1.

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)`

Answer» Correct Answer - A
1/2mv²=E
mv=√2mE
Therefore, 
E=h/mv
= h/√2mE
Therefore, Option A


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