Question

Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.

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 ground state and A be the area of orbit.
Dipole moment of the electron (μ) is given by
μ = niA = qfA
$$\begin{gathered} =e\times \frac{m{e}^4}{4{\in }_0^2{h}^3{n}^3}\times \left(\mathrm{\pi }{r}^2{n}^2\right) \\ =\frac{m{e}^5\times \left(\mathrm{\pi }{r}^2{n}^2\right)}{4{\in }_0^2{h}^3{n}^3} \end{gathered}$$
Here,
h = Planck's constant
e = Charge on the electron
${\epsilon }_0$ = Permittivity of free space
n = Principal quantum number

$$\begin{gathered} \therefore \mu =\frac{\left(9.1\times {10}^{-31}\right){\left(1.6\times {10}^{-19}\right)}^5\times \mathrm{\pi }\times {\left(0.53\times {10}^{-10}\right)}^2}{4\times {\left(8.85\times {10}^{-12}\right)}^2\times {\left(6.64\times {10}^{-34}\right)}^3\times {\left(1\right)}^3} \\ =0.000917\times {10}^{-20} \\ =9.176\times {10}^{-24}\mathrm{A}-{\mathrm{m}}^2 \end{gathered}$$