Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In Eq. 41-6 let, $E - E_{F} = ∆ E = 1.00 eV$ . (a) At what temperature does the result of using this equation differ by 1% from the result of using the classical Boltzmann equation $P (E) = e^{- ∆ E / kT}$ (which is Eq. 41-1 with two changes in notation)? (b) At what temperature do the results from these two equations differ by 10%?

The temperature at which the result of the given equation differs 1% by from the result is 2520 K .
The temperature at which the results from these two equations differ by 10% is $5.28 \times 10^{3} K$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

The given equation, $E - E_{F} = ∆ E = 1.00 eV$

Step 2: Occupancy Probability

The probability that an electron will occupy a certain level, according to Fermi-Dirac statistics and Boltzmann's statistics, is known as occupancy probability. Fo the Fermi level the occupancy probability is 0.5.

The occupancy probability due to Fermi-Dirac statistics,

$P_{FD} = \frac{1}{e^{∆ E / kT} + 1} . . . . . . . . . . . . . . . . . . . . . . . (1)$

The Boltzmann occupation probability,

$P_{B} = e^{- ∆ E / kT} where k = 1.38 \times 10^{- 23} J / K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)$

Step 3: a) Calculation of the temperature of result differing by 1%

Let the fractional difference between the probability differences be f.

Thus, using equations (1) and (2), the fractional difference can be given as follows:

$f = \frac{P_{B} - P_{F D}}{P_{B}} = \frac{1 / e^{∆ E / k T} - 1 / (e^{∆ E / k T} + 1)}{e^{∆ E / k T}} = \frac{(e^{∆ E / k T} + 1) - e^{∆ E / k T}}{(e^{∆ E / k T} + 1) e^{∆ E / k T} . e^{∆ E / k T}} = \frac{1}{(e^{∆ E / k T} + 1)}$

The above equation can also be written as-

$e^{∆ E / k T} = \frac{1}{f} - 1$

For f = 0.01,

$e^{∆ E / k T} = \frac{1}{0.01} - 1 = 99$

Taking logarithm on both sides

$\frac{∆ E}{kT} = In (99) T = \frac{∆ E}{k In (99)} T = \frac{1.0 eV \times (1.6 \times 10^{- 19} J / eV)}{(1.38 \times 10^{- 23} J / K) \times 4.6} T = 2520 K$

Hence, the value of the temperature is 2520 K.

Step 4: b) Calculation of the temperature of result differing by 10%

If f = 0.1

Then, the temperature is given as-

$T = \frac{∆ E}{k In (99)} = \frac{(1.00 eV) (1.6 \times 10^{- 19} J / eV)}{(1.38 \times 10^{- 23} J / K) In (9)} = 5.28 \times 10^{3} K$

Hence, the value of the temperature is $5.28 \times 10^{3} K$ .

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Occupancy Probability

Step 3: a) Calculation of the temperature of result differing by 1%

Step 4: b) Calculation of the temperature of result differing by 10%

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