91Ó°ÊÓ

The magnetic quantum number is $m_l=3$.

The magnetic quantum number is $m_l=-4$.

An electron, along with revolving in its orbit, spins as well along its axis. Thus, it carries angular momentum of its own; this results in the spin magnetic moment of the electron. Its component along the z-axis is given as-

${\mathbf{μ}}_{\mathrm{orb},z}\mathbf{=}\mathbf{-}{\mathbf{m}}_l\frac{\mathrm{eh}}{4\mathrm{Ï€³¾}}$ ..... (1)

Here,

The charge of an electron, role="math" localid="1663056261439" $\mathrm{e}=1.6×{10}^{-19}\text{C}$

Plank’s constant, $\mathrm{h}=6.63×{10}^{-34}\text{J.s}$

The mass of the electron,$\mathrm{m}=9.1×{10}^{-31}\mathrm{kg}$

Substitute known values into equation (1), and you get,

$\begin{array}{rcl}{\mathrm{μ}}_{orb,z}& =& -3×\frac{(1.6×{10}^{-19}\mathrm{C})×(6.63×{10}^{-34}\mathrm{J}.\mathrm{s})}{4×3.14×(9.1×{10}^{-31}\mathrm{kg})}\\ & =& -2.78×{10}^{-23}\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{T}$}\right.\end{array}$

Hence, the measured components of the orbital magnetic dipole moment of an electron with $m_l=3$is ${\mathrm{μ}}_{orb,z}=-2.78×{10}^{-23}\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{T}$}\right.$.

Putting known values into equation (1), and you have

$\begin{array}{rcl}{\mathrm{μ}}_{orb,z}& =& -(-4)×\frac{(1.6×{10}^{-19}\mathrm{C})×(6.63×{10}^{-34}\text{ }\mathrm{J}.\mathrm{s})}{4×3.14×(9.1×{10}^{-31}\mathrm{kg})}\\ & =& +3.71×{10}^{-23}\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{T}$}\right.\end{array}$

Hence, the measured components of the orbital magnetic dipole moment of an electron with $m_l=−4$is .$\begin{array}{rcl}& & +3.71×{10}^{-23}\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{T}$}\right.\end{array}$.