Book edition 1st

Author(s) Daniel V. Schroeder

Pages 356 pages

ISBN 9780201380279

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Short Answer

For a system of fermions at room temperature, compute the probability of a single-particle state being occupied if its energy is
(a) $1 e V$ less than $μ$
(b) $0.01 e V$ less than $μ$
(c) equal to $μ$
(d) $0.01 e V$ greater than $μ$
(e) $1 e V$ greater than $μ$

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

$n_{F D} = \frac{1}{e^{\frac{(ε - μ)}{k T}} + 1}$

Here, $n_{F D}$ is the Fermi-Dirac distribution, $ε$ is the energy, $μ$ 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:

$ε - μ = x$

Here $x$ is the energy of the state.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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

$ε - μ = x$

Substitute $- 1 e V$ for $x$ in the equation $ε - μ = x$ .

$ε = μ - 1 e V ε - μ = - 1 e V$

The room temperature in kelvins is,

$T = 27^{o} C = (27 + 273) K = 300 K$

Calculate the probability of a state being occupied state $1 e V$ less than $μ$ as follows:

$n_{F D} = \frac{1}{e^{\frac{(ε - μ)}{k T}} + 1}$

Substitute $- 1 e V$ for $(ε - μ), 8.617 \times 10^{- 5} e V / K$ for $k$ , and $300 K$ for $T$ in the above equation.

$n_{F D} = \frac{1}{e^{\frac{- 1 e V}{(8.617 \times 10^{- 3} e V / K) (300 K)}} + 1} = 0.9999 n_{F D} \approx 1.0$

Therefore, the probability of a state being occupied state $1 e V$ less than $μ$ is $n_{F D} \approx 1.0$

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

$ε - μ = x$

Substitute $- 0.01 e V$ for $x$ in the equation $ε - μ = x$ .

$ε = μ - 0.01 e V ε - μ = - 0.01 e V$

The room temperature in kelvins is,

$T = 27^{o} C = (27 + 273) K = 300 K$

Calculate the probability of a state being occupied state $0.01 e V$ less than $μ$ as follows:

$n_{F D} = \frac{1}{e^{\frac{(ε - μ)}{k T}} + 1}$

Substitute $- 0.01 e V$ for $(ε - μ)$ , $8.617 \times 10^{- 5} e V / K$ for $k$ , and $300 K$ for $T$ in the above equation.

$n_{F D} = \frac{1}{e^{\frac{- 0.01 e V}{(8.617 \times 10^{- 5} e V / K) (300 K)}} + 1} n_{F D} = = 0.5955$

Therefore, the probability of a state being occupied state $0.01 e V$ less than $μ$ is $n_{F D} = 0.5955$

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

$ε - μ = x$

Substitute $0 e V$ for $x$ in the equation $ε - μ = x$ .

$ε = μ - 0$

The room temperature in kelvins is,

$T = 27^{o} C = (27 + 273) K = 300 K$

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

$n_{F D} = \frac{1}{e^{\frac{^{(ε - μ)}}{k T}} + 1}$

Substitute $0 e V$ for $(ε - μ), 8.617 \times 10^{- 5} e V / K$ for $k$ , and $300 K$ for $T$ in the above equation.

role="math" localid="1647055630164" $n_{F D} = \frac{1}{e^{\frac{0 e V}{(8.617 \times 10^{- 5} e V / K) (300 K)}} + 1} n_{F D} = 0.5$

Therefore, the probability of a state being occupied state equal to $μ$ is $n_{F D} = 0.5$

(d) Calculate the energy for the occupied state 0.01 eV greater than μ as follows:

$ε - μ = x$

Substitute $0.01 e V$ for $x$ in the equation role="math" localid="1647056653513" $ε - μ = x$

role="math" localid="1647056689711" $ε = μ + 0.01 e V ε - μ = 0.01 e V$

The room temperature in kelvins is,

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

Calculate the probability of a state being occupied state $0.01 e V$ less than $μ$ as follows:

role="math" $n_{F D} = \frac{1}{e^{\frac{^{(ε - μ)}}{k T}} + 1}$

Substitute $0.01 e V$ for role="math" localid="1647056852727" $(ε - μ), 8.617 \times 10^{- 5} e V / K$ for $k$ , and $300 K$ for $T$ in the above equation.

role="math" localid="1647056921933" $n_{F D} = \frac{1}{e^{\frac{0.01 e V}{(8.617 \times 10^{- 5} e V / K) (300 K)}} + 1} n_{F D} = 0.4045$

Therefore, the probability of a state being occupied state $0.01 e V$ greater than $μ$ is $n_{F D} = 0.4045$

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

$ε - μ = x$

Substitute $1 e V$ for $x$ in the equation $ε - μ = x$

$ε = μ + 1 e V ε - μ = 1 e V$

The room temperature in kelvins is,

$T = 27^{o} C = (27 + 273) K = 300 K$

Calculate the probability of a state being occupied state role="math" localid="1647056827269" $1 e V$ less than $μ$ as follows:

$n_{F D} = \frac{1}{e^{\frac{^{(ε - μ)}}{k T}} + 1}$

Substitute $1 e V$ for $(ε - μ), 8.617 \times 10^{- 5} e V / K$ for $k$ , and $300 K$ for $T$ in the above equation.

role="math" localid="1647057000902" $n_{F D} = \frac{1}{e^{\frac{1 e V}{(8.617 \times 10^{- 5} e V / K) (300 K)}} + 1} n_{F D} = 1.5852 \times 10^{- 17}$

Therefore, the probability of a state being occupied state $1 e V$ greater than $μ$ is $n_{F D} = 1.5852 \times 10^{- 17}$

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Short Answer

For a system of fermions at room temperature, compute the probability of a single-particle state being occupied if its energy is
(a) $1 e V$ less than $μ$
(b) $0.01 e V$ less than $μ$
(c) equal to $μ$
(d) $0.01 e V$ greater than $μ$
(e) $1 e V$ greater than $μ$

Step by Step Solution

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

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

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

(d) Calculate the energy for the occupied state 0.01 eV greater than μ as follows:

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

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An Introduction to Thermal Physics

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Short Answer

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 μ(b) 0.01 eV less than μ(c) equal to μ(d) 0.01 eV greater than μ(e) 1 eV greater than μ

Step by Step Solution

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

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

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

(d) Calculate the energy for the occupied state 0.01 eV greater than μ as follows:

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

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Turning Points in Physics

For a system of fermions at room temperature, compute the probability of a single-particle state being occupied if its energy is
(a) $1 e V$ less than $μ$
(b) $0.01 e V$ less than $μ$
(c) equal to $μ$
(d) $0.01 e V$ greater than $μ$
(e) $1 e V$ greater than $μ$