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Photoelectric emission is observed from a metallic surface for frequenciesv_(1) and v_(2)of the incident light (v_(1)gtv_(2)).If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio 1:n,then the threshold frequency of the metallic surface is |
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Answer» `(upsilon_2 - upsilon_1)/(K-1)` `(hupsilon_1 - hupsilon_0)/(H upsilon_2 - hupsilon_0) = 1/KimpliesK (upsilon_1 - upsilon_2) = upsilon_2 - upsilon_0` `impliesK upsilon_1 - K upsilon_0 = upsilon_2 - upsilon_0` `impliesupsilon_0 (1-K) = upsilon_2 - K upsilon_1 impliesupsilon_0 = (upsilon_2 - K upsilon_1)/(1-K)` |
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