Question

Starting with Equation 6.80, and using Equations 6.57, 6.61, 6.64, and 6.81, derive Equation 6.82.

Short answer

The equation is derived,$E_{fs}^1=\frac{\left(13.6eV\right)}{n^3}{\alpha }^2\left[\frac{3}{4n}-\left(\frac{I\left(I+1\right)-m_1{m}_s}{I+\left(I+1/2\right)\left(I+1\right)}\right)\right]$ .

Step-by-step solution

Definition of spin-orbit coupling.

The connection between the electron's spin and its orbital motion around the nucleus is known as spin-orbit coupling.

Step 2: Derivation of equation 6.82.

Write the expression for the relativistic correction of the energy levels.

$E_r^1=\frac{{\left(E_n\right)}^2}{2m{c}^2}\left[\frac{4n}{I+1/2}-3\right]$

Write the expression spin- orbit coupling energy.

$$\begin{gathered} H_{so}^{\text{'}}=\frac{e^2}{8{\mathrm{\pi \epsilon }}_0}.\frac{1}{m^2{c}^2{r}^3}.S.L \\ {E}_{so}^1=\frac{e^2}{8{\mathrm{\pi \epsilon }}_0}.\frac{1}{m^2{c}^2{r}^3}.\left(\frac{S.L}{r^3}\right) \end{gathered}$$

It is known that localid="1658141432913" $\left(S.L\right)={ħ}^2{m}_1{m}_s$and $\left(\frac{1}{r^3}\right)=\frac{1}{I\left(I+1/2\right)\left(I+1\right)n^3{a}^3}$ Substitute ${ħ}^2{m}_1{m}_s$for $\left(S.L\right)$and$\frac{1}{I\left(I+1/2\right)\left(I+1\right)n^3{a}^3}$for $\left(\frac{1}{r^3}\right)$ in the above expression.

$E_{so}^1=\frac{e^2}{8{\mathrm{\pi \epsilon }}_0}.\frac{1}{m^2{c}^2{n}^3{a}^3}.\frac{ħm_1{m}_s}{I\left(I+1/2\right)\left(I+1\right)}$

Apply the first-order perturbation theory's fine structure adjustment to energy levels.

$$\begin{gathered} E_{fs}^1=\left(n/m_I{m}_s\left|H_r^{\text{'}}\right|n/m_1{m}_s\right)+\left(n/m_I{m}_s\left|H_{so}^{\text{'}}\right|n/m_1{m}_s\right) \\ =E_r^1+E_{so}^1 \\ =\frac{-{\left(E_n\right)}^2}{2m{c}^2}\left[\frac{4n}{I+1/2}-3\right]+\frac{e^2}{8{\mathrm{\pi \epsilon }}_0}.\frac{1}{m^2{c}^2{n}^3{a}^3}.\frac{{ħ}^2{m}_1{m}_s}{I\left(I+1/2\right)\left(I+1\right)} \end{gathered}$$

Here,$a=\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{ħ}}^2}{m{e}^2}$.

$$\begin{gathered} E_{fs}^1=\frac{-{\alpha }^2}{4n^4}\left(13.6e.V\right)\left[\frac{4n}{I+1/2}-3\right]+{\alpha }^4\left[\frac{m}{2\mathrm{ħ}}{\left(\frac{e^2}{4{\mathrm{\pi \epsilon }}_0}\right)}^2\right]\frac{m_I{m}_s}{n^{3}I\left(I+1/2\right)\left(I+1\right)} \\ =\frac{-{\alpha }^2}{4n^4}\left(13.6e.V\right)\left[\frac{4n}{I+1/2}-3\right]+{\alpha }^2\left(13.6eV\right)\frac{m_I{m}_s}{n^{3}I\left(I+1/2\right)\left(I+1\right)} \\ =\frac{\left(13.6e.V\right)}{n^3}{\alpha }^2\left[\frac{-1}{I+1/2}+\frac{3}{4n}+\frac{m_I{m}_s}{II\left(I+1/2\right)\left(I+1\right)}\right] \end{gathered}$$

Write the expression for the total energy.

role="math" localid="1658144520537" $E_{fs}^1=\frac{\left(13.6eV\right)}{n^3}{\alpha }^2\left[\frac{3}{4n}-\left(\frac{I\left(I+1\right)-m_1{m}_s}{I\left(I+1/2\right)\left(I+1\right)}\right)\right]$

Use the definition of Bohr energy.

role="math" localid="1658144104030" $$\begin{gathered} E_n=\frac{-E_1}{n^2} \\ {E}_n=\frac{-m{c}^2{\alpha }^2}{2n^2} \\ \frac{-{\left(E_n\right)}^2}{2m{c}^2}=\frac{-{\alpha }^2\left(13.6eV\right)}{4n^4} \\ {E}_n^2=\frac{{\left(E_1\right)}^2}{n^4} \end{gathered}$$

The expression becomes,

$$\begin{gathered} E_1=\frac{-m{c}^2{\alpha }^2}{2} \\ \frac{E_1}{n^4}.\frac{-m{c}^2{\alpha }^2}{2}=\frac{-\left(-13.6eV\right)m{c}^2{\alpha }^2}{2n^4} \\ \frac{{\left(E_n\right)}^2}{2m{c}^2}=\frac{\left(-13.6eV\right)m{c}^2{\alpha }^2}{4n^{4}m{c}^2} \\ =\frac{\left(-13.6eV\right)}{4n^4}{\alpha }^2 \end{gathered}$$

Thus, equation 6.82 is derived, that is$E_{fs}^1=\frac{\left(-13.6eV\right)}{n^4}{\alpha }^2\left[\frac{3}{4n}-\left(\frac{I\left(I+1\right)-m_1{m}_s}{I\left(I+1/2\right)\left(I+1\right)}\right)\right]$ .