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When a metallic surface is illuminated with radiation of wavelength `lambda`, the stopping potential is `V`. If the same surface is illuminated with radiation of wavelength `2 lambda` , the stopping potential is `(V)/(4)`. The threshold wavelength surface is :A. ` 3 lambda`B. `4 lambda`C. `5 lambda`D. `(5)/(2) lambda` |
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Answer» Correct Answer - A Einstein `P.E.` equation Case I : `eV = (hc)/(lambda) - (hc)/(lambda_(0))` ….(i) Case II : `eV//4 = (hc)/( 2lambda) - (hc)/(lambda_(0))` …(ii) From equations (i) and (ii) `rArr 4 = ((1)/(lambda))/((1)/(2 lambda)) - ((1)/(lambda_(0)))/((1)/(lambda))` On solving `lambda_(0) = 31` |
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