Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Show that Eq. 41-5 can be written as $N (E) = C E^{1 / 2}$ . (b) Evaluate in terms of meters and electron-volts. (c) Calculate $N (E) for E = 5.00 eV$ for .

The density of the states of conduction electrons is $N (E) = C E^{1 / 2}$ .
The value of the constant in meters and electron-volts is $6.81 \times 10^{27} m^{- 3} . {(eV)}^{- 2 / 3}$ .
The density of states for is $E = 5 eV is 1.52 \times 10^{28} {eV}^{- 1} . m^{- 3}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

The given energy value, $E = 5 e V$

Step 2: Understanding the concept of density of states

Using the formula of the density of states associated with the conduction electrons of metal, we can get the required value of the density of the state for the given energy and constant values.

$N (E) = \frac{8 \sqrt{2} {ττm}^{3 / 2}}{h^{3}} E^{1 / 2}$

Step 3: a) Calculation of the formula of density of states

From the given equation of density of states of the conduction electrons of a metal, we get that $N (E) = C E^{1 / 2} . . . . . . . . . . . . . . . . . . . . (a)$

where

$C = \frac{8 \sqrt{2} {πm}^{3 / 2}}{h^{3}} = \frac{8 \sqrt{2} π {(9.1 \times 10^{- 31} kg)}^{3 / 2}}{{(6.63 \times 10^{- 34} J . s)}^{3}} = 1062 \times 10^{56} {kg}^{3 / 2} J^{3} . s^{3}$

Hence, the density of the states of conduction electrons is $N (E) = C E^{1 / 2}$ .

Step 4: b) Calculation of the value of constant C in meters and electron-volts

We know that $1 J = 1 kg . m^{2} / s^{2}$

So, considering the equation of kinetic energy $(K = \frac{1}{2} m v^{2})$ the unit of mass is $1 kg = 1 J . s^{2} m^{- 2}$ .

Thus, the units of C becomes-

${(J . s^{2})}^{3 / 2} . {(m^{- 2})}^{3 / 2} . J^{- 3} . s^{- 3} = J^{- 3 / 2} m^{- 3}$

Now, the value of C in meters and electron-volts can be given as:

$C = (1.062 \times 10^{56} {kg}^{3 / 2} / J^{3} . s^{3}) {(1.602 \times 10^{- 19} J / eV)}^{3 / 2} = 6.81 \times 10^{27} m^{- 3} . {(eV)}^{- 2 / 3}$

Hence, the value of C is $6.81 \times 10^{27} m^{- 3} . {(eV)}^{- 2 / 3}$ .

Step 5: c) Calculation of the density of states for the given energy

Using the given energy value in equation (a) of density in part (a), we can get the density of the states of the conduction electrons of the metal as follows:

$N (E) = (6.81 \times 10^{27} m^{- 3} . {(eV)}^{- 2 / 3} {(5 eV)}^{1 / 2}) = 1.52 \times 10^{28} {eV}^{- 1} . m^{- 3}$

Hence, the value of the density of electrons is $1.52 \times 10^{28} {eV}^{- 1} . m^{- 3}$ .

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Fundamentals Of Physics

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Short Answer

Show that Eq. 41-5 can be written as $N (E) = C E^{1 / 2}$ . (b) Evaluate in terms of meters and electron-volts. (c) Calculate $N (E) for E = 5.00 eV$ for .

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of density of states

Step 3: a) Calculation of the formula of density of states

Step 4: b) Calculation of the value of constant C in meters and electron-volts

Step 5: c) Calculation of the density of states for the given energy

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Fundamentals Of Physics

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Short Answer

Show that Eq. 41-5 can be written as N(E)=CE1/2. (b) Evaluate in terms of meters and electron-volts. (c) Calculate N(E) for E=5.00eV for .

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of density of states

Step 3: a) Calculation of the formula of density of states

Step 4: b) Calculation of the value of constant C in meters and electron-volts

Step 5: c) Calculation of the density of states for the given energy

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Show that Eq. 41-5 can be written as $N (E) = C E^{1 / 2}$ . (b) Evaluate in terms of meters and electron-volts. (c) Calculate $N (E) for E = 5.00 eV$ for .