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The electronic ground state of neutral sodium atoms in a magnetic field splits into two no longer degenerate energy states. In the thermal equilibrium the occupation probability of the states is the Boltzmann-Distribution, meaning that the occupation probability of a state of energy $E_i$ and temperature T is proportional to

$e^{-\frac{E_i}{k_bT}}$

a) Which orbital angular momentum and spin defines these two energy states?

b) Determine the difference of the occupation number between the upper and the lower states for one mole silver atoms in a magnetic field $B=1T$ for $T= 1K$ and $T=300 K$.

c)What's the magnetic moment (for given temperatures) for one mole sodium atoms in that field?

I'm trying to study ahead in class by reading the chapters in the given textbook and solving the problems associated for the chapters.

This one is one which I can't seem to find an approach to.

For a): How can I deduce the orbital angular momentum and spin of those two states with the given information?

I've tried looking up the formula the orbital angular momentum and in the text book it's given as: $|\textbf{l}|=\sqrt{l(l+1)}\hbar$.

Similarly the spin is given as $|\textbf{s}|=\sqrt{s(s+1)}\hbar$.

I'm guessing that I have to use the boltzmann-distribution to somehow get those two, but I don't know how.

For b): Since we're talking about differences in occupation numbers I suppose that the zeeman-effect comes into play. But I don't know how to get to the difference in occupation numbers from that.

For c): I don't see how the temperature would influence the magnetic moment? My guess would be that an increase in temperature would mean an increase in velocity of the atoms and if we were to look at it as a current loop it would decrease the time it takes to make one period and from that an increase in magnetic moment. But I can't find a formula (neither in my textbook nor on wikipedia) which gives the magnetic moment in terms of the magnetic field and temperature.

I would appreciate any help on this one.

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a) You can get $l$ and $s$ using the shell model. The Total orbital and spin angular momentum for a closed shell is zero. Thus you only have to consider the single electron of sodium in the third shell. The first electron in a shell is always in an $s$ state, so $l=0$; and a single unpaired spin has $s=\frac12$.

b) This is most likely a misprint, as silver has a much more complicated electronic structure than sodium; I'll answer the question for sodium atoms.

As you wrote, the splitting of the ground state in an external magnetic field is referred to as the Zeeman effect. Since $l=0$, there's no spin-orbit coupling and thus no difference between the weak and strong Zeeman effect. In the present case, the energy difference between the two states into which the ground state splits takes the simple form $\mu_Bg_SB$, where $\mu_B$ is the Bohr magneton, $g_S$ is the $g$-factor for the electronic spin, $g_S\approx2.002319$, and $B$ is the magnetic field strength. From the given Boltzmann distribution, the corresponding difference in occupation numbers is

$$ \frac{\mathrm e^{\frac{\mu_Bg_SB}{2kT}}-\mathrm e^{-\frac{\mu_Bg_SB}{2kT}}}{\mathrm e^{\frac{\mu_Bg_SB}{2kT}}+\mathrm e^{-\frac{\mu_Bg_SB}{2kT}}}=\tanh\frac{\mu_Bg_SB}{2kT}\;. $$

c) Each of the two states has a well-defined spin component along the direction of the magnetic field. The difference in occupation numbers results in a corresponding difference in the contributions to the magnetic moment of these two components. Each electron with spin $\pm\frac12$ along the field direction contributes $\pm\frac12g_S\mu_B$ to the magnetic moment. You know the number of atoms in a mole, and each atom has one magnetised single electron, so you can get the magnetic moment from the above difference in occupation numbers.

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