Question

The ratio of the magnetic dipole moment to the angular momentum $(L=mvr)$ 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

Answer: 4.00

$\vec{\mu }=i\vec{A}$ magnetic dipole moment, revolving electron generate current $v=$ frequency of revolving
here $i=ev$
$\vec{\mu }=ev\pi a_0^2$
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 $\frac{\mu }{L}=\frac{\pi ue{a}_0^2}{2\pi m{a}_0^2}$
or $\frac{\mu }{L}=\frac{e}{2m}$ is a constant
$=\frac{1.6\times 10^{-19}C}{2\times 9.1\times 10^{-31}}$
$=0.0879\times 10^{12}C/kg$
$=8.8\times 10^{10}C/kg$