91影视

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 $位$, its energy E determined by formula as follows:

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

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

Spin 鈥搊rbit interaction energy鈥檚 magnitude U is, $U={渭}_{s}B$.

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

Electron's spin dipole moment can be approximated as, ${渭}_s=\frac{e\mathrm{鈩�}}{2m_e}$.

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:

$螖E=E_2-E_1$

That can be expanded by equation (1).

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

Substitute the values of the two wavelengths ofandin above equation.

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

Convertto Joules.

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

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.

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

Replace the by the equation (3).

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

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

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

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