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When a certain metal was irradiated with light of frequency `3.2 xx 10^(16)s^(-1)` the photoelectrons emitted had three twice the KE as did photoelectrons emitted when the same metal was irradited with light of frequency`2.0 xx 10^(16)s^(-1)` .Calculate the thereshold frequency of the metal |
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Answer» `KE = hv - hv_(0)` we have `hv_(1) hv_(0)=2( hv_(2)-hv_(0))` `v_(0) - 2v_(2) - v_(1)` `= 2(2.0 xx 10^(16)s^(-1)) - (3.2 xx 10^(16)s^(-1))` `= 0.8 xx 10^(16)s^(-1))` |
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