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

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(A) According to Bohr's first postulate,the centripetal force required for the circular motion of the electron is provided by the electrostatic force

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

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(A) According to Bohr's first postulate,the centripetal force required for the circular motion of the electron is provided by the electrostatic force of attraction between the nucleus and the electron.The electrostatic force is given by $F_e = \frac{1}{4 \pi \epsilon_0} \frac{(Ze)(e)}{r^2} = K \frac{Ze^2}{r^2}$.The centripetal force is given by $F_c = \frac{mv^2}{r}$.Equating the two forces: $\frac{mv^2}{r} = K \frac{Ze^2}{r^2}$.Multiplying both sides by $\frac{r}{2}$,we get: $\frac{1}{2} mv^2 = \frac{1}{2} K \frac{Ze^2}{r}$.Since kinetic energy $E_k = \frac{1}{2} mv^2$,we have $E_k = \frac{Ze^2}{2r} K$.

If $m$ is the mass of an electron,$v$ is its velocity,$r$ is the radius of a stationary circular orbit around a nucleus with charge $Ze$,then from Bohr's first postulate,the kinetic energy of the electron is (where $K = 1 / 4 \pi \epsilon_0$):

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