Chapter 6: Q22P (page 280)

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

Short Answer

Expert verified

The equation is derived, $E_{f s}^{1} = \frac{(13.6 e V)}{n^{3}} α^{2} [\frac{3}{4 n} - (\frac{I (I + 1) - m_{1} m_{s}}{I + (I + 1 / 2) (I + 1)})]$ .

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.

$E_{r}^{1} = \frac{{(E_{n})}^{2}}{2 m c^{2}} [\frac{4 n}{I + 1 / 2} - 3]$

Write the expression spin- orbit coupling energy.

$H_{s o}^{'} = \frac{e^{2}}{8 {πε}_{0}} . \frac{1}{m^{2} c^{2} r^{3}} . S . L E_{s o}^{1} = \frac{e^{2}}{8 {πε}_{0}} . \frac{1}{m^{2} c^{2} r^{3}} . (\frac{S . L}{r^{3}})$

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

$E_{s o}^{1} = \frac{e^{2}}{8 {πε}_{0}} . \frac{1}{m^{2} c^{2} n^{3} a^{3}} . \frac{ħ m_{1} m_{s}}{I (I + 1 / 2) (I + 1)}$

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

$E_{f s}^{1} = (n / m_{I} m_{s} |H_{r}^{'}| n / m_{1} m_{s}) + (n / m_{I} m_{s} |H_{s o}^{'}| n / m_{1} m_{s}) = E_{r}^{1} + E_{s o}^{1} = \frac{- {(E_{n})}^{2}}{2 m c^{2}} [\frac{4 n}{I + 1 / 2} - 3] + \frac{e^{2}}{8 {πε}_{0}} . \frac{1}{m^{2} c^{2} n^{3} a^{3}} . \frac{ħ^{2} m_{1} m_{s}}{I (I + 1 / 2) (I + 1)}$

Here, $a = \frac{4 {πε}_{0} ħ^{2}}{m e^{2}}$ .

$E_{f s}^{1} = \frac{- α^{2}}{4 n^{4}} (13.6 e . V) [\frac{4 n}{I + 1 / 2} - 3] + α^{4} [\frac{m}{2 ħ} {(\frac{e^{2}}{4 {πε}_{0}})}^{2}] \frac{m_{I} m_{s}}{n^{3} I (I + 1 / 2) (I + 1)} = \frac{- α^{2}}{4 n^{4}} (13.6 e . V) [\frac{4 n}{I + 1 / 2} - 3] + α^{2} (13.6 e V) \frac{m_{I} m_{s}}{n^{3} I (I + 1 / 2) (I + 1)} = \frac{(13.6 e . V)}{n^{3}} α^{2} [\frac{- 1}{I + 1 / 2} + \frac{3}{4 n} + \frac{m_{I} m_{s}}{I I (I + 1 / 2) (I + 1)}]$

Write the expression for the total energy.

role="math" localid="1658144520537" $E_{f s}^{1} = \frac{(13.6 e V)}{n^{3}} α^{2} [\frac{3}{4 n} - (\frac{I (I + 1) - m_{1} m_{s}}{I (I + 1 / 2) (I + 1)})]$

Use the definition of Bohr energy.

role="math" localid="1658144104030" $E_{n} = \frac{- E_{1}}{n^{2}} E_{n} = \frac{- m c^{2} α^{2}}{2 n^{2}} \frac{- {(E_{n})}^{2}}{2 m c^{2}} = \frac{- α^{2} (13.6 e V)}{4 n^{4}} E_{n}^{2} = \frac{{(E_{1})}^{2}}{n^{4}}$

The expression becomes,

$E_{1} = \frac{- m c^{2} α^{2}}{2} \frac{E_{1}}{n^{4}} . \frac{- m c^{2} α^{2}}{2} = \frac{- (- 13.6 e V) m c^{2} α^{2}}{2 n^{4}} \frac{{(E_{n})}^{2}}{2 m c^{2}} = \frac{(- 13.6 e V) m c^{2} α^{2}}{4 n^{4} m c^{2}} = \frac{(- 13.6 e V)}{4 n^{4}} α^{2}$

Thus, equation 6.82 is derived, that is $E_{f s}^{1} = \frac{(- 13.6 e V)}{n^{4}} α^{2} [\frac{3}{4 n} - (\frac{I (I + 1) - m_{1} m_{s}}{I (I + 1 / 2) (I + 1)})]$ .