91Ó°ÊÓ

The orbital magnetic dipole is $m_l=3$.

The orbital magnetic dipole is $m_l=4$.

You have the formula for orbital magnetic dipole moment. From the given data, you can get the required value.

Formula:

The orbital magnetic dipole moment equation is,

${\mathrm{μ}}_{orb}=−m_l{\mathrm{μ}}_B$

Here, ${\mathrm{μ}}_B$is the magnetic moment having a value of $9.27×{10}^{−24}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.$.

First take the orbital magnetic dipole as,

$m_l=3$

So, the orbital magnetic dipole moment will be,

$\begin{array}{rcl}{\mathrm{μ}}_{orb}& =& −m_l{\mathrm{μ}}_B\\ & =& −3×(9.27×{10}^{−24}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.\\ & =& −2.78×{10}^{-23}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.\end{array}$

Hence, the component of the orbital magnetic dipole moment of an electron with the values $m_l=3$ will be $−2.78×{10}^{−23}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.$.

Take $m_l=−4$.

Thus, the orbital magnetic dipole moment is,

$\begin{array}{rcl}{\mathrm{μ}}_{orb}& =& −m_l{\mathrm{μ}}_B\\ & =& −(−4)×\left(9.27×{10}^{−24}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.\right)\\ & =& 3.71×{10}^{-23}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.\end{array}$

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