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The threshold frequency for a certain metal is 3.3xx10^(14)Hz. If light of frequency 8.2xx10^(14)Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission. |
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Answer» Solution :Here `v_(0)=3.3xx10^(14)Hz and V=8.2xx10^(14)Hz` USING Einstein.s RELATION `h(v-v_(0))=eV_(0)`, we have Cut-off voltage `V_(0)=(h)/(e)(v-v_(0))=(6.63xx10^(-34))/(1.6xx10^(-19))(8.2xx10^(14)-3.3xx10^(14))` `=(6.63xx10^(-34)xx4.9xx10^(14))/(1.6xx10^(-19))=2.0V`. |
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