The well-known sodium doublet is two yellow spectral lines of very close wavelength.$\mathbf{589}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{nm}}$and $\mathbf{589}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{nm}}$It is caused by splitting of the $\mathbf{3}\mathit{p}$ energy level. due to the spin-orbit interaction. In its ground state, sodium's single valence electron is in the level. It may be excited to the next higher level. the 3p , then emit a photon as it drops back to the 3s . However. the 3p is actually two levels. in which Land Sare aligned and anti-aligned. (In the notation of Section $\mathbf{8}\mathbf{.}\mathbf{7}$ these are. respectively. the$\mathbf{3}{\mathit{\rho }}_{\mathbf{3}\mathbf{/}\mathbf{2}}$and $3p_{1n}.)$the because the (transitions Stan from slightly different initial energies yet have identical final energies(the $\mathbf{3}\mathit{s}$having no orbital angular momentum to lead to spin-orbit interaction), there are two different wavelengths possible for the emitted photon. Calculate the difference in energy between the two photons. From this, obtain a rough value of the average strength of the internal magnetic field experienced by sodium's valence electron.

Internal magnetic field of sodium having the electrons jumping from$3p_{3/2}$to 3 and$3p_{1/2}$to 3 states emitting photons of wavelengthandrespectively, is determined by equations for energy of photon, spin dipole moment and spin-orbit interaction energy of the electron.

For a photon of wavelength $\lambda$, its energy E determined by formula as follows:

$$\begin{gathered} E=\frac{hc}{\lambda } \\ E=\frac{1240e\text{V}.\text{nm}}{\lambda } \end{gathered}$$

Where h is Planck’s constant and is speed of light in vacuum.

Spin –orbit interaction energy’s magnitude U is, $U={\mu }_{s}B$.

Here ${\mu }_s$represents spin dipole moment of electron and represents magnetic field experienced by the electron.

Electron's spin dipole moment can be approximated as, ${\mu }_s=\frac{e\hslash }{2m_e}$.

In order to get an estimation of the magnetic field felt by sodium's valence electron, it first needs to be found the difference in energies of the photons from the doublet:

$\Delta E=E_2-E_1$

That can be expanded by equation (1).

$\begin{array}{rcl}\Delta E& =& E_2-E_1\\ & =& \frac{1240\text{cV}-\text{nm}}{{\lambda }_2}-\frac{1240\text{cV}-\text{nm}}{{\lambda }_1}\\ & =& (1240\text{eV}-\text{nm})\left(\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}\right)\\ & & \end{array}$

Substitute the values of the two wavelengths ofandin above equation.

$\begin{array}{rcl}\Delta & =& (1240\text{eV}-\text{nm})\left(\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}\right)\\ & =& (1240\text{eV}-\text{nm})\left[\frac{1}{(589.0\text{nm})}-\frac{1}{(589.6\text{nm})}\right]\\ & =& 2.14\times {10}^{-3}\text{eV}\\ & & \end{array}$

Convertto Joules.

$\begin{array}{rcl}\Delta E& =& \left(2.14\times {10}^{-3}\text{eV}\right)\left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)\\ & =& 3.43\times {10}^{-22}\text{J}\\ & & \end{array}$

That energy will be twice the spin-orbit interaction energy as shown in figure 1 with the energy level diagram:

Figure 1

The spin-orbit interaction energy is added or subtracted from the base energy because the electron can be aligned parallel or anti-parallel to the magnetic field.

So, equation (2) can be rewritten with that, and then solve for the magnetic field.

$\begin{array}{rcl}U& =& {\mu }_{s}B\\ \frac{1}{2}\Delta E& =& {\mu }_{s}B\\ B& =& \frac{\Delta E}{2{\mu }_s}\\ & & \end{array}$

Replace the by the equation (3).

$\begin{array}{rcl}B& =& \frac{\Delta E}{2{\mu }_s}\\ & =& \frac{\Delta E}{2\left(\frac{d{m}_e}{2m_e}\right)}\\ & =& \frac{\Delta E}{\left(\frac{\text{ot}}{m_e}\right)}\\ & =& \frac{m_e\Delta E}{e\backslash \mathrm{AA}}\\ & & \end{array}$

Substitute the values of $9.11\times {10}^{-31}\text{kg}$ for $m_e,3.43\times {10}^{-22}\text{J}$for $\Delta E,1.6\times {10}^{-19}$for e , and $1.055\times {10}^{-34}\text{J}.\text{s}$ for$\hslash$in above equation.

$\begin{array}{rcl}B& =& \frac{m_e\Delta E}{e\hslash }\\ & =& \frac{\left(9.1\times {10}^{-31}\text{kg}\right)\left(3.43\times {10}^{-22}\text{J}\right)}{\left(1.6\times {10}^{-19}\text{C}\right)\left(1.055\times {10}^{-34}\text{J}.\text{s}\right)}\\ & =& 18.48\text{T}\\ & =& 18.5\text{T}\\ & & \end{array}$

Therefore, the magnetic field that the valence electron of sodium experiences is $B=18.5\text{T}$.