91Ó°ÊÓ

We have been given that $\stackrel{¯}{n}=\frac{1}{e^{(\mathrm{Ï\mu }-\mathrm{μ})/kT}-1}$

Here

$\mathcal{P}\left(n\right)=e^{-n(\mathrm{Ï\mu }-\mathrm{μ})/kT}\left(1-e^{-(\mathrm{Ï\mu }-\mathrm{μ})/kT}\right)$

$kT=\left(8.62×{10}^{-5}\mathrm{eV}/\mathrm{K}\right)(298\mathrm{K})=0.02569\mathrm{eV}$

to find the average occupancy:

$\stackrel{¯}{n}=\frac{1}{e^{0.001\mathrm{eV}/0.02569\mathrm{eV}-1}}$

$\stackrel{¯}{n}=25.2$

if it contains $n=1$

$\mathcal{P}\left(1\right)=e^{-\left(1\right)(0.001\mathrm{eV})/0.02569\mathrm{eV}}\left(1-e^{-0.001\mathrm{eV}/0.02569\mathrm{eV}}\right)$

$\mathcal{P}\left(1\right)=0.03672$

if it contains $n=3$

$\mathcal{P}\left(3\right)=e^{-\left(3\right)(0.001\mathrm{eV})/0.02569\mathrm{eV}}\left(1-e^{-0.001\mathrm{eV}/0.02569\mathrm{eV}}\right)$

$\mathcal{P}\left(3\right)=0.034$

We have been given that $\mathrm{Ï\mu }-\mathrm{μ}=0.01\mathrm{eV}$

If it contain $n=0$

$\mathcal{P}\left(0\right)=e^{-\left(0\right)(0.01\mathrm{eV})/0.02569\mathrm{eV}}\left(1-e^{-0.01\mathrm{eV}/0.02569\mathrm{eV}}\right)$

$\mathcal{P}\left(0\right)=0.322$

If it contains $n=1$

$\mathcal{P}\left(1\right)=e^{-\left(1\right)(0.01\mathrm{eV})/0.02569\mathrm{eV}}\left(1-e^{-0.01\mathrm{eV}/0.02569\mathrm{eV}}\right)$

$\mathcal{P}\left(1\right)=0.218$

We have been given that $\mathrm{Ï\mu }-\mathrm{μ}=0.1\mathrm{eV}$

If it contains $n=0$

$\mathcal{P}\left(0\right)=e^{-\left(0\right)(0.1\mathrm{eV})/0.02569\mathrm{eV}}\left(1-e^{-0.1\mathrm{eV}/0.02569\mathrm{eV}}\right)$

$\mathcal{P}\left(0\right)=0.9796$

If it contains $n=1$

$\mathcal{P}\left(1\right)=e^{-\left(1\right)(0.1\mathrm{eV})/0.02569\mathrm{eV}}\left(1-e^{-0.1\mathrm{eV}/0.02569\mathrm{eV}}\right)$

$\mathcal{P}\left(1\right)=0.020$

We have been given that $\mathrm{Ï\mu }-\mathrm{μ}=1\mathrm{eV}$

If it contains $n=0$

$\mathcal{P}\left(0\right)=e^{-\left(0\right)\left(1\mathrm{eV}\right)/0.02569\mathrm{eV}}\left(1-e^{-1\mathrm{eV}/0.02569\mathrm{eV}}\right)$

$\mathcal{P}\left(0\right)=1$

If it contains $n=1$

$\mathcal{P}\left(1\right)=e^{-\left(1\right)\left(1\mathrm{eV}\right)/0.02569\mathrm{eV}}\left(1-e^{-1\mathrm{eV}/0.02569\mathrm{eV}}\right)$

$\mathcal{P}\left(1\right)=1.244×{10}^{-17}$