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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 the light of wavelength `3lambda`, the stopping potential is `V_0`. Then, the work function of the metallic surface isA. `hc//6lambda`B. `hc//5lambda`C. `hc//4lambda`D. `2hc//4lambda` |
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Answer» Correct Answer - A `e(5V_0) = (hc)/(lambda) -W….(i)` `eV_0 = (hc)/(3lambda) -W …(ii)` Solving these equations, we get `W =(hc)/(6lambda).` |
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