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If the electron of a hydrogen atom is present in the first orbit, the total energy of the electron is
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The correct answer is:
$\frac{-e^2}{2 r}$
Kinetic energy $=\frac{1}{2} m v^2$
Potential energy $=\frac{-e^2}{r}$
But $m v^2=\frac{e^2}{r}$
$\mathrm{KE}=\frac{1}{2} \frac{e^2}{r}$
Total energy $=\mathrm{KE}+\mathrm{PE}$
$=\frac{1}{2} \frac{e^2}{r}-\frac{e^2}{r}=\frac{e^2}{r}\left(\frac{1}{2}-1\right)=\frac{-e^2}{2 r}$
Potential energy $=\frac{-e^2}{r}$
But $m v^2=\frac{e^2}{r}$
$\mathrm{KE}=\frac{1}{2} \frac{e^2}{r}$
Total energy $=\mathrm{KE}+\mathrm{PE}$
$=\frac{1}{2} \frac{e^2}{r}-\frac{e^2}{r}=\frac{e^2}{r}\left(\frac{1}{2}-1\right)=\frac{-e^2}{2 r}$
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