If an electron in an atom has an orbital angular momentum with$\mathit{m}\mathbf{=}\mathbf{0}$, what are the components (a) ${\mathit{L}}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$and (b)${\mathit{\mu }}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$? If the atom is in an external magnetic field$\overrightarrow{\mathbf{B}}$that has magnitude$\mathbf{35}\mathbf{}\mathbf{mT}$and is directed along the z axis, what are (c) the energy${\mathit{U}}_{\mathbf{o}\mathbf{r}\mathbf{b}}$associated with$\overrightarrow{{\mathbf{\mu }}_{\mathbf{o}\mathbf{r}\mathbf{b}}}$and(d) the energy${\mathit{U}}_{\mathbf{s}\mathbf{p}\mathbf{i}\mathbf{n}}$associated with$\overrightarrow{{\mathbf{\mu }}_{\mathbf{s}}}$? (e) If, instead, the electron has$\mathit{m}\mathbf{=}\mathbf{-}\mathbf{3}$, what are (e)${\mathit{L}}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$? (f) ${\mathit{\mu }}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$(g) ${\mathit{U}}_{\mathbf{o}\mathbf{r}\mathbf{b}}$and (h)${\mathit{U}}_{\mathbf{s}\mathbf{p}\mathbf{i}\mathbf{n}}$?

Magnitude of the external magnetic field,$\left|\overrightarrow{B}\right|=35mT\times \frac{1\text{T}}{{10}^3\text{mT}}=35\times {10}^{-3}\text{T}$

The magnetic field is directed along the z-axis.

The value of the magnetic azimuthally quantum number of the electron,$m=-3$

An electron in an atom has orbital angular momentum and spin angular momentum; the components of the angular momentum are quantized. The angular momentum and orbital angular momentum are specifically due to the results from all the intrinsic properties of the charge that is its spin and its charge. Here, these depend on the magnetic quantum number that represents the shape and spin of the atomic orbital. They depend on the value of the azimuthally quantum number ranging from the value of -l to the +l value of the azimuthally quantum number.

Formulae:

The z component of the orbital angular momentum,$L_{orb,z}=\frac{m_{l}h}{2\pi }$

Where,$m_l$ is the magnetic azimuthally quantum number,$h=6.63\times {10}^{-34 }{\text{m}}^2\text{kg/s}$is the Planck’s constant.

The orbital magnetic dipole moment, ${\mu }_{orb,z}=-m_l{\mu }_B$ (ii)

Where,$m_l$is the magnetic azimuthally quantum number, ${\mu }_B=9.27\times {10}^{-24}\text{J/T}$is the Bohr magneton of an electron.

The potential energy of an atomic orbital,

$\begin{array}{rcl}U& =& -{\overrightarrow{\mu }}_{orb}.{\overrightarrow{B}}_{ext}\\ & =& -{\mu }_{orb,z}{B}_{ext}\end{array}$ (iii)

Where, ${\overrightarrow{\mu }}_{orb}$is the orbital magnetic dipole moment, ${\overrightarrow{B}}_{ext}$is the external magnetic field, ${\mu }_{orb,z}$is the z-component of the orbital magnetic dipole momentum

The z component of the orbital angular momentum for value$(m_l=0)$can be given using equation (i) as follows:

$\begin{array}{rcl}{L}_{orb,z}& =& \frac{\left(0\right)h}{2\pi }\\ & =& 0\end{array}$

Hence, the component$L_{orb,z}$ is zero.

The z component of the orbital angular moment for value$(m_l=0)$can be given using equation (ii) as follows:

$\begin{array}{rcl}{\mu }_{orb,z}& =& -\left(0\right){\mu }_B\\ & =& 0\end{array}$

Hence, the component $m_{orb,z}$is zero

Substituting equation (ii) in equation (iii), the potential energy associated with the value $(m_l=0)$for the z orbital component can be given as follows:

$\begin{array}{rcl}U& =& -m_l{\mu }_B{B}_{ext}\\ & =& \left(0\right){\mu }_B{B}_{ext}\\ & =& 0\end{array}$

Hence, the energy$U_{orb}$ associated with ${\mu }_s$is zero

Now, using the given data in equation (iii), the potential energy associated with the dipole moment due to its spin can be given as follows:

$\begin{array}{rcl}U& =& -{\overrightarrow{\mu }}_{s,z}.\overrightarrow{B}\\ & =& \pm {\mu }_{B}B \\ & =& \pm (9.27\times {10}^{-24}\mathrm{J}/\mathrm{T})\left(35\times {10}^{-3}\text{T}\right)\\ & =& \pm 3.2\times {10}^{-25}\mathrm{J}\end{array}$

Hence, the energy $U_{spin}$associated with ${\mu }_s$is$\pm 3.2\times {10}^{-25}\mathrm{J}$

Now, the value of orbital angular momentum associated with the value$m=-3$can be given using the data in equation (i) as follows:

$\begin{array}{rcl}{L}_{orb,z}& =& \frac{(-3)( 6.63\times {10}^{-34}\mathrm{J}.\mathrm{s})}{2\pi }\\ & =& -3.16\times {10}^{-34} \mathrm{J}.\mathrm{s}\\ & =& -3.2\times {10}^{-34} \mathrm{J}.\mathrm{s}\end{array}$

Hence if, instead, the electron has $m=-3$, the value of$L_{orb,z}$is$-3.2\times {10}^{-34} \mathrm{J}.\mathrm{s}$

Now, the value of orbital dipole moment associated with the value$m=-3$can be given using the data in equation (ii) as follows:

$\begin{array}{rcl}{\mu }_{orb,z}& =& (-3)( 9.27\times {10}^{-24}\mathrm{J}/\mathrm{T})\\ & =& 2.78\times {10}^{-23}J/T\\ & \approx & 2.8\times {10}^{-23}\mathrm{J}/\mathrm{T}\end{array}$

If, instead, the electron has$m=-3$ , the value of $m_{orb,z}$is$2.8\times {10}^{-23}\mathrm{J}/\mathrm{T}$

Using the above value from parts (e) and (f) in equation (iii), the potential energy associated with the value$m=-3$ can be given as follows:

$\begin{array}{rcl}U& =& -(2.78\times {10}^{-23}\mathrm{J}/\mathrm{T})(35\times {10}^{-3}\mathrm{T})\\ & =& -9.7\times {10}^{-25}\mathrm{J}\\ & & \end{array}$

Hence, the value of$U_{orb}$is.$-9.7\times {10}^{-25}\mathrm{J}$

Now, using the given data in equation (iii), the potential energy associated with the dipole moment due to its spin can be given as follows:

$$\begin{gathered} U=-{\overrightarrow{\mu }}_{s,z}.\overrightarrow{B} \\ =\pm {\mu }_{B}B \\ =\pm (9.27\times {10}^{-24}\mathrm{J}/\mathrm{T})\left(35\times {10}^{-3}\text{T}\right) \\ =\pm 3.2\times {10}^{-25}\mathrm{J} \end{gathered}$$

This implies, the potential energy associated with the electron spin is only dependent on the constant Bohr magneton and the external magnetic field and so for a uniform electric field, it is independent of $m_1$ and thus remains the same.

Hence, the value of the energy is $\pm 3.2\times {10}^{-25}\mathrm{J}$.