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When a centimeter thick surface is illuminated with light of wavelength `lamda`, the stopping potential is V. When the same surface is illuminated by light of wavelength `2lamda`, the stopping potential is `(V)/(3)`. Threshold wavelength for the metallic surface isA. `(4 lambda)/3`B. `4 lambda`C. `6 lambda`D. `(8 lambda)/3` |
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Answer» Correct Answer - B `(hC)/(lamda) =phi +eV`....(i) `(hC)/(2 lamda) =phi+ (eV)/(3)`....(ii) 3. II-I `rArr ((3)/(2)-1)(hc)/(lamda)=2 phi rArr phi =(hc)/(4 lamda)` `:. lamda_(th) = 4 lamda` |
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