MSE 200A Survey of Materials Science Fall, 2008 Problem Set No. 2 Problem 1: At high temperature Fe has the fcc structure (called austenite or γ-iron). Would you expect to find C atoms in the octahedral or tetrahedral sites in this structure? Would you expect the solubility of C in γ-iron to be higher or lower than in α-iron? It is visually obvious that the octahedral void is significantly large than the tetrahedral void in fcc, rather measured by free radius or free volume. The free volume of an octahedral void in fcc is 0.35, which is larger than that in bcc. Moreover, the octahedral void is symmetric and requires no elastic distortion to fit a spherical atom. Consequently, we would expect a much higher solubility in fcc, as is observed. When BCC iron has a high carbon content in solution its structure is distorted and its unit cell becomes slightly tetragonal. FCC iron, on the other hand, remains cubic whatever its carbon content. Explain. The octahedral void in fcc is symmetric, so it is undistorted by C. The void in bcc has a short axis, and is distorted if even an approximately spherical interstitial is included. The tetragonal structure is due to a favorable elastic interaction between adjacent atoms when their short axes are aligned. Hence most of the carbon atoms occupy one of the three subsets of voids (say, the O z voids) with the consequence that the overall crystal is elongatted slightly along the z-axis to create a tetragonal cell. Problem 2: Would you expect the element P to be a donor or an acceptor defect in Si? P (valence 5) is a donor in Si (valence 4). Suppose Si were substituted for Ga in the -ZnS compound GaAs. Would you expect the defect to be a donor, an acceptor or an electrically neutral defect? Si (valence 4) is a donor when it replaces Ga (valence 3) in GaAs. Problem 3: Draw an edge dislocation in a simple cubic structure and define its Burgers' vector, b. Figure 2.1 shows an edge dislocation in a simple cubic structure in cross-section; the line of the dislocation, which is designated by the inverted T, is perpendicular to the plane of the figure. Page 1
The Burgers' vector, b, of a dislocation is a vector that measures the crystal discontinuity it carries, or, equivalently, the magnitude and direction of the elementary slip needed to create it. The Burgers' vector can be measured by drawing a circuit around the dislocation that would close on itself if the crystal were perfect. Such a circuit is shown in Fig. 2.3 for the simple edge dislocation. The additional vector that is required to close the circuit in the imperfect crystal is the Burgers' vector, b. Note that the Burgers' vector, b, of the edge dislocation is perpendicular to the direction of the dislocation line (designated by the unit vector l), which is perpendicular to the plane of the Figure. b Fig. 2.1: An edge dislocation in a simple cubic crystal. A Burgers' circuit around the dislocation which defines the Burgers' vector, b. Describe how the dislocation can move by "glide" and "climb". Fig. 2.2: The elementary step in the glide of an edge dislocation in a simple cubic crystal. "Glide" and "climb" motions are distinct when the dislocation has at least some edge character, that is, when it is not a pure screw. They are simplest to illustrate for a pure edge dislocation. Page 2
First consider the glide process, which is the motion of a dislocation in its glide plane, the plane that contains both the Burgers' vector, b, and the dislocation line, L. The glide of an edge dislocation is illustrated in Fig. 2.2. The plane perpendicular to the edge of the extra half-plane that defines the dislocation is the glide plane. The dislocation moves to the right by one step through a simple atom rearrangement at the dislocation core. Only one bond is broken per crystal plane threaded by the dislocation, so the dislocation can be made to glide by a relatively small shear force. Glide motion is conservative in the sense that no atom diffusion is required. The "climb" of a dislocation occurs when a dislocation that has at least some edge character moves in a direction out of its glide plane. Climb is non-conservative; atoms must always be added to or subtracted from the dislocation line to accomplish it. The process of dislocation climb is illustrated for an edge dislocation in a simple cubic crystal in Fig. 2.3. Fig. 2.3: The climb of an edge dislocation in a simple cubic crystal. In the dislocation climbs up by absorbing vacancies. In the dislocation climbs down by absorbing atoms. Fig. 2.3 illustrates the climb of an edge dislocation up from its glide plane. The upward motion of the dislocation requires that atoms be removed from the edge of the extra half-plane. This is accomplished by absorbing vacancies, one for each plane through which the dislocation threads. Fig. 2.3 illustrates the climb of a dislocation down from its glide plane. In this case the dislocation must add atoms, which it can do by liberating vacancies into the surrounding lattice (for example, if the boxed atom in the figure joins the extra half-plane, leaving a vacancy behind). The creation and annihilation of vacancies by the climb of a dislocation shows one important role of dislocation climb; it provides a mechanism to maintain the equilibrium content of vacancies in a material. While Figs. 2.2-3 are drawn for an edge dislocation, note that the distinction between glide and climb always exists unless bxl = 0, that is, unless the dislocation has pure screw character. A pure screw dislocation does not have a defined glide plane, and can move conservatively on any plane. Page 3
