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Question

(D) The electrostatic force provides the necessary centripetal force for the electron in a circular orbit:$F = \frac{k(Ze)(e)}{r^2} = \frac{mv^2}{r}$F

Vedclass · Accepted Answer

$2E = U$

Vedclass · Answer

(D) The electrostatic force provides the necessary centripetal force for the electron in a circular orbit:$F = \frac{k(Ze)(e)}{r^2} = \frac{mv^2}{r}$From this,the kinetic energy $(KE)$ is:$KE = \frac{1}{2} mv^2 = \frac{kZe^2}{2r}$The potential energy $(U)$ is given by:$U = -\frac{kZe^2}{r}$The total energy $(E)$ is the sum of kinetic and potential energy:$E = KE + U = \frac{kZe^2}{2r} - \frac{kZe^2}{r} = -\frac{kZe^2}{2r}$Comparing $E$ and $U$:$E = \frac{1}{2} \left( -\frac{kZe^2}{r} \right) = \frac{U}{2}$Therefore,$2E = U$.

An electron rotates in a circle around a nucleus having positive charge $Ze$. The correct relation between the total energy $(E)$ of the electron and its potential energy $(U)$ is:

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