91影视

We are given a particle that can share a single-particle state with one other particle of the same type but no more and the number of these particles in any state can be $0,1$or $2$.

The grand partition function or Gibbs sum is,

$Z=\underset{n}{鈭�}{e}^{\frac{-n(系-渭)}{kT}}$

According to the given problem, the number of particles of third type in any state can be $0,1$or $2$. That is n can be $0,1\text{or}2$.

$$\begin{gathered} Z=\underset{n=0}{\overset{2}{鈭�}}\exp \left[\frac{-n(蔚-渭)}{kT}\right] \\ Z=\exp \left(0\right)+\exp \left[\frac{-(蔚-渭)}{kT}\right]+\exp \left[\frac{-2(蔚-渭)}{kT}\right] \\ Z=1+\exp \left[\frac{-(蔚-渭)}{kT}\right]+\exp \left[\frac{-2(蔚-渭)}{kT}\right] \end{gathered}$$

Let $x=\exp \left[\frac{-(蔚-渭)}{kT}\right]$,

$Z=1+x+x^2$

The probability of state being occupied by n-particles is,

$$\begin{gathered} \mathrm{P}\left(\mathrm{n}\right)=\frac{1}{Z}\exp \left(\frac{-n(蔚-渭)}{kT}\right) \\ =\frac{1}{Z}{x}^n \\ \mathrm{P}\left(n\right)=\frac{x^n}{1+x+x^2} \end{gathered}$$

The average occupancy of state by particles of this type is,

$$\begin{gathered} \stackrel{炉}{n}=\underset{n=0}{\overset{2}{鈭�}}n\mathrm{P}\left(n\right) \\ =0路P\left(0\right)+1路P\left(1\right)+2路P\left(2\right) \\ =\frac{1路x^1}{1+x+x^2}+\frac{2路x^2}{1+x+x^2} \\ \stackrel{炉}{n}=\frac{x+2x^2}{1+x+x^2} \end{gathered}$$

So, the distribution function for the number of these particles in any state can be $0,1$or $2$is,

$$\begin{gathered} n=\frac{x+2x^2}{1+x+x^2} \\ x=\exp \left(\frac{-(蔚-渭)}{kT}\right) \end{gathered}$$

The graph is as follows,