In the case of the hydrogen molecule, the exchange energy is always negative. Equation 2. Therefore, they are free to migrate between the ion cores. These delocalized valence electrons are involved in the conduction of electricity and are therefore often called conduction electrons.
One can expect metals to form from elements for which the energy cost of removing outer electrons is not too big. Nevertheless, this removal always costs some energy that has to be more than compensated by the bonding. The ultimate reason must be some sort of energy lowering. One energy contribution that is lowered is the kinetic energy of the conduction electrons. For an electron that is localized to an atom, the curvature of the wave function is much higher than for a nearly free electron in a metal and this is where the energy gain comes from.
One should think that the average electrostatic potential of any single electron in a solid is almost zero because there are almost as many other electrons as there are ions with the same amount of charge. But this turns out to be wrong. In fact, the electrons see an attractive potential. The reason is again partly due to the Pauli principle that, loosely speaking, does not allow two electrons with the same spin direction to be at the same place see 2.
We will discuss this in more detail when dealing with magnetism. We can also understand why metals prefer close-packed structures. First of all, the metallic bonding does not have any directional preference. Second, closepacked structures secure the highest possible overlap between the valence orbitals of the atoms, maximizing the delocalization of the electrons and thereby the kinetic energy gain. The structures also maximize the number of nearest neighbors for any given atom, again giving rise to strongly delocalized states.
Typically, metallic bonding is not as strong as covalent or ionic bonding but it amounts to a few electron volts per atom. The explanation for this is that we have a mixed bonding. The s and p electrons turn into delocalized metallic conduction electrons, whereas the d electrons create much more localized, covalent-type bonds.
If the bond is formed with a very electronegative atom like F or O , the electron is mostly located close to that atom and the hydrogen nucleus represents an isolated positive partial charge. This can lead to a considerable charge density because of the small size, and it can therefore attract negative partial charges in other molecules to form an electrostatic bond. This type of bonding is called hydrogen bonding. It is usually quite weak but in some cases, the cohesive energy can be up to several hundred meV per atom. It is responsible for the intermolecular attraction in water ice and for the bonding of the double helix in DNA.
This type of interaction is present in every solid but it is much weaker than ionic, covalent, or metallic bonding. Typical binding energies per atom are in the meV range and, therefore, van der Waals bonding is only observable for solids that do not show other bonding behavior, for example, noble gases. Pure van der Waals crystals can only exist at very low temperatures. They are therefore discussed in great depth in the literature for chemistry and molecular physics. Which elements are likely to form crystals through ionic bonding? What kind of forces are important for ionic bonding?
How does the lattice energy in an ionic crystal depend on the interatomic distance? Which elements are likely to form metals? Where does the energy gain in metallic bonding come from? Why is van der Waals bonding much weaker than most other bonding types? Problems 1 Metallic bonding: The most important contribution to the stability gained by metallic bonding is the lowering of kinetic energy. To see this, consider an electron in a one-dimensional box. Give the result in electron volts. Clearly, this energy is only kinetic energy.
By how much is the kinetic energy lowered when you increase the size of the box by a factor of 10, so that it is roughly the size of the interatomic spacing in a crystal? Also, calculate the cohesive energy in the same units. Compare the result for the one-dimensional case to the analytical value from a. Suppose that one atom forms a spontaneous dipole moment at some time. This can be modeled as two point charges, separated by a distance d. Our discussion is restricted to isotropic solids, that is, solids for which the direction of the applied mechanical stress with respect to the crystal lattice is not important.
Depending on the force direction, one can distinguish between tensile and compressive stress. In the case of the tensile stress applied in Figure 3. It is therefore dimensionless, but in technical texts sometimes the unit meter per meter is found. In most cases, the solid contracts in these directions.
We shall discuss this in more detail below. Again, the material deforms as a consequence of the shear stress. The last situation illustrated in Figure 3. This leads to a reduction of the volume. If we consider only the relation between stress and strain, the typical response of a solid is illustrated in Figure 3. It shows the resulting stress as a function of applied strain. If you think of the strain as a consequence of the applied stress, you might be tempted to draw the curve with swapped axes.
Note that the dark shaded areas represent the area A and that the force is perpendicular and parallel to A in a , c , and b , respectively. The dashed line with the arrows illustrates the behavior when the stress is released and reapplied after increasing the strain into the work hardening region. This means that the deformation is permanent; once the stress is released, the solid does not return to its original shape.
