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The de - Broglie wavelength `lambda` associated with an electron having kinetic energy `E` is given by the expressionA. `(h)/(sqrt(2mE))`B. `(2h)/(mE)`C. `2mhE`D. `(2sqrt(2mE))/(h)` |
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Answer» Correct Answer - A 1/2mv²=E mv=√2mE Therefore, E=h/mv = h/√2mE Therefore, Option A
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