Question

Threshold frequency, `v_(0)` is the minimum frequency which a photon must posses 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 600nm hits the metal surface.

Answer 1

`hv = hv_(0) + K.E. or hv_(0) = hv - K.E.`
or `v_(0) = v - (K.E.)/(h) = 1.0 xx 10^(15) s^(-1) - (1.988 xx 10^(-19) J)/(6.626 xx 10^(-34) Js)`
`= (1.0 xx 10^(15) - 0.30 xx 10^(15)) s^(-1) = 0.97 xx 10^(15) s^(-1)`
`9.7 xx 10^(14) s^(-1)`
When `lamda = 600nm, " i.e., " 600 xx 10^(-9) m or 6.0 xx 10^(-7) m`
`v = (c)/(lamda) = (3.0 xx 10^(8) ms^(-1))/(6.0 xx 10^(-7) m) = 0.5 xx 10^(15) s^(-1) = 5 xx 10^(14) s^(-1)`
Thus, `v lt v_(0)` Hence, no electron will be emitted