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When a metallic surface is illuminated by light of frequency 8xx10^(14)Hz a photoelectron of energy 0.5eV is emitted. When the same surface is illuminated by light of frequency 12xx10^(14)Hz photoelectron of maximum energy 2 eV is emitted. The work function is : |
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Answer» 0.5 eV `12xx10^(14)h=phi_(0)+2` Then `(12)/(8)=(phi_(0)+2)/(phi_(0)+0*5)` SOLVING `phi_(0)=2*5eV` |
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