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:

\frac{1}{2} \frac{k e^{2}}{r}

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- \frac{k e^{2}}{r}

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\frac{m k e^{2}}{r}

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- \frac{1}{2} \frac{k e^{2}}{r}

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Solution

The correct option is D $- \frac{1}{2} \frac{k e^{2}}{r}$
Total energy $= K . E . + P . E .$

K.E. $= - \frac{1}{2} m v^{2}$ = $\frac{1}{2} \frac{Z k e^{2}}{r}$ (cenrtipetal force)

P.E. $= - \frac{Z k e^{2}}{r}$

T.E. $= \frac{1}{2} \frac{Z k e^{2}}{r}$ + $- \frac{Z k e^{2}}{r}$

$= - \frac{1}{2} \frac{Z k e^{2}}{r}$

$= - \frac{1}{2} \frac{k e^{2}}{r}$ ....... ( $Z = 1$ for hydrogen)

Hence, option D is correct.

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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:

The correct option is D −12ke2rTotal energy =K.E.+P.E.K.E. =−12mv2 = 12Zke2r (cenrtipetal force)P.E. =−Zke2rT.E. =12Zke2r + −Zke2r =−12Zke2r =−12ke2r ....... (Z=1 for hydrogen)Hence, option D is correct.