91Ó°ÊÓ

$\mathrm{Î\mu }-\mathrm{μ}=x$

Substitute $-1eV$for $x$in the equation $\mathrm{Î\mu }-\mathrm{μ}=x$.

$$\begin{gathered} \mathrm{Î\mu }=\mathrm{μ}-1eV \\ \mathrm{Î\mu }-\mathrm{μ}=-1eV \end{gathered}$$

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 $\mathrm{μ}$as follows:

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

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

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

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

$\mathrm{Î\mu }-\mathrm{μ}=x$

Substitute $-0.01eV$for $x$in the equation $\mathrm{Î\mu }-\mathrm{μ}=x$.

$$\begin{gathered} \mathrm{Î\mu }=\mathrm{μ}-0.01eV \\ \mathrm{Î\mu }-\mathrm{μ}=-0.01eV \end{gathered}$$

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 $\mathrm{μ}$as follows:

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

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

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

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

$\mathrm{Î\mu }-\mathrm{μ}=x$

Substitute $0eV$for $x$in the equation $\mathrm{Î\mu }-\mathrm{μ}=x$.

$\mathrm{Î\mu }=\mathrm{μ}-0$

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 $\mathrm{μ}$as follows:

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

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

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

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

$\mathrm{Î\mu }-\mathrm{μ}=x$

Substitute $0.01eV$for $x$in the equation role="math" localid="1647056653513" $\mathrm{Î\mu }-\mathrm{μ}=x$

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

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 $\mathrm{μ}$as follows:

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

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

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

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

$\mathrm{Î\mu }-\mathrm{μ}=x$

Substitute $1eV$for $x$in the equation$\mathrm{Î\mu }-\mathrm{μ}=x$

$$\begin{gathered} \mathrm{Î\mu }=\mathrm{μ}+1eV \\ \mathrm{Î\mu }-\mathrm{μ}=1eV \end{gathered}$$

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 $\mathrm{μ}$as follows:

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

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

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

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