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)+K.E. or v^(0)=upsilon-(K.E.)/(h)`
`upsilon^(0)=1.0 10^(15)s^(-1) , K.E. = 1.988xx10^(-19)J, h=6.626xx10^(-34)Js`
`upsilon^(0)=(1.0xx10^(15)s^(-1))-((1.988xx10^(-19)J))/((6.626xx10^(-34)Js))`
`=(1.0xx10^(15)s^(-1))-(0.3xx10^(15)s^(-1))=7.0xx10^(14)s^(-1)(Hz)`
Frequency of striking photon may be calculated as follows :
`upsilon=(c)/(lambda)=((3.0xx10^(8)ms^(-1)))/((600xx10^(-9)m))=5.0xx10^(14)s^(-1)(Hz)`
Since the frequency (upsilon) of the striking photon is less than the threshold frequency `(upsilon^(0))`, an electron will, not be emitted from the metal surface under the influence of the influence of the striking photon.