Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

**Show that P(E), the occupancy probability in Eq. 41-6, is symmetrical about the value of the Fermi energy; that is, show that $P (E_{F} + Δ E) + P (E_{F} - Δ E) = 1$ .**

It is shown that $P (E_{F} + Δ E) + P (E_{F} - Δ E) = 1$ that is the occupancy probability is symmetrical about the value of the Fermi energy.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Understanding the concept of occupancy probability

The probability of a state to be occupied by and electron is referred to as the occupancy probability. At 0 K temperature, the states below the Fermi level have occupancy probability equal to and for the states above the Fermi level, its value is .

Formula:

The occupancy probability of the state with energy E is-

$P (E) = \frac{1}{e^{(E - E_{F}) / K T} + 1}$ ( i )

Here $E_{F}$ is the Fermi energy, $k = 8.62 \times 10^{- 5} e V / K$ and T is the absolute temperature.

Step 2: Calculation of the given symmetrical condition of probability

Upon expansion in view of equation (i), the LHS value of the given equation can be solved as follows:

$L H S = P (E_{F} + ∆ E) + P (E_{F} - ∆ E) = \frac{1}{e^{(E_{F} + ∆ E - E_{F}) / K T} + 1} + \frac{1}{e^{(E_{F} - ∆ E - E_{F}) / K T} + 1} = \frac{1}{e^{∆ E / K T} + 1} + \frac{1}{e^{- ∆ E / / K T} + 1} = \frac{e^{∆ E / K T} + 1 + e^{∆ E / K T} + 1}{(e^{∆ E / K T} + 1) (e^{- ∆ E / K T} + 1)}$

On further solving,

$L . H . S = \frac{e^{∆ E / K T} + 1 + e^{∆ E / K T} + 2}{e^{∆ E / K T} + e^{∆ E / K T} + 2} = 1 = R . H . S$

Hence, the given condition is proved and this implies the symmetrical condition for occupancy probability.

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Fundamentals Of Physics

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Short Answer

**Show that P(E), the occupancy probability in Eq. 41-6, is symmetrical about the value of the Fermi energy; that is, show that $P (E_{F} + Δ E) + P (E_{F} - Δ E) = 1$ .**

Step by Step Solution

Step 1: Understanding the concept of occupancy probability

Step 2: Calculation of the given symmetrical condition of probability

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Fundamentals Of Physics

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Short Answer

Show that P(E), the occupancy probability in Eq. 41-6, is symmetrical about the value of the Fermi energy; that is, show that P(EF+ΔE)+P(EF-ΔE)=1.

Step by Step Solution

Step 1: Understanding the concept of occupancy probability

Step 2: Calculation of the given symmetrical condition of probability

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**Show that P(E), the occupancy probability in Eq. 41-6, is symmetrical about the value of the Fermi energy; that is, show that $P (E_{F} + Δ E) + P (E_{F} - Δ E) = 1$ .**