It only contracts slightly. Eventually, the strain becomes so high that the material fractures.
While the region of elastic deformation is usually quite small, the amount of possible plastic deformation can vary widely. Some materials, such as glass or cast iron, will fracture immediately at the end point of the elastic limit. Such materials 3. Materials that do show plastic deformation before they fracture are called ductile. Most metals are ductile.
Apart from exploring the limits of elastic deformation, an interesting question is how strongly the material resists such a deformation. This is described by the macroscopic elastic constants that we shall introduce now. We will also see that these constants can be connected to the picture of interatomic bonding that we have encountered earlier. Suppose you extend a spring by some small amount.
It is going to respond by a force that is proportional to the extension. The advantage of using Y instead of the spring constant is that it depends only on the material, not on the geometry. The shearing of a solid can also be described by an elastic constant. The minus sign is introduced in order to obtain a positive bulk modulus for a decrease in volume. Note that both G and K have the unit Pascal, just like Y.
The upper limit is caused by the fact that a solid cannot decrease its volume when we pull on one side and that it cannot increase its volume when we press it from one side. So the change in volume is 3. Not surprisingly, they are therefore related to each other. We will, in most cases, restrict ourselves to explaining one type of mechanical property and we are allowed to do so without great loss of generality. In any case, we see from 3. According to Figure 2. Upon the application of a compressive stress, the distance between the atoms is decreased.
This results in a force that presses the atoms away from each other. For tensile stress, it is the other way round. Once the stress is released, the atoms return to their equilibrium distance. This explains why the behavior is elastic, but why is it linear? A linear force for distance changes close to the equilibrium can readily be seen in Figure 2. The third term is responsible for the elastic behavior.
It states that the potential close to the equilibrium is proportional to the square of the distance change, that is, the force depends linearly on the distance change. Moreover, it is the curvature of the potential that gives rise to the interatomic force constant. The fourth and higher order terms are usually neglected.
It turns out, however, that this is very rarely observed.
The interatomic force constant that is calculated from the Taylor series 3. Indeed, we will later see that it is possible to relate the vibrational properties of a solid to its elastic properties see Section 4. As stated earlier, Y is very high for most materials, on the order of many gigapascal. Metals and alloys are all in the range between 15 and GPa. As a tendency, transition metals have a higher Y than simple metals, consistent with a stronger bonding due to localized d electrons.
W and Mo have particularly strong bonding, something that leads to a high Y and high melting points, as we shall see later. Solids with covalent bonding span a much wider range. The sp2 and sp3 bonds in graphite and diamond lead to particular strength, but note that graphite also appears at the lower end of the range. This is because we have neglected the possibility of anisotropy in solids. Graphite is strongly bonded parallel to the sp2 -linked planes but very weakly perpendicular to these planes.
Polymers show low values of Y. The reason is that a reversible length extension in a polymer does not have to be achieved by extending interatomic bonds. We will be able to establish a link to microscopic models, but a detailed understanding of all phenomena cannot be accomplished on the basis of the perfect crystal. The values are merely a guide, as strong variations are possible.
Such imperfections will also be important in our later treatment of electrical resistance. Can we estimate the yield stress or yield strain? On the atomic scale, this is particularly simple for a shear deformation of a crystal. Figure 3. But if the shear stress is increased more and more, the rows of atoms will eventually start to glide over each other. This leads to an unstable equilibrium when x is equal to half the interatomic distance b, as in Figure 3. In fact, the microscopic positions of the atoms with respect to each other in the crystal do now exactly correspond to the starting point before applying any shear stress, but the crystal has undergone plastic deformation.
Consequently, the shear stress must be a periodic function of x with a period of b. Equation 3. But these estimates, however crude, cannot be reconciled with the experimental results. What is the reason for this disagreement? It turns out that there is nothing wrong with the calculation for the simple model here; the problem is that we have assumed a perfect defect-free crystal. Defects are more than small annoying perturbations of the perfect crystal. In the present context, they are essential for explaining the mechanical properties, and later we will see that they are also indispensable for phenomena like electrical resistance or for the electronic properties of semiconductors.
We broadly distinguish between very localized point defects and extended defects such as grain boundaries or dislocations. There can be an atom missing in the otherwise perfect crystal structure. Such a defect is called a vacancy.
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These defects are called substitutional and interstitial, respectively. The former play an important role for changing the conductivity of semiconductors; the latter are often used to design alloys with improved mechanical properties see below. Dislocations are line-type defects and therefore much more extended.
