InterviewSolution
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InterviewSolution
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Two identical photo-cathodes receive light of frequencies `v_1` and `v_2`. If the velocities of the photoelectrons (of mass m) coming out are `v_1` and `v_2` respectively, thenA. `v_1- v_2 = [((2h)/(m))(v_1-v_2))]^(1//2)`B. `v_1^2 - v_2^2=(2h)/(m) (v_1 - v_2)`C. `v_1- v_2 = [((2h)/(m))(v_1-v_2))]^(1//2)`D. `v_1^2 - v_2^2=(2h)/(m) (v_1 - v_2)` |
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Answer» Correct Answer - B `(1)/(2) mv_1^2=hv_1 - W` `(1)/(2) mv_2^2 = hv_2 -W` From these two equation, we can see that `V_1^2 -v_2^2= (2h)/(m) (v_1 - v_2)` |
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