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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 cutoff voltage for the photoelectric emission. |
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Answer» Solution :Here `V_(0)=3.3xx10^(14)HZ,h=6.63xx10^(-34)Js` `V=8.2xx10^(14)Hz` `V_(0)=?` `implies` Maximum kinetic energy of photon electron, `(1)/(2)mv_(MAX)^(2)=hv-phi_(0)` `therefore eV_(0)=hv-hv_(0)` `therefore V_(0)=(h)/(e )(v-v_(0))` `=(6.63xx10^(-34)XX(8.2xx10^(14)-3.3xx10^(14)))/(1.xx10^(-19))` `=20.30xx10^(-1)` `therefore V_(0)=2.03 V` |
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