The Zeeman effect occurs in sodium just as in hydrogen-sodium's lone $\mathbf{3}\mathit{s}$valence electron behaves much as hydrogen's 1.5. Suppose sodium atoms are immersed in a$\mathbf{0}\mathbf{.}\mathbf{1}\mathit{T}$magnetic field.

(a) Into how many levels is the$\mathbf{3}{\mathit{P}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$level split?

(b) Determine the energy spacing between these states.

(c) Into how many lines is the$\mathbf{3}{\mathit{P}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$to$\mathbf{3}{\mathit{s}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$spectral line split by the field?

(d) Describe quantitatively the spacing of these lines.

(e) The sodium doublet $\left(589.0nmand589.6nm\right)$is two spectral lines.$\mathbf{3}{\mathit{P}}_{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{\to }\mathbf{3}{\mathit{s}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$and$\mathbf{3}{\mathit{P}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{\to }\mathbf{3}{\mathit{s}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$. which are split according to the two differentpossible spin-orbit energies in the 3Pstate (see Exercise 60). Determine the splitting of the sodium doublet (the energy difference between the two photons). How does it compare with the line splitting of part (d), and why?

The magnitude of the external magnetic field is$B=0.1T$ .

The Lande factoris given by

$\mathit{g}\mathbf{=}\frac{\mathbf{3}\mathbf{j}\left(j+1\right)\mathbf{-}\mathbf{l}\left(l+1\right)\mathbf{+}\mathbf{s}\left(s+1\right)}{\mathbf{2}\mathbf{j}\left(j+1\right)}$ …(1)

The orientation energy Uis given by

$\mathit{U}\mathbf{=}\mathit{g}\frac{\mathbf{e}}{\mathbf{2}\mathbf{m}}{\mathit{m}}_{\mathbf{j}}\sout{\mathbf{h}}\mathit{B}$ …(2)

Whereis the charge of an electron andis the mass of electron.

(a)

Here, the value of the j quantum number is$j=1/2$ .

Thus, the given level is split into $2j+1=2\times \frac{1}{2}+1=2$levels.

(b)

$$\begin{gathered} g_p=\frac{3\times {\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}+1-1\times \left(1+1\right)+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}+1}{2\times {\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}+1} \\ =\frac{2}{3} \end{gathered}$$

The energy for the$m_j=\frac{1}{2}$ state is given by Equation (2)

$E_{\frac{1}{2}}=\frac{g_{p}e}{4m}\sout{h}B$ …(3)

For the$m_j=-\frac{1}{2}$ , the energy is the same as the one given by Equation (3) but with a minus sign. Thus, the energy difference between these states$∆E$is two times greater than the one given by Equation (3)

role="math" localid="1660048069510" $$\begin{gathered} ∆E=\frac{g_{p}e}{2m}\sout{h}Bn \\ =\frac{{\displaystyle \frac{2}{3}}\times 1.6\times {10}^{-19}kg}{2\times 9.11\times {10}^{-31}}\times 1.055\times {10}^{-34}js\times 0.1T\times 6.24\times {10}^{18}e \\ ∆E=3.9\times {10}^{-6}eV \end{gathered}$$

Hence the energy spacing between the two $3p_{\frac{1}{2}}$states is$∆U=3.9\times {10}^{-6}eV$ .

(c)

In this case, we are considering the transition$3p_{\frac{1}{2}}\to 3s_{\frac{1}{2}}$ . For the s state, the j quantum number is$j=\frac{1}{2}$ . Thus, the possible values for$m_j$ are$m_j=\pm \frac{1}{2}$ . As we've already noted, the possible values for$m_j$ for the p state are also$m_j=\pm \frac{1}{2}$ . We see that every choice for the difference in the quantum number $∆m_j$is possible; since there are 4 possible choices, we conclude that the line is split into 4 lines.

(d)

The difference in the energies of these states is obtained by using Equation (2). In order to calculate it, we firstly need to calculate the Lande factor for the lower states,

$$\begin{gathered} g_s=\frac{3\times {\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}+1-0\left(0+1\right)+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}+1}{2\times {\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}+1} \\ =2 \end{gathered}$$

Now, by using Equation (2) we find the energy differences$∆E$as

$$\begin{gathered} \mathrm{\Delta }E=\frac{2}{3}\times \frac{e}{2m}\pm \frac{1}{2}ħB-2\times \frac{e}{2m}\pm \frac{1}{2}ħB \\ \mathrm{\Delta }E=\frac{eħB}{2m}\left(\pm \frac{1}{3}\pm 1\right) \\ \mathrm{\Delta }E=\frac{2}{3}\times \frac{eħB}{2m} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} ∆\mathrm{E}=\frac{2}{3}\times \frac{1.6\times {10}^{-19}\mathrm{kg}\times 1.055\times {10}^{-34}\mathrm{js}\times 0.1\times 6.24\times {10}^{18}\mathrm{e}}{2\times 9.11\times {10}^{-31}\mathrm{kg}} \\ ∆\mathrm{E}=3.9\times {10}^{-6}\mathrm{eV} \end{gathered}$$

Which is the same as the spacing of the two $3P_{\frac{1}{2}}$levels:$3.9\times {10}^{-6}\mathrm{eV}$ .

(e)

The sodium splitting is,

$E=\frac{hc}{\lambda }$

Substitute all the value in the above equation.

For $\mathrm{\lambda }=589.0\mathrm{nm}$

$$\begin{gathered} E=\frac{4.1357\times {10}^{-15}\mathrm{eVs}\times 3\times {10}^8\mathrm{m}/\mathrm{s}}{589.0\mathrm{nm}} \\ =\frac{1.2407\times {10}^{-6}\mathrm{eVm}}{589.0\mathrm{nm}} \\ =\frac{1240.7\mathrm{eVnm}}{589.0\mathrm{nm}} \\ E=2.106\mathrm{eV} \end{gathered}$$

And

For$\lambda =589.6nm$

$\begin{array}{l}E=\frac{1240.7eVnm}{589.6nm}\\ E=2.104eV\end{array}$

The sodium doublet splitting is about $2.1x{10}^{3}eV$. This is much less than in the case of the external magnetic field $\overrightarrow{B}$. This is due to the fact that the magnitude of the external magnetic field B is much less than the magnitude of the internal field.