91Ó°ÊÓ

$2{\mathrm{p}}_{3/2}$ State of hydrogen and$2{\mathrm{p}}_{1/2}$ state of hydrogen atom.

When L and S are aligned, they look like in figure 1.

Figure 1.

Here, $\mathbf{θ}$is the angle between and .

The state is for when and are anti-aligned, look like in figure 2.

Figure 2.

Use the law of cosines in order to find angle, by magnitude of the vectors.

$$\begin{gathered} c^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\mathrm{³¦´Ç²õθ} \\ {\mathrm{J}}^2={\mathrm{L}}^2+{\mathrm{S}}^2-2\mathrm{LS}\cos \\ {\mathrm{J}}^2={\mathrm{L}}^2+{\mathrm{S}}^2-2\mathrm{LS}\cos \left({180}^{\mathrm{o}}-\mathrm{Ï•}\right) \end{gathered}$$

Simplify further as shown below.

$\begin{array}{rcl}{\mathrm{J}}^2& =& {\mathrm{L}}^2+{\mathrm{S}}^2-2\mathrm{LS}\cos \mathrm{Ï•}\\ {\mathrm{J}}^2-{\mathrm{L}}^2-{\mathrm{S}}^2& =& 2\mathrm{LS}\mathrm{³¦´Ç²õÏ•}\\ \mathrm{³¦´Ç²õÏ•}& =& \frac{{\mathrm{J}}^2-{\mathrm{L}}^2-{\mathrm{S}}^2}{2\mathrm{LS}}\\ \mathrm{Ï•}& =& {\cos }^{-1}\left(\frac{{\mathrm{J}}^2-{\mathrm{L}}^2-{\mathrm{S}}^2}{2\mathrm{LS}}\right)\end{array}$ ……. (1)

To find the magnitude of vectors L, S, and J, for which find quantum numbers l, s, and j.

For p shell, l=1.

The electron's spin s is 1/2 and 3/2 from the $2{\mathrm{p}}_{3/2}$provide j.

Use these values in the equations.

$$\begin{gathered} \mathrm{L}=\sqrt{\mathcal{l}\left(\mathcal{l}+1\right)}\sout{\mathrm{h}},\mathrm{S}=\sqrt{\mathrm{s}\left(\mathrm{s}+1\right)}\sout{\mathrm{h}},\mathrm{J}=\sqrt{\mathrm{j}\left(\mathrm{j}+1\right)}\sout{\mathrm{h}} \\ \mathrm{L}=\sqrt{\left(1\right)\left[\left(1\right)+1\right]}\sout{\mathrm{h}},\mathrm{S}=\sqrt{\left(\frac{1}{2}\right)\left[\left(\frac{1}{2}\right)+1\right]}\sout{\mathrm{h}},\mathrm{J}=\sqrt{\left(\frac{3}{2}\right)\left[\left(\frac{3}{2}\right)+1\right]}\sout{\mathrm{h}} \\ \mathrm{L}=\sqrt{2}\sout{\mathrm{h}},\mathrm{S}=\frac{\sqrt{3}}{2}\sout{\mathrm{h}},\mathrm{J}=\frac{\sqrt{15}}{2}\sout{\mathrm{h}} \end{gathered}$$

Substitute the values in equation (1).

$\begin{array}{rcl}\mathrm{φ}& =& {\cos }^{-1}\left(\frac{{\mathrm{J}}^2-{\mathrm{L}}^2-{\mathrm{S}}^2}{2\mathrm{LS}}\right)\\ & =& {\cos }^{-1}\left[\frac{{\left({\displaystyle \frac{\sqrt{15}}{2}}\sout{\mathrm{h}}\right)}^2-{\left(\sqrt{2}\sout{\mathrm{h}}\right)}^2-{\left({\displaystyle \frac{\sqrt{3}}{2}}\sout{\mathrm{h}}\right)}^2}{2\left(\sqrt{2}\sout{\mathrm{h}}\right)\left({\displaystyle \frac{\sqrt{3}}{2}}\sout{\mathrm{h}}\right)}\right]\\ & =& {\cos }^{-1}\left(\frac{1}{\sqrt{6}}\right)\\ & =& 65.9^0\end{array}$

Therefore, the angle between L and S when they're aligned is role="math" localid="1658381059167" ${66}^{\mathrm{o}}$.

(b)

Find the value as follows:

$\begin{array}{rcl}\mathrm{J}& =& \sqrt{\mathrm{j}\left(\mathrm{j}+1\right)\sout{\mathrm{h}}}\\ & =& \sqrt{\frac{1}{2}\left(\frac{1}{2}+1\right)\sout{\mathrm{h}}}\\ & =& \frac{\sqrt{3}}{2}\sout{\mathrm{h}}\end{array}$

Then that, along with the same L and S as before can be inserted into equation (2).

$\begin{array}{rcl}\mathrm{φ}& =& {\cos }^{-1}\left(\frac{{\mathrm{J}}^2-{\mathrm{L}}^2-{\mathrm{S}}^2}{2\mathrm{LS}}\right)\\ & =& {\cos }^{-1}\left[\frac{{\left({\displaystyle \frac{\sqrt{3}}{2}}\sout{\mathrm{h}}\right)}^2-{\left(\sqrt{2}\sout{\mathrm{h}}\right)}^2-{\left({\displaystyle \frac{\sqrt{3}}{2}}\sout{\mathrm{h}}\right)}^2}{2\left(\sqrt{2}\sout{\mathrm{h}}\right)\left({\displaystyle \frac{\sqrt{3}}{2}}\sout{\mathrm{h}}\right)}\right]\\ & =& {\cos }^{-1}\left(-\frac{2}{\sqrt{6}}\right)\\ & =& 144.7^{\mathrm{o}}\end{array}$

Therefore, the angle between L and S when they're anti-aligned is $\mathrm{φ}={145}^{\mathrm{o}}$.