Q.7.11 - An Introduction to Thermal Physics

Question

For a system of fermions at room temperature, compute the probability of a single-particle state being occupied if its energy is

(a) $1 eV$ less than $\mu$

(b) $0.01 eV$ less than $\mu$

(c) equal to $\mu$

(d) $0.01 eV$ greater than $\mu$

(e) $1 eV$ greater than $\mu$

Step-by-Step Solution

Answer

According to the Fermi-Dirac distribution, the probability of a state being occupied is given below:

${\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{(\epsilon -\mu )}{kT}}}+1}$

Here, ${\overline{n}}_{FD}$ is the Fermi-Dirac distribution, $\epsilon$ is the energy, $\mu$ is the chemical potential, $k$ is the Boltzmann constant, and $T$ is the absolute temperature.

Formula to energy for the occupied state is given below:

$\epsilon -\mu =x$

Here $x$ is the energy of the state.

(a) Calculate the energy for the occupied state 1   e V less than than μ as follows:

$\epsilon -\mu =x$

Substitute $-1 eV$ for $x$ in the equation $\epsilon -\mu =x$.

$$\begin{gathered} \epsilon =\mu -1 eV \\ \epsilon -\mu =-1 eV \end{gathered}$$

The room temperature in kelvins is,

$$\begin{gathered} T={27}^{o}C \\ =(27+273)K \\ =300 K \end{gathered}$$

Calculate the probability of a state being occupied state $1 eV$less than $\mu$ as follows:

${\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{(\epsilon -\mu )}{kT}}}+1}$

Substitute $-1 eV$ for $(\epsilon -\mu ), 8.617\times {10}^{-5} eV/K$ for $k$, and $300 K$ for $T$ in the above equation.

$$\begin{gathered} {\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{-1eV}{(8.617\times {10}^{-3}eV/K)\left(300K\right)}}}+1} \\ =0.9999 \\ {\overline{n}}_{FD}\approx 1.0 \end{gathered}$$

Therefore, the probability of a state being occupied state $1 eV$ less than $\mu$ is ${\overline{n}}_{FD}\approx 1.0$

(b) Calculate the energy for the occupied state 0 . 01   e V less than μ as follows:

$\epsilon -\mu =x$

Substitute $-0.01 eV$ for $x$ in the equation $\epsilon -\mu =x$.

$$\begin{gathered} \epsilon =\mu -0.01 eV \\ \epsilon -\mu =-0.01 eV \end{gathered}$$

The room temperature in kelvins is,

$$\begin{gathered} T={27}^{o}C \\ =(27+273)K \\ =300 K \end{gathered}$$

Calculate the probability of a state being occupied state $0.01 eV$ less than $\mu$ as follows:

${\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{(\epsilon -\mu )}{kT}}}+1}$

Substitute $-0.01 eV$ for $\left(\epsilon -\mu \right)$, $8.617\times {10}^{-5} eV/K$ for $k$, and $300 K$ for $T$ in the above equation.

$$\begin{gathered} {\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{-0.01 eV}{(8.617\times {10}^{-5} eV/K)(300 K)}}}+1} \\ {\overline{n}}_{FD}==0.5955 \end{gathered}$$

Therefore, the probability of a state being occupied state $0.01 eV$ less than $\mu$ is ${\overline{n}}_{FD}=0.5955$

(c) Calculate the energy for the occupied state energy is equal to μ as follows:

$\epsilon -\mu =x$

Substitute $0 eV$ for $x$ in the equation $\epsilon -\mu =x$.

$\epsilon =\mu -0$

The room temperature in kelvins is,

$$\begin{gathered} T={27}^{o}C \\ =(27+273)K \\ =300 K \end{gathered}$$ 

Calculate the probability of a state being occupied state equal to $\mu$ as follows:

${\overline{n}}_{FD}=\frac{1}{e^{\frac{{}^{\left(\epsilon -\mu \right)}}{kT}}+1}$

Substitute $0 eV$ for $\left(\epsilon -\mu \right), 8.617\times {10}^{-5} eV/K$ for $k$, and $300 K$ for $T$ in the above equation.

$$\begin{gathered} {\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{0 eV}{(8.617\times {10}^{-5} eV/K)(300 K)}}}+1} \\ {\overline{n}}_{FD}=0.5 \end{gathered}$$

Therefore, the probability of a state being occupied state equal to $\mu$ is  ${\overline{n}}_{FD}=0.5$

(d) Calculate the energy for the occupied state 0 . 01   e V greater than μ as follows:

$\epsilon -\mu =x$

Substitute $0.01 eV$ for $x$ in the equation $\epsilon -\mu =x$

$$\begin{gathered} \epsilon =\mu +0.01 eV \\ \epsilon -\mu =0.01 eV \end{gathered}$$

The room temperature in kelvins is,

$$\begin{gathered} T={27}^{o}C \\ =(27+273) K \\ =300 K \end{gathered}$$

Calculate the probability of a state being occupied state $0.01 eV$ less than $\mu$ as follows:

${\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{{}^{\left(\epsilon -\mu \right)}}{kT}}}+1}$

Substitute $0.01 eV$ for $\left(\epsilon -\mu \right), 8.617\times {10}^{-5 }eV/K$ for $k$, and $300 K$ for $T$in the above equation.

$$\begin{gathered} {\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{0.01 eV}{(8.617\times {10}^{-5 }eV/K)(300 K)}}}+1} \\ {\overline{n}}_{FD}=0.4045 \end{gathered}$$

Therefore, the probability of a state being occupied state $0.01 eV$greater than $\mu$ is ${\overline{n}}_{FD}=0.4045$

(e) Calculate the energy for the occupied state 1   e V greater than μ as follows:

$\epsilon -\mu =x$

Substitute $1 eV$ for $x$ in the equation $\epsilon -\mu =x$

$$\begin{gathered} \epsilon =\mu +1 eV \\ \epsilon -\mu =1 eV \end{gathered}$$

The room temperature in kelvins is,

$$\begin{gathered} T={27}^{o}C \\ =(27+273) K \\ =300 K \end{gathered}$$

Calculate the probability of a state being occupied state $1 eV$ less than $\mu$ as follows:

${\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{{}^{\left(\epsilon -\mu \right)}}{kT}}}+1}$

Substitute $1 eV$ for $\left(\epsilon -\mu \right), 8.617\times {10}^{-5 }eV/K$ for $k$, and $300 K$ for $T$in the above equation.

$$\begin{gathered} {\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{1 eV}{(8.617\times {10}^{-5 }eV/K)(300 K)}}}+1} \\ {\overline{n}}_{FD}=1.5852 \times {10}^{-17} \end{gathered}$$

Therefore, the probability of a state being occupied state $1 eV$ greater than $\mu$ is ${\overline{n}}_{FD}=1.5852 \times {10}^{-17}$