The angular momentum of an electron in the third orbit of a hydrogen atom is

(Given $h = 6.6 × 10^{-34} Js$)

$3.15 × 10^{-34} Js$

The correct answer is Option (1) → $3.15 × 10^{-34} Js$

The angular momentum of an electron in the nth orbit of a hydrogen atom is

$L = n \frac{h}{2\pi}$

For $n = 3$:

$L = 3 \frac{6.6 \times 10^{-34}}{2 \pi}$

$L = \frac{19.8 \times 10^{-34}}{6.283}$

$L \approx 3.15 \times 10^{-34} \, \text{Js}$

$L = 3.15 \times 10^{-34} \, \text{Js}$