These lines can extend through the whole crystal, or they can have the shape of a loop. For the mechanical properties of a solid, the edge dislocation shown in Figure 3. This type of dislocation is caused by one extra sheet of atoms in the crystal. It can move within the slip plane as we will explain next. This is illustrated in Figure 3. As the solid with a dislocation is exposed to shear stress, a plastic yielding can be 41 42 3 Mechanical Properties Slip plane Extra sheet of atoms Figure 3.
The dislocation can move in the slip plane. The dislocation moves through the solid by breaking only one row of bonds at a time. It is immediately evident why this is much easier than the process shown in Figure 3. Since edge dislocations are always present in real materials, the observed yield stress is the stress at which dislocations start to move. It is thus far lower than the yield stress from 3. The yield stress of materials can therefore be increased by hindering the movement of dislocations.
In fact, impurities often gather in the extra space available in dislocations and simultaneously hinder their movement. Impurities are therefore frequently introduced into real materials. Examples are carbon, turning iron into steel, or beryllium that can stop dislocation movement in copper. Once the yield stress is overcome, dislocation-assisted glide sets in. This is the so-called easy-glide region. The next part of the curve is called the work hardening region. The meaning of the term work hardening becomes obvious when we consider what happens as the stress is released see Figure 3.
But this point is reached at a higher stress than for the original material. This means that the yield stress is higher and hence the term work hardening. The microscopic picture behind the work hardening is that the number of dislocations increases for higher strain values, for reasons not discussed here. Work hardening can be a useful technique to increase the strength of materials by pre-straining them. At elevated temperature, the role of the entropy in a crystal becomes more important and defects are generated in order to minimize the Gibbs free energy.
In addition to this, activation barriers, such as the one needed to move a dislocation across a point defect, can be overcome more easily. Immediately before fracturing, the stress might even decrease. This is due to a phenomenon called necking, in which the material narrows somewhere between the points at which the stress is applied.
The narrower cross section means that the local stress is even higher than elsewhere. What happens in brittle materials that do not show any plastic deformation at all, but fracture at the end of the elastic part of the curve?
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For a ductile material, such local stress can be relieved by a plastic deformation that work hardens the material in this area. Then, the crack is stopped and cannot proceed further. If such a plastic deformation cannot happen, the crack will propagate through the whole material. Indeed, as the stress is increasing for deeper cracks, 43 44 3 Mechanical Properties Figure 3. The stress per atom is encoded as grayscale. Bright corresponds to high stress. Image courtesy of James Kermode www. The image encodes the local stress in the grayscale of the atoms, and it is easy to see what drives the propagation of the crack: The presence of the crack completely relaxes the stress in the crystal on the left-hand side, above, and below the crack.
However, it also leads to a strongly increased stress near the tip of the crack, causing fracture there and thus further propagation of the crack. Again, temperature is important for the behavior of materials. At elevated temperatures, the propagation of dislocations is easier and materials that behave brittle at low temperature can be ductile as the temperature is raised.
A prominent example is glass, which is usually brittle, but at high temperature, it is so ductile that its shape can be changed by blowing. Why and how? Explain why. Problems 1 Elastic constants: a Consider a cube of isotropic material and show that the bulk modulus 3. Use 3. Are these two aspects of the same thing? The restoring force can be derived from the interatomic potential, as expressed in the Taylor series 3. This leads to a description of the lattice vibrations as harmonic oscillators and is therefore called the harmonic approximation.
While this is clearly much too simple the oscillators are all coupled to each other , it is surprising how far one gets with this picture. Here, we have two squared coordinates and, therefore, the mean energy of the oscillator in contact with a heat bath is kB T. This is still very simple but now the oscillators are coupled to each other. If we restrict ourselves to longitudinal vibrations in this chain, the model is only one-dimensional. However, we can learn a lot about vibrations in three-dimensional solids already from this.
The atoms can move out of their equilibrium position along the direction of the chain, as shown in Figure 4. The solutions given by 4. They describe waves propagating along the chain. Relations of the type of 4. We will encounter them many more times, for example, in connection with electronic states. Note that such a vibration is not localized to one particular atom in the chain. How do the atoms actually move? For a small k, the wavelength of the mode must be much longer than the lattice constant. Therefore, atoms that are close to each other must very nearly move in phase.