(c) Show how the glide of an edge dislocation leads to plastic deformation. The plastic deformation of a crystal by the simple glide of a dislocation is illustrated in Fig. 2.4. The first figure shows an edge dislocation that has just penetrated the crystal. The material near the left edge is slipped by the vector b of the dislocation. If the dislocation moves all the way through the crystal it creates the configuration shown at the right, where the material above the slip plane has been displaced with respect to the material below by the vector b. Fig. 2.4: Illustration of the plastic deformation of a crystal by the glide of an edge dislocation. Problem 4: Describe how you might create a screw dislocation in a crystal, and show how its Burgers vector, b, is related to its line length. A screw dislocation is a dislocation whose Burgers vector is parallel to its line. To create such a dislocation, imagine a planar cut in the crystal that terminates in a straight line, as in the notes. Displace the part of the crystal above the cut by the Burgers vector, b, parallel to the line, making sure that b is a vector that connects identical atoms in the perfect crystal. If you then rejoin the surfaces across the crack, they fit together perfectly except immediately around the line of the dislocation. The result is a line dislocation with a Burgers vector parallel to the line: a screw dislocation. Show how the motion of a screw dislocation can lead to plastic deformation. Let the dislocation move in a direction perpendicular to its line. The material above the plane of the dislocation is slipped by the vector, b, with respect to that below the plane as the dislocation moves. If the dislocation is passed completely through the crystal, the whole block of material above the plane moves by the vector b with respect to that below, shearing the crystal in a direction parallel to the dislocation line. Page 4
Problem 5: Consider a dislocation loop: a circular dislocation in a plane. The Burgers vector is the same everywhere inside the loop. Assume it lies in the plane of the loop. Show that the dislocation is an edge dislocation at precisely two points on its circumference, and is a screw dislocation at precisely two other points. Let b point toward 12 o clock on the circle. The dislocation has edge character where b is perpendicular to the dislocation line, that is, at precisely 6 and 12 o clock. It has screw character where b is parallel to the line, that is, at precisely 3 and 9 o clock. It is a mixed dislocation everywhere else on the line. Describe how the homogeneous expansion of the loop leads to plastic deformation. The material in the cylinder above the dislocation loop has slipped by b with respect to that below. As the loop expands, it creates more and more slipped area. If it passes all the way our of the crystal, the body of crystal above the plane of the loop is slipped by b with respect to that below it. Problem 6: The elemental metals are all crystalline. Why don t they look like crystals (i.e., polygonal shapes with flat faces)? The shape of a crystal will tend to be one that minimizes its total surface energy. If σ is the surface tension, the total surface energy is E = σ( n)da 6.1 where the interfacial tension is written σ(n) to indicate that it is a function of the orientation of the interface, which is specified by the normal vector, n. If σ is not a strong function of n, then the energy is minimized if the shape minimizes the surface area, that is, if it is nearly spherical. If, on the other hand, σ is a strong function of n, the energy is minimized by a polygonal shape that uses surfaces of minimum energy; the surface area is larger in this case, but the total surface energy is less. Ordinary metals do not look crystalline because their surface energies do not vary greatly with orientation. Their preferred shapes are nearly spherical. Grains of table salt (NaCl) look like small cubes. Why would you expect that? An examination of the NaCl structure shows that, of the low-index faces, only the {100} is electrically neutral. Other low-index surfaces have a preponderance of one type Page 5
of ion. Electrical neutrality leads to a very low interfacial tension for the {100} surfaces. Hence the preferred shape of an ionic material with the NaCl structure is a cube. Page 6