Question

Calculate the magnetic dipole moment corresp onding to the motion of the electron in the ground state of a hydogen atom.

Answer 1

Dipole moment $\left(\mu \right)=niA=qfA$

= $e\times \frac{m{e}^4}{4{ϵ}_0^2{h}^2{n}^3}\times \left(\pi r_0^2{n}^2\right)$

$\frac{m{e}^5\times \left(\pi r^3{n}^3\right)}{4{ϵ}_0^2{h}^2{n}^3}$

$\frac{(9.1\times {10}^{-31})(1.6\times {10}^{-19}{)}^5\times \pi \times (0.53)\times {10}^{-20}\times 1}{4\times (8.85\times 1^{-12}{)}^2\times (6.64\times {10}^{-34}\left)\right(1{)}^3}$

= $\frac{m{e}^5\times \left(\pi r^2{n}^2\right)}{4{ϵ}_0^2{h}^2{n}^2}$

= $0.000917\times {10}^{-20}$

= $9.176\times {10}^{-23}A-m^2$