Classical and Quantum Statistics

[atarr3]

Homework Statement

Consider an atom with a magnetic moment $$\mu$$ and a total spin of ½. The atom is placed in a uniform magnetic field B at temperature T. (a) Assuming Maxwell-Boltzmann statistics are valid at this temperature, find the ratio of atoms with spins aligned with the field to those aligned opposite the field.

Homework Equations

$$n\left(E\right)=g\left(E\right)F$$ and the corresponding equation for F depending on Maxwell-Boltzmann, Fermi-Dirac, or Bose-Einstein

The Attempt at a Solution

I'm having a little trouble figuring out which distribution function to use. I want to say that it's Fermi-Dirac because of the half-integer spin, but the question states that Maxwell-Boltzmann statistics are valid, which throws me off. And I'm still a little confused about the density of states. And where the magnetic field factors into my equation.

 

[Ben Niehoff]

As the question states, you can use classical Boltzmann statistics. What you need is an expression for the chemical potential. Treat up spins and down spins as two species of "particle". In an external B field, these species have different energies, and so there will be a chemical potential for one of them to change into the other. Once you figure that out, you just need the Boltzmann factor.

 

[atarr3]

So are you saying we have 2 possible states, making the density of states two (either up spin or down spin)?

I'm not sure what you mean by chemical potential, but I'm assuming you're talking about the potential due to intrinsic spin, so $$V=-\vec{\mu_{s}}\bullet\vec{B}$$, which would make the potential either $$\pm\mu_{B}B$$. And that would make the energy in the distribution function $$E_{0}+V$$. Is that right?

 

[atarr3]

Ok so my equation for the ratio is $$\frac{n\left(E_{2}\right)}{n\left(E_{1}\right)}=\frac{g\left(E_{2}\right)}{g\left(E_{1}\right)}e^{\beta\left(E_{1}-E_{2}\right)}$$ with $$E_{1}-E_{2}=2\mu_{B}B$$. I still don't know how to find the density of states. I want to say for each one it is two because the energies depend on the direction of the spin, so for each energy state it is either an +1/2 or -1/2 spin.

 

[Ben Niehoff]

That's good so far. In this particular problem, does g(E) actually depend on energy?

 

[atarr3]

Well the energy for each state depends on the direction of the spin, +1/2 or -1/2, so I feel like there is only one orientation for each state. At the very least, the density of states for each one should be equal. I think?