According to Bohr’s theory, the expressions for the kinetic and potential energy of an electron revolving in an orbit is given respectively by

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$+ \frac{e^{2}}{8 π ε_{o} r} and - \frac{e^{2}}{4 π ε_{o} r}$

$+ \frac{8 π ε_{o} e^{2}}{r} and - \frac{4 π ε_{o} e^{2}}{r}$

$- \frac{e^{2}}{{8πε}_{o} r} and - \frac{e^{2}}{{4πε}_{o} r}$

$+ \frac{e^{2}}{8 π ε_{o} r} and + \frac{e^{2}}{{4πε}_{o} r}$

answer is A.

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Detailed Solution

$P.E. = - \frac{{ke}^{2}}{r} = - \frac{e^{2}}{4 π ε_{0} r}$

$K.E. = - \frac{1}{2} (P . E,) = \frac{e^{2}}{8 π ε_{0} r}$

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