Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Calculate $N_{0} (E)$ the density of occupied states, for copper at 10000K for energy E of (a)4.00eV , (b) 6.75eV, (c) 7.00eV, (d) 7.25eV, and (e) 9.00eV. Compare your results with the graph of Fig. 41-8b. The Fermi energy for copper is 7.00eV.

a) The density of occupied states for energy $4 eV is 1.36 \times 10^{28} m^{- 3} . {eV}^{- 1}$ .

b) The density of occupied states for energy $6.75 eV is 1.68 \times 10^{28} m^{- 3} . {eV}^{- 1}$ .

c) The density of occupied states for energy $7 eV is 9.01 \times 10^{28} m^{- 3} . {eV}^{- 1}$ .

d) The density of occupied states for energy $7.25 eV is 9.56 \times 10^{28} m^{- 3} . {eV}^{- 1}$ .

e) The density of occupied states for energy $9 eV is 1.71 \times 10^{28} m^{- 3} . {eV}^{- 1}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

a) The Fermi energy for copper, $E_{F} = 7 eV$

b) The temperature of the energy state of copper, E = 1000 K

c) The energy states are 4 eV , 6.75 eV , 7 eV ,7.25 eV and 9 eV .

Step 2: Understanding the concept of density of occupied states

Using the concept of probability occupancy and the density of states, the required value of the densities of the conduction electrons for copper at different energy levels can be calculated substituting the given data.

Formulae:

The density of states associated with the conductions electrons of a metal $N (E) = {CE}^{1 / 2}, where C = \frac{8 \sqrt{2} {ττm}^{3 / 2}}{h^{3}}$ (i)

The probability of the condition that a particle will have energy E, (ii) $P (E) = \frac{1}{e^{(E - E_{F}) / kT} + 1}, where k = 8.62 \times 10^{- 5} eV$

The density of states for the condition of occupancy by the conduction electrons,

$N_{0} (E) = N (E) P (E)$ (iii)

Step 3: a) Calculation of the density of occupied states for 4 eV

At first, the density of the occupied states can be given using equations (i) and (ii) in equation (iii) can be given as follows:

$N_{0} (E) = C E^{1 / 2} {[e^{(E - E_{F}) / k T} + 1]}^{- 1} . . . . . . . . . . . . . . . . . . . . . . (a)$

where,

$C = \frac{8 \sqrt{2} π m^{3 / 2}}{h^{3}} = \frac{8 \sqrt{2} π {(9.1 \times 10^{- 31} kg)}^{3 / 2}}{{(6.63 \times 10^{- 34} J . s)}^{3}} = 1.062 \times 10^{56} {kg}^{3 / 2} J^{3} . s^{3} = 6.81 \times 10^{27} m^{- 3} . {(eV)}^{- 2 / 3}$

Thus, the density of occupied states at E = 4eV is given using equation (a) as follows:

role="math" localid="1661929491044" $N_{0} (E) = \frac{(6.81 \times 10^{27} m^{- 3} . {(eV)}^{- 2 / 3}) {(4 eV)}^{1 / 2}}{[\exp ((4 eV - 7 eV) / (8.62 \times 10^{- 5} eV) (1000 K)) + 1]} = 1.36 \times 10^{28} m^{- 3} . {eV}^{- 1}$

Hence, the value of the density of states is $1.36 \times 10^{28} m^{- 3} . {eV}^{- 1}$ .

Step 4: b) Calculation of the density of occupied states for 6.75 eV

Thus, the density of occupied states at E = 6.75 eV is given using equation (a) as follows:

$N_{0} (E) = \frac{(6.81 \times 10^{27} m^{- 3} . {(eV)}^{- 2 / 3}) {(6.75 eV)}^{1 / 2}}{[\exp ((6.75 eV - 7 eV) / (8.62 \times 10^{- 5} eV) (1000 K)) + 1]} = 1.68 \times 10^{28} m^{- 3} . {eV}^{- 1}$

Hence, the value of the density of states is $1.68 \times 10^{28} m^{- 3} . {eV}^{- 1}$ .

