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78E

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Found in: Page 344

### Modern Physics

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

# In its ground state, carbon's 2p electrons interact to produce ${{\mathbf{j}}}_{{\mathbf{T}}}{\mathbf{=}}{\mathbf{0}}$. Given Hund's rule. what does this say about the total orbital angular momentum of these electrons?

Total orbital angular momentum is $\mathrm{L}=\sqrt{2}\overline{)\mathrm{h}}$.

See the step by step solution

## Step 1: Total spin of two electrons

The total spin of the two electrons ${{\mathbf{S}}}_{{\mathbf{T}}}$ is

${{\mathbf{S}}}_{{\mathbf{T}}}{\mathbf{=}}\sqrt{{\mathbf{S}}_{\mathbf{T}}\left({S}_{T}+1\right)}\overline{)\mathbf{h}}$ ….(1)

## Step 2: total spin quantum number

The quantum number jT is zero ${\mathrm{j}}_{\mathrm{T}}=0$.

The spin quantum number of each electron is ${\mathrm{s}}_{1}={\mathrm{s}}_{2}=\frac{1}{2}$.

By using Hund's rule, we conclude that the total spin quantum number st is given by

$\begin{array}{rcl}{\mathrm{S}}_{\mathrm{T}}& =& {\mathrm{s}}_{1}+{\mathrm{s}}_{2}\\ & =& \frac{1}{2}+\frac{1}{2}\\ {\mathrm{S}}_{\mathrm{T}}& =& 1\end{array}$

## Step 3: total spin of electrons

By using Eq. (I), we find the total spin of the electrons sT as

$\begin{array}{rcl}{\mathrm{S}}_{\mathrm{T}}& =& \sqrt{{\mathrm{s}}_{\mathrm{T}}\left({\mathrm{s}}_{\mathrm{T}}+1\right)}\overline{)\mathrm{h}}\\ & =& \sqrt{1.\left(1+1\right)}\overline{)\mathrm{h}}\\ {\mathrm{S}}_{\mathrm{T}}& =& \sqrt{2}\overline{)\mathrm{h}}\end{array}$

## Step 4: The total orbital angular momentum LT

Now, since ${\mathrm{j}}_{\mathrm{T}}=0$, we conclude that the total orbital angular momentum LT must be equal to the total spin ST

${\mathrm{L}}_{\mathrm{T}}={\mathrm{S}}_{\mathrm{T}}=\sqrt{2}\overline{)\mathrm{h}}$

Hence the total orbital angular momentum is ${\mathrm{L}}_{\mathrm{T}}=\sqrt{2}\overline{)\mathrm{h}}$.

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