A) \[{{E}_{1}}>{{E}_{3}}>{{E}_{2}}\]
B) \[{{E}_{2}}>{{E}_{3}}>{{E}_{1}}\]
C) \[{{E}_{1}}>{{E}_{2}}>{{E}_{3}}\]
D) \[{{E}_{1}}={{E}_{2}}={{E}_{3}}\]
Correct Answer: A
Solution :
Since K.E. \[\frac{1}{2}m{{v}^{2}}\]and \[\lambda =\frac{h}{mv}.\] \[\therefore \] \[K.E.=\frac{1}{2}m.\frac{{{h}^{2}}}{{{m}^{2}}{{\lambda }^{2}}}=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}.\]As \[\lambda \]is the same. \[\therefore \] \[K.E.\propto \frac{1}{m}\]
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