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Q.

If the electron of a hydrogen atom is present in the first orbit, the total energy of the electron is:

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a

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

b

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

c

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

d

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

answer is C.

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Detailed Solution

Total energy = PE + KE

$TE = - \frac{{Ze}^{2}}{r} + \frac{{Ze}^{2}}{2 r} = - \frac{{Ze}^{2}}{2 r} For Hydrogen Z = 1 TE = - \frac{e^{2}}{2 r}$

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