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Photoelectric effect is obtained on metal surface for a light having frequencies f1 & f2 where f1 > f2. If ratio of maximum kinetic energy of emitted photo electrons is 1:K, so threshold frequency for metal surface is…………

(A) {(f1 – f2) / (K – 1)}

(B) {(Kf1 – f2) / (K – 1)}

(C) {(K f2 – f1) / (K – 1)}

(D) {(f2 – f1) / K}

1 Answer

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Best answer

The correct option is (B) {(Kf1 – f2) / (K – 1)}.

Explanation:

Kinetic energymax = EK = hf – hfo

∴ EK1 = h(f1 – fo)

and EK2 = h(f2 – fo)

Given: (EK1 / EK2) = (1/K) hence [{h(f1 – fo)} / {h(f2 – fo)}] = (1/K)

 ∴ Kf1 – Kfo = f2 – fo

 ∴ Kf1 – Kfo = f2 – fo

fo = {(Kf1 – f2) / (K – 1)}

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