InterviewSolution
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InterviewSolution
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In a photoemissive cell, with exciting wavelength `lambda`, the faster electron has speed v. If the exciting wavelength is changed to `3lambda//4`, the speed of the fastest electron will beA. `v(3//4)^(1//2)`B. `v(4//3)^(1//2)`C. Less than `v(4//3)^(1//2)`D. Greater than `v(4//3)^(1//2)` |
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Answer» Correct Answer - D |
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