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Two identical metal plates show photoelectric effect by a light of wavelength `lambda_(A)` falls on plate `A` and `lambda_(B)` on plate `B(lambda_(A) = 2 lambda_(B))`. The maximum kinetic energy isA. `2K_(A) = K_(B)`B. `K_(A) lt K_(B)//2`C. `K_(A) = 2K_(B)`D. `K_(A) gt K_(B)//2` |
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Answer» Correct Answer - B `(hc)/(lambda_(A)) =phi+K_(A)` ...(i) `(hc)/(lambda_(B)) =phi+K_(B)` …(ii) Let `lambda_(A) = lambda, lambda_(B) =lambda//2` `(K_(B))/(K_(A)) =((hc)/(lambda//2)-phi)/((hc)/(lambda)-phi) =((hc)/(lambda)-phi)/((hc)/(lambda)-phi) +((hc)/(lambda))/((hc)/(lambda)-phi)` `=1 +(gt1) = gt 2` `K_(B)gt2 K_(A)` `K_(A)lt(K_(B))/(2)` |
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