If we pick a certain instant in time, and the leftmost atom of the chain in Figure 4. A small k also allows us to replace the sine in 4. In fact, the situation is very similar to long-wavelength sound propagation in air, but since the atoms in a solid are much closer packed than in air, the speed of sound is considerably higher. The limit of short wavelengths is also very instructive. If we again assume that the leftmost atom in the chain of Figure 4. The atom on the neighboring lattice site, on the other hand, is half a wavelength away and must therefore move in the opposite direction.
Note that the wave is transverse for illustrative purposes. Otherwise, we have only considered longitudinal waves in one-dimensional chains. According to 4. Amazingly, this is really the case: It is illustrated in Figure 4. We see that this is so because the wave 4. Any wave that has an even shorter wavelength can be equivalently described by one with a longer wavelength.
It does not matter if the wave travels to the left or to the right, that is, if k is positive or negative. The calculation is very similar to the case of one atom per unit cell. We write down the forces on each atom in a similar manner as above and obtain two equations of motion, one for each type of atom. What is new is that we have two branches of solutions.
The solution that goes to zero for small k is called the acoustic branch. As before, it corresponds to the propagation of sound waves through the crystal. For this wave vector, the two atoms in the unit cell vibrate exactly out of phase, as shown in Figure 4. Figure 4. This situation is discussed in detail in Chapter 9.
This can be done by limiting the length and introducing boundary conditions. Although this approach would not 53 54 4 Thermal Properties of the Lattice be wrong, it would be more convenient to start with traveling wave solutions if we are to describe phenomena such as heat transport by lattice vibrations later. The most convenient boundary conditions solving the problem of what to do at the ends of the chain have been introduced in by M.
Born and T. Therefore, the conditions are also called cyclic boundary conditions or periodic boundary conditions. First of all, it is clear that the longest possible wavelength for a chain of N atoms with a spacing of a is Na. The black markers represent the vibrational frequencies that are actually allowed. For one harmonic oscillator, as described by 4. These energy levels are displayed in Figure 4. The notation in 4. So far, we have viewed k as the one-dimensional wave vector.
But here it becomes apparent that k also takes the role of a quantum number, just as l. The combination of k and l describes one vibrational excitation of the chain, the normal mode k that is excited to the level l. The interpretation of k as a quantum number has a lot to do with the symmetry of the system. In atoms, we have spherical symmetry that gives rise to the quantum numbers l and m. In solids, we have translational symmetry and the appropriate quantum number is k. Depending on the type of experiment, photons can have wave character as well as particle character and the same is true for phonons.
The quantization of the excitation energies and the Bose—Einstein statistics for phonons will become important when we evaluate the heat capacity of solids. Such an excitation involves the creation of a phonon and the annihilation of a photon. For this to be allowed, the phonon and photon must have the same energy and wave vector.
We do not go into much detail because little additional physical insight is gained and the equations get quite messy because one has to keep track of many indices. The wave-type ansatz 4. However, the notation has to be more complex in order to accommodate not only more atoms per unit cell but also more directions of motion.
For instance, for a three-dimensional solid with two atoms per unit cell i. There will be three acoustic branches, one with longitudinal polarization as in one dimension and two with transverse polarization. Similarly, there will also be three optical branches. If the crystal has a basis containing only one atom, there will only be three acoustic branches.follow site
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For a simple cubic crystal with a lattice spacing a and N atoms in every direction, the generalization of the periodic boundary conditions 4. For a face-centered cubic fcc crystal, it is the truncated octahedron that is shown in the inset of Figure 4. In the Brillouin zone and on the Brillouin zone boundary, points of high symmetry are abbreviated by certain letters. The reason for this is simple: Both materials have an fcc Bravais lattice but Al can be described with only one atom as basis, whereas two atoms are needed for diamond.
Consider a simple cubic crystal with a lattice constant a and imagine that we cut a rod from this crystal with a side length of just one lattice constant, as shown in Figure 4. When we 4. The result can be compared to the experimental values determined by other techniques. Table 4. We estimate the atomic force constant for diamond and lead from 4. While this estimate is admittedly rather crude, ignoring the true three-dimensional nature of the problem and the existence of optical phonons, it gives the right order of magnitude. We also see that diamond, which has light atoms and strong bonds, has much higher vibrational frequencies than lead, which has heavy atoms and weak bonds.
