A) decreases.
B) increases.
C) remains same.
D) upto 100 keV increases and then decreases.
Correct Answer: A
Solution :
Option [a] is correct. Explanation: \[\lambda =\frac{h}{\sqrt{2me\operatorname{V}}}\] So, as V increases, \[\lambda \] decreases.You need to login to perform this action.
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