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Q.

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

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a

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

b

$\begin{aligned} \frac{1}{2} \frac{J^{2}}{m r^{2}} \end{aligned}$

c

$\begin{aligned} \frac{J v}{r} \end{aligned}$

d

$\frac{J^{2}}{2 π}$

answer is A.

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Detailed Solution

It is known that;

$\begin{array}{l} m v r = J \end{array}$

$\Rightarrow P = \frac{J}{r}; K E = \frac{P^{2}}{2 m} K E = \frac{J^{2}}{2 m r^{2}}$

Correct option is A.

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