A) decrease
B) increase
C) remains unchanged
D) None of these
Correct Answer: A
Solution :
We know that,\[\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}\] \[\therefore \,\,\lambda \propto \frac{1}{\sqrt{E}}\](as \[h\] and \[m\] will be constants) So, when the kinetic energy of electron increases, the wavelength associated with it decreases.
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