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 (3) → $3.15 × 10^{-34}\, Js$

Angular momentum of an electron in the n-th orbit of a hydrogen atom is given by Bohr's quantization condition:

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

For the third orbit, $n = 3$:

$L = 3 \frac{h}{2 \pi}$

Substitute Planck's constant $h = 6.626 \times 10^{-34} \, \text{Js}$:

$L = 3 \frac{6.626 \times 10^{-34}}{2 \pi}$

$L = \frac{19.878 \times 10^{-34}}{6.2832}$

$L \approx 3.163 \times 10^{-34} \, \text{Js}$

Answer: $L \approx 3.16 \times 10^{-34} \, \text{Js}$