Q. 7.13 - An Introduction to Thermal Physics

Question

For a system of bosons at room temperature, compute the average occupancy of a single-particle state and the probability of the state containing $0,1,2,3$ bosons, if the energy of the state is 

(a) $0.001\mathrm{eV}$ greater than $\mu$

(b) $0.01\mathrm{eV}$ greater than $\mu$

(c)  $0.1\mathrm{eV}$ greater than $\mu$

 (d) $1\mathrm{eV}$greater than $\mu$

Step-by-Step Solution

Answer

Result is:

The energuy state is: (a).$\overline{n}=25.2$

The energuy state is: (b).$\overline{n}=2.10$

The energuy state is: (c).$\overline{n}=0.0208$

The energuy state is: (d).$\overline{n}=1.244\times {10}^{-17}$

Part(a) Step 1:Given information

We have been given that $\overline{n}=\frac{1}{e^{(ϵ-\mu )/kT}-1}$

Part(a) Step 2: Simplify

Here

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

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

to find the average occupancy:

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

$\overline{n}=25.2$

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 $n=3$

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

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

Part(b) Step 1:Given information

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

Part(b) Step 2: Simplify

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 $n=1$

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

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

Part(c) Step 1: Given information

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

Part(c) Step 2: Simplify

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 $n=1$

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

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

Part(d) Step 1:Given information

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

Part(d) Step 2: Simplify

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 $n=1$

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

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