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When a metallic surface is illuminated with monochromatic light of wavelength `lambda`, the stopping potential is `5 V_0`. When the same surface is illuminated with light of wavelength `3lambda`, the stopping potential is `V_0`. Then the work function of the metallic surface is:A. `(hc)/(6 lambda)`B. `(hc)/(5lambda)`C. `(hc)/(4lambda)`D. `(2hc)/(4lambda)` |
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Answer» Correct Answer - A `(hc)/(lambda) =5 eV_(0)+phi` `(hc)/(3 lambda) =eV_(0) +phirArr (2hc)/(3lambda) = 4eV_(0) rArr phi=(hc)/(6 lambda) ` |
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