If m is the mass of an electron,v is its velo | vedclass

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(A) In the $C.G.S.$ system,the electrostatic force of attraction between the nucleus (charge $Ze$) and the electron (charge $e$) provides the necessar

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$\frac{1}{2} \frac{Ze^2}{r}$

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(A) In the $C.G.S.$ system,the electrostatic force of attraction between the nucleus (charge $Ze$) and the electron (charge $e$) provides the necessary centripetal force for circular motion.The electrostatic force is given by $F_e = \frac{(Ze)(e)}{r^2} = \frac{Ze^2}{r^2}$.The centripetal force required for circular motion is $F_c = \frac{mv^2}{r}$.Equating the two forces: $\frac{mv^2}{r} = \frac{Ze^2}{r^2}$.Multiplying both sides by $r$,we get $mv^2 = \frac{Ze^2}{r}$.The kinetic energy $K$ is defined as $K = \frac{1}{2}mv^2$.Substituting the expression for $mv^2$: $K = \frac{1}{2} \left( \frac{Ze^2}{r} \right) = \frac{Ze^2}{2r}$.

If $m$ is the mass of an electron,$v$ is its velocity,and $r$ is the radius of a stationary circular orbit around a nucleus with charge $Ze$,then from Bohr's postulate,the kinetic energy $K = \frac{1}{2}mv^2$ of the electron in the $C.G.S.$ system is equal to:

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