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

`hv = hv_0 + 1/2 m u^2`
or `hv_0 = hv - 1/2 m u^2`
`= 6.63 xx 10^(-34) xx 1.0 xx 10^(15) - 1.988 xx 10^(-19)`
`= 6.63 xx 10^(-19) - 1.988 xx 10^(-19)`
`= 4.642 xx 10^(-19) J`
`:. v_0 = (4.642 xx 10^(-19) J)/(6.63 xx 10^(-34) Js) = 6.988 xx 10^(14) s^(-1)` Now frequency of photon of wavelength 600 nm = `600 xx 10^(-9) m`
`v = (3.0 xx 10^8 ms^(-1))/(600 xx 10^(-9) m) = 5.0 xx 10^(14) s^(-1)`
Since frequency of photon is less than `v_0` , electron will not be emitted.