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.

Thershold frequency, `v_(0)` is the minimum frequency which a photon m

1.	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.
Answer» 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.

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