Question

Calculate the angular orbital momentum of electron in first shell of hydrogen atom.

• A. $\frac{4.4}{\pi }\times {10}^{-34}{\text{kg-m}}^2\text{/s}$
• B. $\frac{1.1}{\pi }\times {10}^{-34}{\text{kg-m}}^2\text{/s}$
• C. $\frac{2.2}{\pi }\times {10}^{-34}{\text{kg-m}}^2\text{/s}$
• D. $\frac{3.3}{\pi }\times {10}^{-34}{\text{kg-m}}^2\text{/s}$

Answer 1

The correct option is D $\frac{3.3}{\pi }\times {10}^{-34}{\text{kg-m}}^2\text{/s}$

Given, $n=1$

We know that,

$L_n=mv{r}_n=\frac{nh}{2\pi }$

$\Rightarrow L_1=\frac{1\times 6.6\times {10}^{-34}}{2\pi }=\frac{3.3}{\pi }\times {10}^{-34}{\text{kg-m}}^2\text{/s}$

Hence, option $\left(D\right)$ is the correct answer.