# Metals

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**Metals**

The feature that defines a metal is that, as shows, ~ highest occupied energy level falls Somewhere nellie “middle of an .energy band. If we’ apply a potential difference across a-metal, a current can exist because there are plenty of vacant levels at nearby higher energies into which electrons (the charge carriers in a metal) can jump. Thus, a metal can conduct electricity because ‘electrons in its highest occupied band can easily move into higher energy levels within that band.In Section 27-6 we introduced the free-electron model of a metal, in which the conduction electrons are free to move throughout the volume of the sample like the molecules of a gas in a closed container. We used this model to derive an expression for the resistivity of a metal, assuming that the electrons follow the laws

of Newtonian mechanics. Here we use that same model to explain the behavior of the electrons-called the conduction electrons-in the partially filled band of Fig. 2-4b. However, we follow the laws of quantum physics by assuming the energies of these electrons to be quantized and the Pauli exclusion principle to hold. e assume too that the electric potential energy of a conduction electron has the same constant value at all points within the lattice. If we choose this value’ of the potential energy to be zero, as we are free to do, then the mechanical energy E of the conduction electrons is entirely kinetic. The level at the bottom of the partially filled band of Fig. 42-4b corresponds to

E = O. The highest occupied level in this band at absolute zero (T = 0 K) is called the Fermi level, and the energy corresponding to it is called the Fermi energy EF; for copper, EF = 7.0 eV. The electron speed corresponding to the Fermi energy is called the Fermi speed VF’ For copper the Fermi speed is 1.6 X 106 m/s. This fact should.be enough to shatter the popular misconception that all motion ceases at absolute zero; at that temperature-and solely because of the Pauli exclusion principle-the conduction electrons an: stacked up in the partially filled band of Fig. 42-4b with energies that range from zero to the Fermi energy.

**How Many Conduction Electrons Are There?**

If we could bring individual atoms together to form a sample of a metal, we would find that the conduction electrons in the metal are the valence electrons of the atoms (the electrons in the outer shells of the individual atoms). A monovalent atom contributes one such electron to the conduction electrons in a metal a bivalent atom contributes two such electrons. Thus, the total number of conduction electrons

(In this chapter, we shall write several equations largely in words because the symbols we have previously used for the quantities in them now represent other quantities.) The number density n of conduction electrons in a sample is the number of conduction electrons per unit volume:

We can relate the number of atoms in a sample to various other properties of the sample and the material making up the sample with the following equation.

where the molar mass is the,ass of one mole of the material in the sample and NA is Avogadro’s number (6.02;k 1023 mol .

**Sample Problem**

How many conduction electrons are in a cube of magnesium with a volume of 2.00 x 10 -6 m3? Magnesium atoms are bivalent,

**SOLUTION:** The Ke, Ideas here are these: (1.738 X 103 kg/m3)(2.00 X 10-6 m3)(6.02 X 1023mol “) = 2.0926 X 1021kg/mol.

I. Because magnesium atoms are bivalent, each magnesium atom contributes two conduction electrons.

2. The number of conduction electrons in the cube is related to the number of magnesium atoms in the cube by Eq. 42-2.

3. We can find the number of atoms with Eq. 42-4 and known data about the cube’s volume and magnesium’s properties

We can write Eq. 42-4 as

**Conductivity at T> 0**

Our practical interest in the conduction of electricity in metals is at temperatures above absolute zero. What happens to the electron distribution of at such higher temperatures? As we shall see, surprisingly little. Of the electrons in the partially filled band of only those that are close to the Fermi energy find unoccupied levels above them, and only those electrons are free to be boosted to these higher levels by thermal agitation. Even at T = 1000 K, a temperature at which copper would glow brightly in a dark room, the distribution of electrons among the available levels does not differ much from the distribution at T.= 0 K.Let us see why. The quantity kT, where k is the Boltzmann constant, is a convenient measure of the energy that may be given to a conduction electron by the random thermal motions of the lattice. At T ~ 1000 K, we have kT = 0.086 eY. No electron can hope to have its energy changed by more than a few times this relatively small amount by thermal agitation alone, so at best, only those few conduction electrons whose energies are close to the Fermi energy are likely to jump to higher energy levels due to’ thermal agitation. Poetically stated, thermal agitation normally causes only ripples on the surface of the Fermi sea of electrons; the vast depths of that sea lie undisturbed.

**How Many Quantum States Are There?**

The ability of a metal to conduct electricity depends on how many quantum states are available to its electrons and what the energies of those states are. Thus, a question arises: What are the energies of the individual states in the partially filled band of This question is tQOdifficult to answer because we cannot possibly list the energies of so many states individually. We ask instead: How many states in a unit volume of a sample have energies in the energy range E to E + dE? We write this number as N(E) dE, where N(E) is called the density of states at energy E. The conventional unit for N(E) dE is states per cubic meter (states/m”, or simply m=’): the unit for N(E) is states per cubic meter per electron-volt (m ? ey-t).We can find an expression for the density of states by counting the number of standing electron matter waves that can fit into a box the size of the metal sample we are considering. This is analogous to counting the number of standing waves of sound that can exist in a closed organ pipe. The differences are that our problem is three-dimensional (the organ pipe problem is one-dimensional) and the waves are atter waves (the organ-pipe waves are sound waves). The result of such counting can be shown to be

CHECKPOINT : (a) Is the spacing between adjacent energy levels at E = 4 eV in cop- , per larger than, the same as, or smaller than the spacing at E = 6 eV? (b) Is the spacing between adjacent energy levels at E = 4 eV in copper larger than, the same as, or smaller than the spacing for an identical volume of aluminum at that same energy?

**Sample Problem**

**The Occupancy Probability P(**E}

The ability of a metal to conduct electricity depends on the probability that available vacant levels will actually be occupied. Thus, another question arises: If an energy level is available at energy E, what is the probability P(E) that it is actually occupied by an electron? At T = 0 K, we know that for all levels with energies below theFermi energy, P(E) = I, corresponding to a certainty that the level is occupied. We also know that, at T = 0 K, for all levels with energies above the Fermi energy, P(E) = 0, corresponding to a certainty that the level is not occupied. illustrates this situation. To find P(E) at temperatures above absolute zero, we must use a set of quantum counting rules called Fermi – Dirac statistics, named for the physicists who introduced them. Using these rules, it is possible to show that the occupancy probabiUty P(E) is .

in which EF is the Fermi energy. Note that P(E) depends not on the energy E of the level but only on the difference E – EF’ which may be positive or negative. To see whether Eq. 42-6 describes , we substitute T = 0 K in it. Then, For E < EF’ the exponential term in Eq. 42-6 is e:”, or zero, so P(E) = I, in agreement with For E > EF’ the exponential term is e+”‘, so P(E) = 0, again in agreement.Figure 42-6b is a plot of P(E) for T = 1000 K. It shows that, as stated above, changes in the distribution of electrons among the available states involve only states energies are near the Fermi energy EF. Note that if E = EF (no matter what the temperature T), the exponential term in is eO = I and P(E) = 0.5. This leads us to a more useful definition of the Fermi energy:

**Sample Problem**