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A metal surface having ` v_0` as threshode frequency is incident by light of frequency (v) then select the correct :A. ` u = sqrt(2 h(lambda_0 - lambda))`B. ` u = sqrt(2 h(lambda_0 - lambda))`C. ` u = sqrt(2 h(lambda_0 - lambda)/(m))`D. ` u = sqrt (2 h(lambda_0 - lambda)/(m))` |
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Answer» Correct Answer - A::B::D `hv = hv_0 + 1/2 mu^2` ` :. U = sqrt(2h (v-v_0)/m)0 = sqrt((2hc (lambda_0 - lambda)/(mx lambda_0 xx lambda))` ` hv = w + 1/2 mu^2` ` :. U = sqrt((2 (hv-w))/m)`. |
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