In the beginning of the last century, the situation was extremely puzzling. Classical statistical mechanics explained the heat capacity of insulators at room temperature fairly well, but it failed for lower temperatures, and it totally failed for metals. We shall see later why this is so, and we focus on the lattice now. Experimentalists usually like to measure the heat capacity at constant pressure, for example, at ambient pressure.
Theorists, on the other hand, prefer calculations at constant volume because otherwise all the quantum mechanical eigenvalues have to be recalculated for every volume. The worrying thing is that they do not give the same answer. The agreement with the Dulong—Petit value is rather good at room temperature but less good at the boiling point of nitrogen where the heat capacity is generally smaller.
The Dulong—Petit law predicts a temperature-independent heat capacity, whereas the heat capacity has to vanish at zero temperature according to classical thermodynamics. A vanishing heat capacity at zero temperature is also supported by 5 While we have seen that the atoms do not behave as independent oscillators, we also know that the number of independent normal modes is still three times the number of atoms.
Data from Desnoyers and Morrison , Victor Already Table 4. Another example is shown in Figure 4. At high temperatures, the heat capacity approaches the Dulong— Petit value but at lower temperatures it drops to zero. The low-temperature limit is a line in this plot, suggesting a power law behavior. A microscopic theory of the heat capacity should be able to reproduce this behavior.
His idea was to approach the problem using quantum mechanics to describe the oscillators in the solid. This is given by 3NA times the mean energy for one oscillator: 1 4. In the high-temperature limit, the Einstein model correctly reproduces the Dulong— Petit value. The heat capacity also drops to zero at lower temperatures, in agreement with the experimental data.
The only problem is that it drops too quickly to zero. On the linear temperature scale, this does not show up too clearly but on the double log scale, the problem is evident. Can we understand this behavior in simple terms?
This is 4. The only condition is that the temperature must be higher than the Einstein temperature. Debye noticed that the problem can be cured by using a more realistic model for the lattice vibrations. The basic assumption in the Debye model is now that this dispersion holds for all values of k. This is clearly inaccurate for the excitations at higher k, as we have seen in Figure 4. It is totally incorrect for a unit cell containing more than one atom because it ignores the existence of the optical branches see Figure 4.
However, these modes are not excited at low temperatures anyway. At high temperatures, they may be excited but this does not matter so much. From the Einstein model, we have seen that the high-temperature heat capacity approaches the classical value, independent of the actual oscillator frequencies. For low temperatures, the Debye assumption is appropriate and it leads to good results as we shall see now.
We need to calculate the mean thermal energy for a set of oscillators with frequencies given by 4. This factor is called the density of states, for obvious reasons. We consider a cube of solid with a macroscopic side length L and use the periodic boundary conditions 4. If nmax is large, this comes down to a simple geometrical problem as illustrated in Figure 4. With this 4. Instead of writing down an expression for the heat capacity, we focus on the low- and high-temperature limits. For high temperatures, x in 4. For 1 mol of atoms, this is equal to 3RT, that is, it leads to the Dulong—Petit result.
The less obvious limit is that for low temperatures. The reason that the Debye model works so well at low temperatures has been mentioned in the preceding text: It provides a good description of the vibrational modes at low energies and long wavelengths while it is inaccurate for higher energies.
But at low temperatures, only the low-energy modes are excited anyway, and therefore it works well. The Debye temperatures and frequencies for a number of selected materials are given in Table 4. The Debye temperatures follow a tendency that is consistent with intuition: Heavy atoms and weak i. Daily experience tells us that metals are usually much better thermal conductors than insulators. Therefore, the contributions of the free electrons to the thermal conductivity could be thought to be much more important than the lattice contribution. This, however, is not always the case.
A classic example is the insulator diamond that has one of the highest thermal conductivities of all materials at room temperature. It may not be close to daily experience, but it would not be good to make teaspoons out of diamond. Fortunately, the two contributions just add. Suppose we have a rod with a cross-sectional area A. It is not straightforward to describe thermal conduction using the wave picture of phonons we have discussed so far. This can be achieved by the superposition of normal modes to generate wave packets that then travel through the crystal with a certain group velocity.
For describing the thermal conductivity due to the gas of phonons, we adopt a result from kinetic gas theory: 1 4. The only quantity that is unknown is the mean free path of the phonons. We discuss it in the following. As the phonons propagate through the crystal, they can be scattered by imperfections of the lattice, such as point defects, dislocations, and the like.
Then, the scattering of the phonons at the sample boundaries can be observed. Scattering from crystal imperfection is the dominant mechanism at low temperatures. At high temperatures, another scattering process becomes important: The number of phonons increases and phonons can be scattered from other phonons.
