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Calcualte the maximum kinetic energy of photoelectrons emitted where as light of freqency `2 xx 10^(16)Hz` is irradiated on a metal surface with threshold frequency `(v_(0))` equal to`8.68 xx10^(15)Hz`. |
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Answer» `hv= hv_(0)+KE` Threshold frequency `(v_(0)) = 8.68 xx 10^(15) Hz ( s^(-1))` Frequency of light `(v) = 2 xx 10^(16) Hz` KE `= h(v-v_(0))` `= 6.626 xx 10^(-34) ( 2 xx 10^(16) - 8.68 xx 10^(-15))` `KE = 7.5 xx 10^(-18)J` Maximum kinetic energy of photoelectrons `= 7.5 xx 10^(-18) J` |
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