Question

In its ground state, carbon's 2pelectrons 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?

Short answer

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

Step-by-step solution

Total spin of two electrons

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

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

total spin quantum number

The quantum number j_T 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 s_t is given by

$\begin{array}{rcl}{\mathrm{S}}_{\mathrm{T}}& =& {\mathrm{s}}_1+{\mathrm{s}}_2\\ & =& \frac{1}{2}+\frac{1}{2}\\ {\mathrm{S}}_{\mathrm{T}}& =& 1\end{array}$

total spin of electrons

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

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

The total orbital angular momentum LT

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

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

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