Question

Show that the ratio of the magnetic dipole moment to the angular momentum (l = mvr) is a universal constant for hydrogen-like atoms and ions. Find its value.

Answer 1

Mass of the electron, m = 9.1×10^$-$31kg
Radius of the ground state, r = 0.53×10^$-$10m
Let f be the frequency of revolution of the electron moving in the ground state and A be the area of orbit.
Dipole moment of the hydrogen like elements (μ) is given by
μ = niA = qfA
$$\begin{gathered} =e\times \frac{m{e}^4}{4{\in }_0^2{h}^3{n}^3}\times \left(\mathrm{\pi }{r}_0^2{n}^2\right) \\ =\frac{m{e}^5\times \left(\mathrm{\pi }{r_0}^2{n}^2\right)}{4{\in }_0^2{h}^3{n}^3} \end{gathered}$$
Here,
h = Planck's constant
e = Charge on the electron
${\epsilon }_0$ = Permittivity of free space
n = Principal quantum number

Angular momentum of the electron in the hydrogen like atoms and ions (L) is given by
$L=mvr=\frac{nh}{2\mathrm{\pi }}$
Ratio of the dipole moment and the angular momentum is given by
$\frac{\mu }{L}=\frac{e^5\times m\times \mathrm{\pi }{r}^2{n}^2}{4{\in }_0{h}^3{n}^3}\times \frac{2\mathrm{\pi }}{nh}$
$$\begin{gathered} \frac{\mu }{L}=\frac{{\left(1.6\times {10}^{-19}\right)}^5\times \left(9.10\times {10}^{-31}\right)\times {\left(3.14\right)}^2\times {\left(0.53\times {10}^{-10}\right)}^2}{2\times {\left(8.85\times {10}^{-12}\right)}^2\times {\left(6.63\times {10}^{-34}\right)}^3\times 1^2} \\ \frac{\mu }{L}=3.73\times {10}^{10}\mathrm{C}/\mathrm{kg} \end{gathered}$$

Ratio of the magnetic dipole moment and the angular momentum do not depends on the atomic number 'Z'.
Hence, it is a universal constant.