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A stopping potential of 0.82 volt is required to stop the photoelectrons emitted from a metallic surface by light of wavelength 4000 Å. The stopping potential for wavelength 3000 Å will be : |
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Answer» Solution :`0.82eV=(12375)/(4000)-phi_(0)` `:. Phi_(0)=2*27eV` Also `eV_(S)=(12375)/(3000)-2.27` Then `V_(S)=1*85V`. |
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