A) \[\,\frac{{{J}^{2}}}{2m{{r}^{2}}}\]
B) \[\,\frac{Jv}{r}\]
C) \[\,\frac{{{J}^{2}}}{2m}\]
D) \[\,\frac{{{J}^{2}}}{2\pi }\]
Correct Answer: A
Solution :
Angular momentum\[=mrv=J\text{ }\therefore v=\frac{{{J}^{2}}}{2m{{r}^{2}}}\] \[\text{K}\text{.E}\text{. of electron=}\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\left( \frac{J}{mr} \right)}^{2}}=\frac{{{J}^{2}}}{2m{{r}^{2}}}\]You need to login to perform this action.
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