a metal surface, an electron having `1.988 xx 10^(-19)J` of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.

Explanation

We know that hv =`hv_(0)+ke` or
`hv-ke =hv_(0)=-1.988xx10^(-19)`
`hv_(0)=6.626 xx10^(-19)-1.988xx10^(-19)j`
`hv_(0)=4.638xx10^(-19)j`
`v_(0)=(4.638 xx10j)/(6.626xx10^(-34)=0.699xx10^(15)s^(-1)`
when `lambda=600 nm =600 xx10^(-19) m`
thus `v lt v_(0)` hence no electro will be emitted