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When a certain metal was irradiated with light of frequency `1.6xx10^(16)Hz`, the photoelectrons emitted had twice the kinetic energy as photoelectrons emitted when the same metal was irradiated with light of frequency `1.0xx10^(6)Hz`. Calculate the threshold frequency `(v_(0))` for the metal. |
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Answer» Correct Answer - `4xx10^(15)Hz` According to available information : `KE_(1)=h(v_(1)-v_(0))=h(1.6xx10^(16)-v_(0))` `KE_(2)=h(v_(2)-v_(0))=h(1.0xx10^(16)-v_(0))` Dividing eqn. (ii) by (i), we have `(KE_(2))/(KE_(1))=((1.0xx10^(16)-v_(0)))/((1.6xx10^(16)-v_(0)))` or `(KE_(1))/(2xxKE_(1))=((1.0xx10^(16)-v_(0)))/((1.6xx10^(16)-v_(0)))(KE_(2)=1//2KE_(1))` or` ((1.0xx10^(16)-v_(0)))/((1.6xx10^(16)-v_(0)))=(1)/(2) , 2.0xx10^(16)-2v_(0)=1.6xx10^(16)-v_(0)` or ` v_(0)=(2.0-1.6)xx10^(16)=0.4xx10^(16)=0.4xx10^(16)=4xx10^(15)`Hz. |
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