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When metal surface with 2.1 eV work function is irridated with photon of 6.0 eV energy.Value of stopping potential will be…. |
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Answer» `+8.1` `therefore eV_(0)hv-phi_(0)[because (1)/(2)mv_(max)^(2)=eV_(0)`] `therefore eV_(0)=(6.0-2.1)eV` `therefore eV_(0)=3.9 eV` Stopping POTENTIAL is NEGATIVE VOLTAGE. `therefore V_(0)=-3.9V` |
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