Question

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.

Short answer

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

Step-by-step solution

Given data

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.

Concept of spin orbit interaction energy and electron dipole moment

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, ${\mu }_s=\frac{e\hslash }{2m_e}$.

Find the difference in energies of the photons from the doublet

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 wavelengths

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.

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

Sketch the energy level diagram

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

Figure 1

Solve for the magnetic field

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).

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

Find the magnetic field

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 $B=18.5\text{T}$.