Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

The Fermi energy of aluminum is 11.6 eV; its density and molar mass are $2.70 g / {cm}^{3}$ and $2.70 g / mol$ , respectively. From these data, determine the number of conduction electrons per atom.

The number of conduction electrons contributed by an atom of aluminum is 3.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

a) Fermi energy of aluminum, $E_{F} = 11.6 e V$

b) Density of aluminum, $d = 2.7 g / {cm}^{3}$

c) Molar mass of aluminum, $A = 27 g / mol$

Step 2: Understanding the concept of conduction electrons per atom

The electrons that jump from the valence band to the conduction band, by absorbing energy from the surrounding, are called conduction electrons. These electrons are now free to move within the walls of the sample of a substance.

Formulae:

The Energy of Fermi level of a metal, $E_{F} = A n^{2 / 3} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)$

where, is the number density of conduction electrons and $A = 3.65 \times 10^{- 19} m^{2} e V$

The number of atoms per unit volume, $N = d / M \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)$

Where, d = density and M is the mass per atom $M = A / N_{A} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)$

The mass of a substance per atom,

$w h e r e N_{A} = 6.022 \times 10^{23} / m o l$ and A is the molar mass.

Step 3: Calculation of the number of conduction electrons per atom

Let N be the number of atoms per unit volume and n be the number of conduction electrons per unit volume.

At first, the number of conduction electrons per unit volume of aluminum can be calculated using equation as follows:

$n = {(\frac{E_{F}}{A})}^{3 / 2} = {(\frac{11.6 e V}{3.65 \times 10^{- 19} m^{2} e V})}^{3 / 2} = 1.79 \times 10^{29} m^{- 3} . . . . . . . . . . . . . . . . . . . . . . . . (a)$

Now, the mass of aluminum per atom can be calculated using equation and the given data, as follows:

$M = \frac{27 g / m o l}{6.022 \times 10^{23} / m o l} = 4.48 \times 10^{- 23} g$

Now, using this mass value in equation , we can get the value of the number of atoms per unit volume as follows:

$N = \frac{2.7 g / c m^{3}}{4.48 \times 0^{- 23} g} = 6.03 \times 10^{22} / c m^{3} = 6.03 \times 10^{28} / m^{3} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (b)$

Now, the number of conduction electrons per atom can be calculated by dividing equation (a) by equation (b) as follows:

$\frac{n}{N} = \frac{1.79 \times 10^{29} m^{- 3}}{6.03 \times 10^{28} / m^{3}} = 2.97 \approx 3$

Hence, the required number of conduction electrons per atom is 3.

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Fundamentals Of Physics

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Short Answer

The Fermi energy of aluminum is 11.6 eV; its density and molar mass are $2.70 g / {cm}^{3}$ and $2.70 g / mol$ , respectively. From these data, determine the number of conduction electrons per atom.

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of conduction electrons per atom

Step 3: Calculation of the number of conduction electrons per atom

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Fundamentals Of Physics

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Short Answer

The Fermi energy of aluminum is 11.6 eV; its density and molar mass are2.70 g/cm3 and 2.70 g/mol, respectively. From these data, determine the number of conduction electrons per atom.

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of conduction electrons per atom

Step 3: Calculation of the number of conduction electrons per atom

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The Fermi energy of aluminum is 11.6 eV; its density and molar mass are $2.70 g / {cm}^{3}$ and $2.70 g / mol$ , respectively. From these data, determine the number of conduction electrons per atom.