Question

The angles between S and ${\mathbf{\mu }}_{\mathbf{S}}$and between L and ${\mathbf{\mu }}_{\mathbf{L}}$are 180^o. What is the angle between J and ${\mathbf{\mu }}_{\mathbf{J}}$ in a$\mathbf{2}{\mathbf{p}}_{\frac{\mathbf{3}}{\mathbf{2}}}$state of hydrogen?

Short answer

Angle between J and ${\mathrm{\mu }}_{\mathrm{J}}$in a $2{\mathrm{p}}_{\frac{3}{2}}$ state of hydrogen is 167⁰.

Step-by-step solution

Magnitude of orbital, spin, total angular momentum of  L,S,J.

Magnitudes of orbital angular momentum Lis given by-

$\mathbf{L}\mathbf{=}\sqrt{\mathbf{l}\left(l+1\right)}\sout{\mathbf{h}}$

Magnitudes of spin angular momentum Sis given by-

$\mathbf{S}\mathbf{=}\sqrt{\mathbf{s}\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}}\sout{\mathbf{h}}$ …(1)

And Magnitudes of total angular momentum Jis given by

$\mathbf{J}\mathbf{=}\sqrt{\mathbf{j}\mathbf{(}\mathbf{j}\mathbf{+}\mathbf{1}\mathbf{)}}\sout{\mathbf{h}}$

Magnetic dipole vector

${\mathbf{\mu }}_{\mathbf{J}}\mathbf{=}\frac{\mathbf{e}}{\mathbf{2}{\mathbf{m}}_{\mathbf{e}}}\left(L+2S\right)$ …(2)

 Determination of angle.

We know that

$$\begin{gathered} \mathrm{J}.{\mathrm{\mu }}_{\mathrm{J}}=\left|\mathrm{J}\right|\left|{\mathrm{\mu }}_{\mathrm{J}}\right|\mathrm{cos\theta } \\ \mathrm{cos\theta }=\frac{\mathrm{J}.{\mathrm{\mu }}_{\mathrm{J}}}{\left|\mathrm{J}\right|\left|{\mathrm{\mu }}_{\mathrm{J}}\right|} \\ \mathrm{\theta }={\cos }^{-1}\left(\frac{\mathrm{J}.{\mathrm{\mu }}_{\mathrm{J}}}{\left|\mathrm{J}\right|\left|{\mathrm{\mu }}_{\mathrm{J}}\right|}\right) \end{gathered}$$

Rewritten in terms of L and S using equations (2).

$$\begin{gathered} \mathrm{\theta }={\cos }^{-1}\left[\frac{\mathrm{L}+\mathrm{S}.\left(-{\displaystyle \frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}}\left(\mathrm{L}+2\mathrm{S}\right)\right)}{\left|\mathrm{J}\right|\left|{\mathrm{\mu }}_{\mathrm{J}}\right|}\right] \\ \mathrm{\theta }={\cos }^{-1}\left[-\frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}\frac{\left(\mathrm{L}+\mathrm{S}\right).\left(\mathrm{L}+2\mathrm{S}\right)}{\left|\mathrm{J}\right|\left|{\mathrm{\mu }}_{\mathrm{J}}\right|}\right] \\ \mathrm{\theta }={\cos }^{-1}\left[-\frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}\frac{\mathrm{L}.\mathrm{L}+2\mathrm{S}.\mathrm{S}+3\mathrm{L}.\mathrm{s}}{\left|\mathrm{J}\right|\left|{\mathrm{\mu }}_{\mathrm{J}}\right|}\right] \end{gathered}$$

