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A certain metallic surface is illuminated with monochromatic light of wavelength lambda.The stopping potentail for photelectric current for this light is 3V_(0).If the same surface is illuminated with light of wavelength 2lambda,the stopping potential is V_(0).The threshold wavelength for this surface for photoelectric effect is |
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Answer» `4LAMBDA` `eV_(s)=E-phi` `THEREFORE V_(s)=(hc)/(lambda_(e))-(hc)/(lambda_(0)e)` `therefore 3V_(0)=(hc)/(lamdbae)-(hc)/(lambda_(o)e)`…..(1) and `therefore V_(o)=(hc)/(2lambdae)-(hc)/(lambda_(0)e)`…….(2) By multiplying EQ. (1) with (2) and then substracting from (1), `3V_(0)=(hc)/(lambdae)-(hc)/(lambda_(0)e)` `3V_(0)=(3hc)/(2lambdae)-(3hc)/(lambda_(0)e)` `(--+)/(0=-(hc)/(2lambdae)+(2hc)/(lambda_(0)e))` `0=-(hc)/(2lambdae)+(2hc)/(lambda_(0))` `therefore (2)/(lambda_(0))=(1)/(2lambda)` `therefore lambda_(0)=4lambda` |
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