Question

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

(a) $1eV$less than $\mu$

(b) $0.01eV$less than $\mu$

(c) equal to $\mu$

(d) $0.01eV$greater than $\mu$

(e) $1eV$greater than$\mu$

Short 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.

Step-by-step solution

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

$\epsilon -\mu =x$

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

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

The room temperature in kelvins is,

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

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

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

Substitute $-1eV$for $(\epsilon -\mu ),8.617\times {10}^{-5}eV/K$for $k$, and $300K$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$1eV$less than$\mu$is${\overline{n}}_{FD}\approx 1.0$

(b) Calculate the energy for the occupied state 0.01 eV less than μ as follows:

$\epsilon -\mu =x$

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

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

The room temperature in kelvins is,

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

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

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

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

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

Therefore, the probability of a state being occupied state$0.01eV$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 $0eV$for $x$in the equation $\epsilon -\mu =x$.

$\epsilon =\mu -0$

The room temperature in kelvins is,

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

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

role="math" localid="1647055630164" $$\begin{gathered} {\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{0eV}{(8.617\times {10}^{-5}eV/K)(300K)}}}+1} \\ {\overline{n}}_{FD}=0.5 \end{gathered}$$

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 eV greater than μ as follows:

$\epsilon -\mu =x$

Substitute $0.01eV$for $x$in the equation role="math" localid="1647056653513" $\epsilon -\mu =x$

role="math" localid="1647056689711" $$\begin{gathered} \epsilon =\mu +0.01eV \\ \epsilon -\mu =0.01eV \end{gathered}$$

The room temperature in kelvins is,

role="math" localid="1647056731594" $$\begin{gathered} T={27}^{o}C \\ =(27+273)K \\ =300K \end{gathered}$$

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

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

Substitute $0.01eV$for role="math" localid="1647056852727" $\left(\epsilon -\mu \right),8.617\times {10}^{-5}eV/K$for $k$, and $300K$for $T$in the above equation.

role="math" localid="1647056921933" $$\begin{gathered} {\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{0.01eV}{(8.617\times {10}^{-5}eV/K)(300K)}}}+1} \\ {\overline{n}}_{FD}=0.4045 \end{gathered}$$

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

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

$\epsilon -\mu =x$

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

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

The room temperature in kelvins is,

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

Calculate the probability of a state being occupied state role="math" localid="1647056827269" $1eV$less than $\mu$as follows:

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

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

role="math" localid="1647057000902" $$\begin{gathered} {\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \frac{1eV}{(8.617\times {10}^{-5}eV/K)(300K)}}}+1} \\ {\overline{n}}_{FD}=1.5852\times {10}^{-17} \end{gathered}$$

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