Question

In hydrogen atom, the electron is at a distance of $4.768\mathring{A}$ from the nucleus. The angular momentum of the electron is:

• A. $\frac{3h}{2\pi }$
• B. $\frac{h}{2\pi }$
• C. $\frac{h}{\pi }$
• D. $\frac{3h}{\pi }$

Answer 1

The correct option is A $\frac{3h}{2\pi }$

Bohr radius of $n^{th}$ orbit($R_n$) = $R_0\frac{n^2}{Z}$

Where $R_0=0.529A$

For hydrogen, $Z=1$

GIven , $R_n=4.768\mathring{A}$

$⟹\frac{4.768}{0.529}=n^2$

$n=3$

The angular momentum (mvr) = $\frac{nh}{2\pi }⟹\frac{3h}{2\pi }$ of the electron.