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Photoelectric emission is observed from a metallic surface for frequencies `v_(1) and v_(2)` of the incident light. If the maximum value of kinetic energies of the photoelectrons emitted in the two cases are in the ratio `n:1` then the threshold frequency of the metallic surface isA. `(v_(2)-v_(1))/(n-1)`B. `(nv_(2)-v_(1))/(n-1)`C. `(nv_(1)-v_(2))/(n-1)`D. `(v_(2)-v_(1))/(n)` |
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Answer» Correct Answer - b `(E_(1))/(E_(2))=(h(v_(1)-v_(0)))/(h(v_(2)-v_(0)))=n/1` On solving, `v_(0)=(nv_(2)-v_(1))/(n-1)` |
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