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The threshold frequency for a certain me...

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.

Text Solution

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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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