Question

What is the angle between $\mathbf{L}$and $\mathbf{S}$in a (a) $\mathbf{2}{\mathbf{p}}_{\mathbf{3}\mathbf{/}\mathbf{2}}$and(b) $\mathbf{2}{\mathbf{p}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$ state of hydrogen?

Short answer

(a) The angle between L and S when they're aligned is $\mathrm{\phi }={66}^{\mathrm{o}}$.

(b) The angle between L and S when they're anti-aligned is $\mathrm{\phi }={145}^{\mathrm{o}}$.

Step-by-step solution

Given data

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

Concept used

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

Figure 1.

Here, $\mathbf{\theta }$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

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

$$\begin{gathered} c^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\mathrm{cos\theta } \\ {\mathrm{J}}^2={\mathrm{L}}^2+{\mathrm{S}}^2-2\mathrm{LS}\cos \\ {\mathrm{J}}^2={\mathrm{L}}^2+{\mathrm{S}}^2-2\mathrm{LS}\cos \left({180}^{\mathrm{o}}-\mathrm{\varphi }\right) \end{gathered}$$

Simplify further as shown below.

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

Find the magnitude of vectors L, S, and J  

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).

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

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

Find the value of J 

(b)

Find the value as follows:

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

Find the angle between L and S when they're anti-aligned

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

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

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