Step 5: c) Calculation of the density of occupied states for 7 eV

Thus, the density of occupied states at E = 7 eV is given using equation (a) as follows:

$N_{0} (E) = \frac{(6.81 \times 10^{27} m^{- 3} . {(eV)}^{- 2 / 3}) {(7 eV)}^{1 / 2}}{[\exp ((7 eV - 7 eV) / (8.62 \times 10^{- 5} eV) (1000 K)) + 1]} = 9.01 \times 10^{28} m^{- 3} . {eV}^{- 1}$

Hence, the value of the density of states is $9.01 \times 10^{28} m^{- 3} . {eV}^{- 1}$ .

Step 6: d) Calculation of the density of occupied states for 7.25 eV

Thus, the density of occupied states at E = 7.25 eV is given using equation (a) as follows:

$N_{0} (E) = \frac{(6.81 \times 10^{27} m^{- 3} . {(eV)}^{- 2 / 3}) {(7.25 eV)}^{1 / 2}}{[\exp ((7.25 eV - 7 eV) / (8.62 \times 10^{- 5} eV) (1000 K)) + 1]} = 9.56 \times 10^{28} m^{- 3} . {eV}^{- 1}$

Hence, the value of the density of states is $9.56 \times 10^{28} m^{- 3} . {eV}^{- 1}$ .

Step 7: e) Calculation of the density of occupied states for 9 eV

Thus, the density of occupied states at E = 9 eV is given using equation (a) as follows:

$N_{0} (E) = \frac{(6.81 \times 10^{27} m^{- 3} . {(eV)}^{- 2 / 3}) {(9 eV)}^{1 / 2}}{[\exp ((9 eV - 7 eV) / (8.62 \times 10^{- 5} eV) (1000 K)) + 1]} = 1.71 \times 10^{28} m^{- 3} . {eV}^{- 1}$

Hence, the value of the density of states is $1.71 \times 10^{28} m^{- 3} . {eV}^{- 1}$ .

All the values are compared to the graph and consistent with the graph values.

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Fundamentals Of Physics

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Short Answer

Calculate $N_{0} (E)$ the density of occupied states, for copper at 10000K for energy E of (a)4.00eV , (b) 6.75eV, (c) 7.00eV, (d) 7.25eV, and (e) 9.00eV. Compare your results with the graph of Fig. 41-8b. The Fermi energy for copper is 7.00eV.

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of density of occupied states

Step 3: a) Calculation of the density of occupied states for 4 eV

Step 4: b) Calculation of the density of occupied states for 6.75 eV

Step 5: c) Calculation of the density of occupied states for 7 eV

Step 6: d) Calculation of the density of occupied states for 7.25 eV

Step 7: e) Calculation of the density of occupied states for 9 eV

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Fundamentals Of Physics

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Short Answer

Calculate N0(E) the density of occupied states, for copper at 10000K for energy E of (a)4.00eV , (b) 6.75eV, (c) 7.00eV, (d) 7.25eV, and (e) 9.00eV. Compare your results with the graph of Fig. 41-8b. The Fermi energy for copper is 7.00eV.

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of density of occupied states

Step 3: a) Calculation of the density of occupied states for 4 eV

Step 4: b) Calculation of the density of occupied states for 6.75 eV

Step 5: c) Calculation of the density of occupied states for 7 eV

Step 6: d) Calculation of the density of occupied states for 7.25 eV

Step 7: e) Calculation of the density of occupied states for 9 eV

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Calculate $N_{0} (E)$ the density of occupied states, for copper at 10000K for energy E of (a)4.00eV , (b) 6.75eV, (c) 7.00eV, (d) 7.25eV, and (e) 9.00eV. Compare your results with the graph of Fig. 41-8b. The Fermi energy for copper is 7.00eV.