Question

The ratio of the magnetic dipole moment to the angular momentum $(L=m v r)$ is a universal constant for hydrogen-like atoms and ions. Its value is $a \times 2.2 \times 10^{10} C / kg$. Find the value of $a$.

Answer 1

$\vec{\mu}= i \vec{ A }$ magnetic dipole moment, revolving electron generate current $v=$ frequency of revolving
here $\quad i=e v$
$\vec{\mu}= ev \pi a _{0}^{2}$
and angular momentum
$L = m v _{1} a _{0}$
$\left(\because v _{1}=2 \pi a _{0}\right)$
$L = m \left(2 \pi a _{0}\right) a _{0}$
$L =2 \pi m a _{0}^{2}$
Ratio $\frac{\mu}{ L }=\frac{\pi uea _{0}^{2}}{2 \pi ma _{0}^{2}}$
or $\quad \frac{\mu}{L}=\frac{e}{2 m}$ is a constant
$=\frac{1.6 \times 10^{-19} C }{2 \times 9.1 \times 10^{-31}}$
$=0.0879 \times 10^{12} C / kg$
$=8.8 \times 10^{10} C / kg$