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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