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If $m$ and $e$ are the mass and charge of the revolving electron in the orbit of radius $r$ for hydrogen atom, the total energy of the revolving electron will be:
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Verified Answer
The correct answer is:
$-\frac{1}{2} \frac{e^2}{r}$
$-\frac{1}{2} \frac{e^2}{r}$
Total energy of a revolving electron is the sum of its kinetic and potential energy. Total energy $=$ K.E. $+$ P. E.
$$
=\frac{e^2}{2 r}+\left(-\frac{e^2}{r}\right)=-\frac{e^2}{2 r}
$$
$$
=\frac{e^2}{2 r}+\left(-\frac{e^2}{r}\right)=-\frac{e^2}{2 r}
$$
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