At low temperatures, on the other hand, the heat capacity in 4. An example is the thermal conductivity of silicon given in Figure 4. Some numerical values for the thermal conductivity of solids at room temperature are given in Table 4. Evidently, not only metals show high thermal conductivity but also some insulators, especially diamond.
Diamond has a nearly perfect crystal structure such that defect scattering of phonons is unimportant. Problem 6. Reproduced from Glassbrenner and Slack Phonon—phonon scattering deserves a comment: In the case of purely harmonic vibrations and waves, this cannot happen. Consider, for example, water waves or low-intensity light waves i. There, the principle of superposition holds.
Phonon—phonon scattering can only happen in the anharmonic case, that is, if the amplitude of the oscillations 69 70 4 Thermal Properties of the Lattice becomes so large that the fourth and higher order terms in 3. This may seem problematic because our whole treatment so far is based on the assumption that the vibrations are harmonic. In fact, the whole concept of a phonon only makes sense for a harmonic solid. In this section, we will encounter another result of anharmonic vibrations, which is well known from daily experience: the thermal expansion of solids.
We restrict our treatment to the linear expansion of isotropic solids. It is also temperature-dependent and can be shown to vanish at zero temperature, something that limits the validity of 4. Its discovery by C. Guillaume around was a major technical breakthrough because it permitted the construction of highly stable measurement instruments, clocks, and the like. We can understand the thermal expansion of solids by an inspection of the Taylor series for the interatomic potential 3.
This is easily seen in Table 4. The gray line marks the temperaturedependent mean interatomic distance.
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We choose the energy scale such that the potential minimum is at zero. According to the equipartition theorem, the mean energy of the oscillator at a temperature T is kB T. For a low temperature T1 , the oscillation thus takes place between the positions r1 and r2. As the potential is roughly symmetric around its minimum, the average interatomic spacing is equal to the equilibrium distance. For a higher temperature T2 , the oscillation takes place between r3 and r4. The potential is not symmetric anymore, and the interatomic distance expands slightly on average.
This picture is not changed qualitatively in a quantum mechanical treatment. The energy levels are then discrete. For a purely harmonic oscillator, they are equidistant and the mean interatomic distance is the same for all levels. For a nonharmonic oscillator, the energy level separation is not constant and the mean interatomic distance depends on the energy level. This is only true at zero temperature. Here, we discuss two types of structural changes at higher temperatures caused by this: allotropic phase transitions in which a crystal structure is transformed to another crystal structure and melting.
We start with a thermodynamic picture. In most cases, we are interested in the structure at a given temperature, pressure, and particle number. At the temperature TC a phase transition occurs. For low temperatures, the A phase has the lower Gibbs free energy, but at high temperatures, the phase B has. There must be a phase transition between the two structures at a transition temperature TC. This is the idea behind the so-called allotropic phase transitions from one crystal structure to another. As an example, iron crystallizes in a body-centered cubic bcc structure at low temperatures but transforms into a fcc structure at K and again into a bcc structure at K.
Sometimes, structures with a high Gibbs free energy exist even though alternative structures with a lower G are possible under the same conditions. An example is diamond, which has a higher Gibbs free energy than graphite. Still, diamond exists under normal conditions because there is a high activation barrier for the transformation to graphite. Diamond is thus a metastable structure. The melting of a crystal can be described in the same picture, only that the B phase is taken to be the liquid. If we want to predict the melting temperature of a solid, we therefore have to consider the energy and entropy of both the liquid and the solid phase as a function of temperature, a formidable task.
First of all, one might suspect that Tm is related to the cohesive energy of a solid. There is indeed a strong correlation, as shown in Figure 4. This results in a low cohesive energy and melting temperature. On the other extreme, we have the refractory transition metals such W or Mo for which covalent bonding is important, as well as covalent materials such as Si. Simple metals like the alkali metals are found in the middle of the range. An alternative idea was developed by F. Lindemann in He suggested that melting would occur when the amplitude of the interatomic vibration xmax becomes too large, that is, when it reaches a certain fraction of the interatomic spacing.
Using 4. References Desnoyers, J. Glassbrenner, C. Grabowski, B. B, 76, Mounet, N. B, 71, Victor, A. These books all include a more detailed discussion of the phonon dispersion for three-dimensional solids. Why does one speak of optical and acoustic branches? What about the amplitude of the vibrations? How does it compare to the experiment? How do its predictions compare to the experiment?
