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Photons of energy 1 eV and 2.5 eV successively illuminate a metal whose work function is 0.5 eV, The ratio of maximum speed of emitted electron is (A) 1 : 2 (B) 2 : 1 (C) 3 : 1 (D) 1: 3 |
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Answer» udent,◆ ANSWER -(A) 1 : 2 ● EXPLANATION -For 1ST photon,KE1 = hν1 - Φmv1²/2 = hν1 - Φmv1²/2 = 1 - 0.5v1² = 0.5 × 2/mv1² = 1/mv1 = 1/√mFor 2nd photon,KE2 = hν1 - Φmv2²/2 = hν1 - Φmv2²/2 = 2.5 - 0.5v2² = 2 × 2/mv2² = 4/mv2 = 2/√mRatio of MAXIMUM speeds is -v1/v2 = (1/√m) / (2/√m)v1/v2 = 1/2Therefore, ratio of maximum speeds is 1 : 2.Hope this helps you... |
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