Question

Magnitude of angular magnetic moment associated with a revolving electron in a hydrogen atom will be

• A. $\mu_{l}=\frac{e}{2 m_{e}}$
• B. $\mu_{l}=\frac{-e l^{2}}{2 m_{e}}$
• C. $\mu_{l}=\frac{-e}{4 m_{e}}$
• D. $\mu_{l}=\frac{-e l}{2 m_{e}}$

Answer 1

Answer: (D)

There will be a magnetic moment, usually denoted by $\mu_{i}$ associated with this circulating current.
Its magnitude is $\mu_{i}=I \pi r^{2}=e v r / 2$.
Multiplying and dividing the right hand side of the above expression by the electron mass $m_{e}$,
We have $\quad \mu_{l}=\frac{e}{2 m_{e}}\left(m_{e} v r\right)=\frac{e}{2 m_{e}} l$
Here, $l$ is the magnitude of the angular momentum of the electron about the central nucleus (orbital angular momentum). Vectorially,
$\mu_{l}=-\frac{e}{2 m_{e}} l$
The negative sign indicates that the angular momentum of the electron is opposite in direction to the magnetic moment.