A) 41%
B) 100%
C) 144%
D) 200%
Correct Answer: B
Solution :
If KE of a particle reduced to one fourth, then \[\lambda =\frac{h}{\sqrt{8\,m\,(KE)/4}}\] \[=\frac{2h}{\sqrt{2m\,KE}}\] Hence, increase in the de-Broglie wavelength is 100%.You need to login to perform this action.
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