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The threshold frequcney for a certain metal is 3.3xx10^(14) Hz. If light of frequency 8.2xx10^(14) Hz is incidenton the metal, predict the cut-off voltage for the photoelectric emission. |
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Answer» Solution :`upsilon_(0)=3.3xx10^(14)HZ, upsilon=8.2xx10^(14)Hz` `(1)/(2)mv_("MAX")^(2)=h(upsilon-upsilon_(0))`, But `eV_(0)=(1)/(2)mv_("max")^(2)` `eV_(0)=h(upsilon-upsilon_(0))=6.6xx10^(-34)[8.2xx10^(14)-3.3xx10^(14)]` `V_(0)=(6.6xx10^(34)xx4.9xx10^(14))/(1.6xx10^(-19))=20.2xx10^(-1)V=2V` |
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