Squared $\left(\mathrm{A}.\mathrm{A}={\mathrm{A}}^2\right)$,

Then, Equation can be simplified:

role="math" localid="1658403367529" $\begin{array}{rcl}\mathrm{\theta }& =& {\cos }^{-1}\left[-\frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}\frac{\mathrm{L}.\mathrm{L}+2\mathrm{S}.\mathrm{S}+3\mathrm{L}.\mathrm{S}}{\left|\mathrm{J}\right|\left|{\mathrm{\mu }}_{\mathrm{J}}\right|}\right]\\ & =& {\cos }^{-1}\left[-\frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}\frac{{\mathrm{L}}^2+2{\mathrm{S}}^2+3\mathrm{L}.\mathrm{S}}{\mathrm{J}\sqrt{{\mathrm{\mu }}_{\mathrm{J}}.{\mathrm{\mu }}_{\mathrm{J}}}}\right]\\ \mathrm{\theta }& =& {\cos }^{-1}\left[-\frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}\frac{{\mathrm{L}}^2+2{\mathrm{S}}^2+3\mathrm{L}\times \mathrm{S}}{\left|\mathrm{J}\right|\left|{\mathrm{\mu }}_{\mathrm{J}}\right|}\right]\\ & =& {\cos }^{-1}\left[-\frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}\frac{{\mathrm{L}}^2+2{\mathrm{S}}^2+3\mathrm{L}.\mathrm{S}}{\mathrm{J}\sqrt{\left[-{\displaystyle \frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}}\left(\mathrm{L}+2\mathrm{S}\right)\right].\left[-{\displaystyle \frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}}\left(\mathrm{L}+2\mathrm{S}\right)\right]}}\right]\\ & =& {\cos }^{-1}\left[-\frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}\frac{{\mathrm{L}}^2+2{\mathrm{S}}^2+3\mathrm{L}.\mathrm{S}}{\mathrm{J}\sqrt{{\left({\displaystyle \frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}}\right)}^2\left(\mathrm{L}.\mathrm{L}+4\mathrm{S}.\mathrm{S}+4\mathrm{L}.\mathrm{S}\right)}}\right]\\ & =& {\cos }^{-1}\left[-\frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}\frac{{\mathrm{L}}^2+2{\mathrm{S}}^2+3\mathrm{L}.\mathrm{S}}{\mathrm{J}\left({\displaystyle \frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}}\right)\sqrt{\left(\mathrm{L}.\mathrm{L}+4\mathrm{S}.\mathrm{S}+4\mathrm{L}.\mathrm{S}\right)}}\right]\\ \mathrm{\theta }& =& {\cos }^{-1}\left[-\frac{{\mathrm{L}}^2+2{\mathrm{S}}^2+3\mathrm{L}.\mathrm{S}}{\mathrm{J}\sqrt{\left({\mathrm{L}}^2+4{\mathrm{S}}^2+4\mathrm{L}.\mathrm{S}\right)}}\right]\end{array}$…(3)

Spin-Orbit Coupling L.S.

$\begin{array}{rcl}J& =& L+S\\ \mathrm{J}.\mathrm{J}& =& \left(\mathrm{L}+\mathrm{S}\right).\left(\mathrm{L}+\mathrm{S}\right)\\ \mathrm{J}.\mathrm{J}& =& \mathrm{L}.\mathrm{L}+\mathrm{S}.\mathrm{S}+2\mathrm{L}.\mathrm{S}\\ 2\mathrm{L}.\mathrm{S}& =& \mathrm{J}.\mathrm{J}-\mathrm{L}.\mathrm{L}-\mathrm{S}.\mathrm{S}\\ 2\mathrm{L}.\mathrm{S}& =& {\mathrm{J}}^2-{\mathrm{L}}^2-{\mathrm{S}}^2\\ \mathrm{L}.\mathrm{S}& =& \frac{{\mathrm{J}}^2-{\mathrm{L}}^2-{\mathrm{S}}^2}{2}\\ & & \end{array}$

