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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. `4lambda`B. `5lambda`C. `5lambda//2`D. `3lamda` |
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Answer» Correct Answer - d Case(i), `eV=(hc)/(lambda)-(hc)/(lambda_(0))..........(i)` Case (ii), `(eV)/4=(hc)/(2lambda)-(hc)/(lambda_(0))` or `eV=(4hc)/(2lambda)-(4hc)/(lambda_(0)).......(ii)` From (i) and (ii), `(hc)/(lambda)-(hc)/(lambda_(0))=(4hc)/(2lambda)-(4hc)/(lambda_(0))` `1/lambda-1/(lambda_(0))=2/lambda-4/(lambda_(0)) or 3/(lambda_(0)) =1/lambda or lambda_(0)=3lambda` |
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