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

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

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