The kinetic energy of an electron in a given orbit in a hydrogen atom is:

(symbols have their usual meanings)

$\frac{e^2}{8πε_0r}$

The correct answer is Option (3) → $\frac{e^2}{8πε_0r}$

For a hydrogen atom, the centripetal force is provided by the Coulomb force:

$\frac{mv^2}{r} = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{r^2}$

$mv^2 = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{r}$

Kinetic energy:

$K = \frac{1}{2} mv^2 = \frac{1}{8 \pi \epsilon_0} \frac{e^2}{r}$

Final Answer:

$K = \frac{e^2}{8 \pi \epsilon_0 r}$