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Q.7.11

Expert-verifiedFound in: Page 269

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

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 $

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

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

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.

$\epsilon -\mu =x$

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

$\epsilon =\mu -1eV\phantom{\rule{0ex}{0ex}}\epsilon -\mu =-1eV$

The room temperature in kelvins is,

$T={27}^{o}C\phantom{\rule{0ex}{0ex}}=(27+273)K\phantom{\rule{0ex}{0ex}}=300K$

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

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

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

${\overline{)n}}_{FD}=\frac{1}{{e}^{\frac{-1eV}{(8.617\times {10}^{-3}eV/K)\left(300K\right)}}+1}\phantom{\rule{0ex}{0ex}}=0.9999\phantom{\rule{0ex}{0ex}}{\overline{)n}}_{FD}\approx 1.0$

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

$\epsilon -\mu =x$

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

$\epsilon =\mu -0.01eV\phantom{\rule{0ex}{0ex}}\epsilon -\mu =-0.01eV$

The room temperature in kelvins is,

$T={27}^{o}C\phantom{\rule{0ex}{0ex}}=(27+273)K\phantom{\rule{0ex}{0ex}}=300K$

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

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

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.

${\overline{)n}}_{FD}=\frac{1}{{e}^{\frac{-0.01eV}{(8.617\times {10}^{-5}eV/K)(300K)}}+1}\phantom{\rule{0ex}{0ex}}{\overline{)n}}_{FD}==0.5955$

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

$\epsilon -\mu =x$

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

$\epsilon =\mu -0$

The room temperature in kelvins is,

$T={27}^{o}C\phantom{\rule{0ex}{0ex}}=(27+273)K\phantom{\rule{0ex}{0ex}}=300K$

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" ${\overline{)n}}_{FD}=\frac{1}{{e}^{\frac{0eV}{(8.617\times {10}^{-5}eV/K)(300K)}}+1}\phantom{\rule{0ex}{0ex}}{\overline{)n}}_{FD}=0.5$

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

$\epsilon -\mu =x$

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

role="math" localid="1647056689711" $\epsilon =\mu +0.01eV\phantom{\rule{0ex}{0ex}}\epsilon -\mu =0.01eV$

The room temperature in kelvins is,

role="math" localid="1647056731594" $T={27}^{o}C\phantom{\rule{0ex}{0ex}}=(27+273)K\phantom{\rule{0ex}{0ex}}=300K$

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

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

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" ${\overline{)n}}_{FD}=\frac{1}{{e}^{\frac{0.01eV}{(8.617\times {10}^{-5}eV/K)(300K)}}+1}\phantom{\rule{0ex}{0ex}}{\overline{)n}}_{FD}=0.4045$

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

$\epsilon -\mu =x$

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

$\epsilon =\mu +1eV\phantom{\rule{0ex}{0ex}}\epsilon -\mu =1eV$

The room temperature in kelvins is,

$T={27}^{o}C\phantom{\rule{0ex}{0ex}}=(27+273)K\phantom{\rule{0ex}{0ex}}=300K$

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

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

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" ${\overline{)n}}_{FD}=\frac{1}{{e}^{\frac{1eV}{(8.617\times {10}^{-5}eV/K)(300K)}}+1}\phantom{\rule{0ex}{0ex}}{\overline{)n}}_{FD}=1.5852\times {10}^{-17}$

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

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