91影视

Maxwell-Boltzmann statistics:

$\mathcal{N}\left(E\right)=\frac{B}{e^{E/k_{B}T}},$

(a)

To find B, we will make use of the normalization condition defined as:

$N={鈭�}_0^{鈭�}\mathcal{N}\left(E\right)D\left(E\right)dE$ 鈥︹��. (1)

Where , is the occupation number following the Maxwell-Boltzmann statistics:

$\mathcal{N}\left(E\right)=\frac{B}{e^{E/k_{B}T}},$ 鈥︹��.. (2)

The density of states for this system of harmonic oscillators with intrinsic spin:

$D\left(E\right)=\frac{2s+1}{\mathrm{鈩�}{蝇}_0}$ 鈥︹��. (3)

Put the equation (2) and(3) in equation (1), we get

$\begin{array}{rcl}N& =& {鈭�}_0^{鈭�}\frac{B}{e^{E/k_{B}T}}路\frac{2s+1}{\mathrm{鈩�}{蝇}_0}dE\\ & =& B\frac{(2s+1)}{\mathrm{鈩�}{蝇}_0}{鈭�}_0^{鈭�}{e}^{-E/k_{B}T}dE\\ & =& B\frac{(2s+1)}{\mathrm{鈩�}{蝇}_0}\left({\left.e^{-E/k_{B}T}\left(-k_{B}T\right)\right|}_0^{鈭�}\right)\\ & =& B\frac{(2s+1)}{\mathrm{鈩�}{蝇}_0}\left(k_{B}T\right)\\ & & \end{array}$

Solving for B we obtain:

$B=\frac{N\mathrm{鈩�}{蝇}_0}{2s+1}路\frac{1}{k_{B}T}$

According to the problem:

$\mathcal{E}=\frac{N\mathrm{鈩�}{蝇}_0}{2s+1}$

Thus, the expression for B is:

$B=\frac{\mathcal{E}}{k_{B}T}$

Put this expression for B into eq. (2), the occupation number is thus:

$\mathcal{N}(E{)}_{\text{boltz}}=\frac{\mathcal{E}}{k_{B}T}\frac{1}{e^{E/k_{B}T}}$

(b)

Assuming that,$k_{B}T>>蔚$ the denominator of N becomes extremely large and the magnitude of$\mathcal{E}$ pales in comparison with the size of $k_{B}T$. As a result, the occupation number is much less than 1 for all possible values of E, and that $\mathcal{N}鈫�0$holds true if$k_{B}T鈫�鈭�$

In high temperature systems, the harmonic oscillators are uniformly distributed across individual energy levels. In such cases, the occupation number is effectively less than 1.

$B=\frac{\mathcal{E}}{k_{B}T}$