A) zero
B) less than that of a proton
C) more than that of a proton
D) equal to that of a proton
Correct Answer: C
Solution :
Given, \[\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mKE}}\] \[\Rightarrow \] \[KE=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\] As \[\lambda \] is same for both electron and proton So, \[KE\propto \frac{1}{m}\] Hence, kinetic energy will be maximum for particle with lesser mass, ie, electron.
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