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65E

Expert-verified
Found in: Page 343

### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

# 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?

(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 $\mathrm{\phi }={145}^{\mathrm{o}}$.

See the step by step solution

## Step 1: Given data

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

## Step 2: 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.

## Step 3: 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.

${c}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\mathrm{cos\theta }\phantom{\rule{0ex}{0ex}}{\mathrm{J}}^{2}={\mathrm{L}}^{2}+{\mathrm{S}}^{2}-2\mathrm{LS}\mathrm{cos}\phantom{\rule{0ex}{0ex}}{\mathrm{J}}^{2}={\mathrm{L}}^{2}+{\mathrm{S}}^{2}-2\mathrm{LS}\mathrm{cos}\left({180}^{\mathrm{o}}-\mathrm{\varphi }\right)$

Simplify further as shown below.

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

## Step 4: 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.

$\mathrm{L}=\sqrt{\mathcal{l}\left(\mathcal{l}+1\right)}\overline{)\mathrm{h}},\mathrm{S}=\sqrt{\mathrm{s}\left(\mathrm{s}+1\right)}\overline{)\mathrm{h}},\mathrm{J}=\sqrt{\mathrm{j}\left(\mathrm{j}+1\right)}\overline{)\mathrm{h}}\phantom{\rule{0ex}{0ex}}\mathrm{L}=\sqrt{\left(1\right)\left[\left(1\right)+1\right]}\overline{)\mathrm{h}},\mathrm{S}=\sqrt{\left(\frac{1}{2}\right)\left[\left(\frac{1}{2}\right)+1\right]}\overline{)\mathrm{h}},\mathrm{J}=\sqrt{\left(\frac{3}{2}\right)\left[\left(\frac{3}{2}\right)+1\right]}\overline{)\mathrm{h}}\phantom{\rule{0ex}{0ex}}\mathrm{L}=\sqrt{2}\overline{)\mathrm{h}},\mathrm{S}=\frac{\sqrt{3}}{2}\overline{)\mathrm{h}},\mathrm{J}=\frac{\sqrt{15}}{2}\overline{)\mathrm{h}}$

Substitute the values in equation (1).

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

## Step 5: Find the value of J

(b)

Find the value as follows:

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

## Step 6: 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).

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