role="math" localid="1658404178766" $\begin{array}{rcl}\mathrm{\theta }& =& {\cos }^{-1}\left[-\frac{{\mathrm{L}}^2+2{\mathrm{S}}^2+3\mathrm{L}.\mathrm{S}}{\mathrm{J}\sqrt{\left({\mathrm{L}}^2+4{\mathrm{S}}^2+4\mathrm{L}.\mathrm{S}\right)}}\right]\\ & =& {\cos }^{-1}\left[-\frac{{\mathrm{L}}^2+2{\mathrm{S}}^2+3\left({\displaystyle \frac{{\mathrm{J}}^2-{\mathrm{L}}^2-{\mathrm{S}}^2}{2}}\right)}{\mathrm{J}\sqrt{\left({\mathrm{L}}^2+4{\mathrm{S}}^2+4\left({\displaystyle \frac{{\mathrm{J}}^2-{\mathrm{L}}^2-{\mathrm{S}}^2}{2}}\right)\right)}}\right]\\ & =& {\cos }^{-1}\left[-\frac{{\mathrm{L}}^2+2{\mathrm{S}}^2+{\displaystyle \frac{3}{2}}\left({\displaystyle {\mathrm{J}}^2-{\mathrm{L}}^2-{\mathrm{S}}^2}\right)}{\mathrm{J}\sqrt{\left({\mathrm{L}}^2+4{\mathrm{S}}^2+2\left({\displaystyle {\mathrm{J}}^2-{\mathrm{L}}^2-{\mathrm{S}}^2}\right)\right)}}\right]\\ & =& {\cos }^{-1}\left[-\frac{{\mathrm{L}}^2+2{\mathrm{S}}^2+{\displaystyle \frac{3}{2}}{\mathrm{J}}^2-{\displaystyle \frac{3}{2}}{\mathrm{L}}^2-{\displaystyle \frac{3}{2}}{\mathrm{S}}^2}{\mathrm{J}\sqrt{\left({\mathrm{L}}^2+4{\mathrm{S}}^2+2{\mathrm{J}}^2-2{\mathrm{L}}^2-2{\mathrm{S}}^2\right)}}\right]\\ \mathrm{\theta }& =& {\cos }^{-1}\left[-\frac{3{\mathrm{J}}^2-{\mathrm{L}}^2+{\mathrm{S}}^2}{2\mathrm{J}\sqrt{\left(2{\mathrm{J}}^2-{\mathrm{L}}^2+2{\mathrm{S}}^2\right)}}\right]\end{array}$

role="math" localid="1658404753019" $\begin{array}{rcl}\mathrm{\theta }& =& {\cos }^{-1}\left[-\frac{3{\left({\displaystyle \frac{\sqrt{15}}{2}}\sout{\mathrm{h}}\right)}^2-{\left(\sqrt{2}\sout{\mathrm{h}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\sout{\mathrm{h}}\right)}^2}{2\left({\displaystyle \frac{\sqrt{15}}{2}}\sout{\mathrm{h}}\right)\sqrt{\left(2{\left({\displaystyle \frac{\sqrt{15}}{2}}\sout{\mathrm{h}}\right)}^2-{\left(\sqrt{2}\sout{\mathrm{h}}\right)}^2+2{\left({\displaystyle \frac{\sqrt{3}}{2}}\sout{\mathrm{h}}\right)}^2\right)}}\right]\\ & =& {\cos }^{-1}\left[-\frac{{\displaystyle \frac{45}{4}}{\mathrm{h}}^2-2{\mathrm{h}}^2+{\displaystyle \frac{3}{4}}{\mathrm{h}}^2}{\sqrt{15}\mathrm{h}\sqrt{{\displaystyle \frac{15}{2}}{\mathrm{h}}^2-2{\mathrm{h}}^2+{\displaystyle \frac{3}{2}}{\mathrm{h}}^2}}\right]\\ & =& {\cos }^{-1}\left[-\frac{{\displaystyle 10{\mathrm{h}}^2}}{\sqrt{15}\mathrm{h}\sqrt{{\displaystyle 7{\mathrm{h}}^2}}}\right]\end{array}$

$\begin{array}{rcl}\mathrm{\theta }& =& {\cos }^{-1}\left[-\frac{{\displaystyle 10{\mathrm{h}}^2}}{{\mathrm{h}}^2\sqrt{{\displaystyle 105}}}\right]\\ & =& {\cos }^{-1}\left[-\frac{{\displaystyle 10}}{\sqrt{{\displaystyle 105}}}\right]\\ & =& 167.4^0\end{array}$

Hence the angle is 167.4⁰.