In which respect does it work better than the Einstein model and why? Why does it decrease for lower temperatures? Why does it decrease for higher temperatures? Why is it called so? The result is 4. Show this formally using 4. The solutions are given by 4. Consider a linear chain of copper atoms. The length of the chain should be 1 cm, the lattice spacing should be 0.
What is the vibrational angular frequency for this wave vector? What is the corresponding energy in electron volts and temperature in Kelvin? What would the dispersion look like in the Debye model? How does it depend on the temperature for the two-dimensional 2D graphene? Hint: In order to give a correct answer to this question, you have to know a curious fact about the phonon dispersion in graphene.
For very low temperatures, this is the important branch and you should base your calculation on this dispersion only. The reason for this unusual behavior is that graphene may be 2D, but it exists in a 3D world. At K, the same peak is observed at an angle of Consider a couple of simple possibilities: One could argue that metals are good conductors of heat and electricity, whereas semiconductors and insulators are not. In the case of heat conduction, we have already seen that diamond, which is an insulator, conducts even better than most metals.
Electrical conductivity is not of much help either: Some semiconductors such as silicon conduct electricity reasonably well. Here we start out with a classical description of metals. Thomson, P. Drude suggested a simple model to explain many of the observed properties of metals. He did this by combining the existence of electrons as charge carriers with the highly successful kinetic gas theory.
We will later see that the Drude model has many shortcomings, but it is still of fundamental importance for the concepts associated with electrical conductivity. They do not interact with each other at all: There is no Coulomb interaction and, as opposed to a classical gas model, they do not collide with each other either. This is known as the independent electron approximation. We will later see that this approximation is quite a good one: The electrons do indeed not interact much with each other.
The electrons can collide with the ion cores. These collisions instantaneously change their velocity. However, in between collisions, the electrons do not interact with the ions either. This is known as the free electron approximation. We will see that this approximation is not very good. Indeed, the whole picture of the electrons colliding with the ions is problematic. In a perfect crystalline solid at low temperatures, the electrons do not collide with the ions at all, as we shall see later. For the description of almost all properties within the Drude model, it is essential to know the density of the gas formed by the free electrons.
This is known as the conduction electron density n, that is, the number of conduction electrons per volume. The core electrons remain bound to the metal ions. For the alkali metals, ZV is 1, for the alkaline earth metals, it is 2, and so on. Values of n for selected metals are given in Table 5.
In Section 5. This is very slow compared to the thermal movement of the electrons. Having the drift velocity, we can calculate the conductivity. One could be tempted to think that it is too high by a factor of 2. See Further Reading. Consider now the explicit expressions for the conductivity and resistivity, which we have obtained. We will come back to this when discussing semiconductors where positively charged carriers do in fact appear. The concept of mobility can be useful for solids in which the electron concentration can be changed by some external parameter without changing the scattering mechanism inside the solid, that is, without changing the relaxation time.
We can also perform a quantitative comparison of the predicted and measured conductivities. Figure 5. One could be tempted to conclude that the Drude model does not reproduce the details, but the general trend is correct anyway. For lower temperatures, the situation becomes more problematic. At 77 K, the calculated conductivity increases because vt gets smaller, but the measured conductivity increases much more. At even lower temperature, the comparison becomes increasingly unfavorable.
It is illustrated in Figure 5. This is explained quite easily in the steady state see Figure 5. The result is shown in Table 5. For the alkali metals, the result is close to the expected value of 1, and for the noble metals, the agreement is also acceptable. It is very bad for Bi. The very high value means that for some reason the true conduction electron density must be much smaller than the calculated value. It appears therefore that the current in Be, Mg, and Al is carried by positive charges: Imagine that we had positive charge carriers with a density p see Figure 5.
The notion of positive carriers does not make sense in the Drude model, but we will see that the quantum model of the electronic states is able to give an intuitive picture of positive carriers. We shall need these equations again when we discuss the optical properties of insulators in Chapter 9. Some of the concepts used here are explained in more detail in the beginning of Chapter 9. With this, 5. We can then ignore the collisions with the ions altogether and treat the electrons as completely free.
By inserting 5. To see this, consider 5. The transition happens at the plasma frequency. The plasma frequency can be calculated solely from the conduction electron density of the metal. Calculated and measured values are given in Table 5. In , G. Wiedemann and R.
