Q. The angular momentum of an electron in a given orbit is J. Its kinetic energy will be

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A

$\frac{1}{2}\frac{J^{2}}{m r^{2}}$

B

$\frac{J v}{r}$

C

$\frac{J^{2}}{2 m}$

D

$\frac{J^{2}}{2 n}$

Solution:

$mvr=J$
$mv=\frac{J}{r}$
$\left(\right.mv\left(\left.\right)^{2}=\frac{J^{2}}{r^{2}}$
$mv^{2}=\frac{J^{2}}{m r^{2}}$
$K.E.=\frac{1}{2}mv^{2}=\frac{1}{2}\frac{J^{2}}{m r^{2}}$

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