Ideal Bose and Fermi Gases Questions and Answers

Answer: C

The chemical potential at T = 0 is \(\frac{3}{2}\hbar\omega\).

⁴He atoms are bosons. For an ideal Bose gas the chemical potential must stay below the lowest single-particle energy, and as T falls to zero all the atoms condense into that ground state and \(\mu\) rises to meet it. The ground state of the 3D isotropic oscillator has zero-point energy \(\frac{1}{2}\hbar\omega\) in each of the three directions:

\[ \varepsilon_{0} = \tfrac{3}{2}\hbar\omega \]

So \(\mu \to \frac{3}{2}\hbar\omega\). Zero would be the answer only if energies were measured from the ground state, \(\frac{1}{2}\hbar\omega\) counts one direction only, and \(3\hbar\omega\) doubles the zero-point energy.

Answer: B

The Fermi wave number is \( k_{F} = \frac{\pi N}{2L} \).

In a box of length \( L \) the allowed wave numbers are \( k = n\pi/L \), \( n = 1, 2, 3, \ldots \). Each level holds two spin-1/2 particles (spin up and spin down), so N particles fill the levels up to \( n_{F} = N/2 \):

\[ k_{F} = \frac{n_{F}\pi}{L} = \frac{\pi N}{2L} \]

\( \frac{\pi N}{L} \) is what results from ignoring spin and putting one particle per level. The options without \( \pi \) drop the \( \pi/L \) spacing of the allowed \( k \) values.

Answer: A

The density of states is proportional to \(\varepsilon^{1/2}\).

For free particles in three dimensions, the number of states with momentum below p is proportional to the volume of a sphere in momentum space, \(p^{3}\). With \(\varepsilon = p^{2}/2m\), that number goes as \(\varepsilon^{3/2}\), and its derivative gives the density of states:

\[ \begin{aligned} g(\varepsilon) &= \frac{V}{2\pi^{2}}\left(\frac{2m}{\hbar^{2}}\right)^{3/2} \\ &\quad\times\varepsilon^{1/2} \end{aligned} \]

(times 2 for spin). In two dimensions the density of states is constant (\(\varepsilon^{0}\)), and in one dimension it goes as \(\varepsilon^{-1/2}\), so those options describe lower-dimensional gases. \(\varepsilon^{2}\) belongs to no free-particle gas; photons, with \(\varepsilon \propto p\), give \(\varepsilon^{2}\) dependence, not a non-relativistic Fermi gas.

Answer: B

The peak wavelength is 1000 nm.

Wien's displacement law gives the wavelength of maximum emission of a black body:

\[ \begin{aligned} \lambda_{max} &= \frac{b}{T} = \frac{2.898 \times 10^{-3}\ \text{m K}}{2898\ \text{K}} \\ &= 10^{-6}\ \text{m} \end{aligned} \]

That is 1000 nm, in the near infrared. The temperature was chosen so the numbers cancel.

289.8 nm would need a star ten times hotter, and 10000 nm a body ten times cooler; 550 nm is the peak of the Sun, at about 5800 K.

Answer: D

The pressure ratio is half the ratio of the energy densities.

Pressure and energy density are related differently for radiation and for a non-relativistic ideal gas:

\[ \begin{aligned} P_{\text{photon}} &= \frac{u_{\text{photon}}}{3} \\ P_{\text{gas}} &= \frac{2}{3}u_{\text{gas}} \end{aligned} \]

Dividing,

\[ \begin{aligned} &\frac{P_{\text{photon}}}{P_{\text{gas}}} \\ &= \frac{1}{2}\cdot\frac{u_{\text{photon}}}{u_{\text{gas}}} \end{aligned} \]

The 1/3 for photons comes from their energy being \(pc\), linear in momentum; the 2/3 for the gas from \(p^{2}/2m\). 'Twice' inverts the factor, 'equal' would need the same equation of state for both, and 'thrice' uses only the photon's 1/3.