Franz found that the ratio of thermal to electrical conductivity is constant for all metals at a given temperature. Later, it was found by L. In the Drude model, the ratio of thermal and electrical conductivity is readily calculated. The thermal conductivity is that of a classical gas and can be described by an equation similar to 4. The electrical conductivity is given by 5. L, as calculated here, is roughly a factor of 2 smaller than the value obtained by experiments or by a proper quantum mechanical calculation, see 6.
Drude, however, had made a mistake of a factor of 2 in his calculation such that L came out almost correct. Therefore, his theory was in impressive quantitative agreement with the experimental data. We discuss several of them here to motivate the quantum treatment of metals in the next chapter. Even before starting our work on the Drude model, several assumptions could have raised suspicion. Take, for example, the nature of the scattering. In addition to this, the de Broglie wavelength for electrons with a thermal energy is on the order of nanometers.
The criterion for treating the electrons as classical particles, however, is that their de Broglie wavelength is much smaller than the typical dimensions of the structures they are moving in. As for a comparison to experimental data, we have already seen that the predicted conductivity at low temperatures is not high enough. In fact, at low temperatures, the mean free path of electrons in very pure and perfect crystals can become macroscopic, micrometers, or even millimeters. Apparently, the electrons manage to sneak past all other electrons and all ions as well.
This appears quite mysterious, but we will be able to explain it in the next chapter. Another problem is that the Drude model cannot explain the conductivity of alloys. Alloying a small amount of impurities into an otherwise pure metals can drastically reduce the conductivity. This happens even if the impurity atoms are quite similar to the host and would be expected to give rise to a similar electron concentration e. The historically most important issue associated with the classical treatment of electrons in a metal is that these electrons should give a considerable contribution to the heat capacity, but this is not observed.
In the previous chapter, we have seen that the experimentally determined heat capacity of most solids, including metals, agrees with the Dulong—Petit value at room temperature see Table 4. If the metal has one 5. This is significantly higher than the Dulong—Petit value that is actually observed. For metals with more conduction electrons per atom, the agreement with the experimental result would be even poorer.
The fact that the Dulong—Petit value is observed for many metals therefore suggests that the electrons do not contribute to the heat capacity, even though they are free to move and they do contribute to the conduction of electricity. This puzzle can only be resolved by a quantum mechanical treatment of metals. A particularly good and in-depth description, including the issue of the factor of 2 in 5. Explain the relaxation time and the mean free path of the electrons. Where does the electrical resistance come from? How is the quantitative agreement? Is this so for all light frequencies?
Problems 1 Classical versus quantum description of metals: Calculate the classical mean kinetic energy for the electrons in Na at room temperature. Show that this is not the case. What properties would you require the coating material to have? All the rest of the crystal is ignored. Show that the restoring force is proportional to the magnitude of 5. We have seen the success and the limitations of this approach. We will also see that, going beyond the assumption of free electrons, it is possible to explain not only metals but also nonmetallic solids.
We will, however, retain the approximation that the electrons move independently from each other. This works surprisingly well for many solids, and we will try to understand why. Finding the quantum mechanical eigenstates for the electrons in the solid is a formidable problem: We would have to construct a wave function that depends on the coordinates of all the electrons and also of all the ions, which make up the positive part of the potential.
This is clearly hopeless! However, it turns out to work rather well. Suppose that the ions are in some given position and the electrons are in their ground state. When the ions move back, the electrons adjust themselves to the old ground state. This is called the Born—Oppenheimer approximation, and it is also often used for treating molecules. This gives the correct occupation of the states but only for zero temperature. At higher temperatures, the statistical occupation of the states is given by the Fermi—Dirac distribution.
Before we start with a detailed quantum mechanical description of solids along these lines, we consider very simple models for the electronic states in solids. This approach is intuitive and would give the correct results, but it is not very practical for treating crystals, in particular because their high degree of symmetry is not exploited. The course will be pitched at the level of a first year course in MSc in Physics. For this course, an exposure to undergraduate level of: a quantum mechanics, b statistical and thermal physics and c electromagnetic theory, is expected for this course.
It is hoped that through this course, the student will understand the quantum theory of solids which is used to describe the thermal and electrical properties of a solid. The student will explore the interaction of a solid with electromagnetic radiation like X-rays and how this can be used to understand the atomic crystal structure of the solid. During the course the student will be exposed to some experiments related to probing various properties of solid.
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Related Solid State Physics: An Introduction
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