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The critical frequency for emitting photoellectrons from a metal surface is ` 5 xx 10^(14) sec ^(-1)`. What should be the frequency fo radiation to produce photoelectons having twice the kinetic energy of those produced by the radiation fo frequency ` 10^(15) sec ^(-1)` ? |
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Answer» Given critical frequency `= 5 xx 10^(14) sec^(-1)` When surface is exposed to frequency of ` 10^(15) sec^(-1)` Energy fo photon ` = hv_0 + K.E.` ` h xx 10^(15) = h xx 5 xx 10^(14) + K.E` ` therefore K.E. = 10 xx h xx 10^(14)` `= 5 xx h xx 10^(14)` ` = 5 xx h xx 10^(14)` Now for (II) case . if K.E. becomes twice of this then New ` K.E, = 2 xx 5 xx h xx 10^(14)` The energy of photon ` = hv_0 + K. E` ` hv= 5xx 10^(14) xx h + 2 xx 5 xx h xx 10^(14)` ` therefore v= 15 xx 10 ^(14) sec^(-1)` . |
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