E 45 - Fall 2010 - Gronsky - Midterm 1 (solution)

# D in a diffraction experiment designed to utilize

This preview shows pages 9–11. Sign up to view the full content.

Note the charge on the ions and the need to preserve charge neutrality. d . In a diffraction experiment designed to utilize Bragg's Law ( n λ = 2 d sin θ ), a single crystal of this nuclear fuel pellet is oriented so an incident beam of Cu K α radiation ( λ = 0.154 nm) makes an angle of 14.22° with the (040) family of planes. At what angle should the detector be positioned to capture the 040 diffraction peak? In all Bragg experiments, the scattering condition is established to mimic "reflection" from the diffracting planes, so the incident angle is equal to the diffraction angle. The detector should therefore be positioned at 14.22° from the sample surface, opposite the source. ! "# \$ %& ! " # E45 Fall 10 Midterm 01 Solutions Professor R. Gronsky page9 of 11

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
5. Crystal Defects a . It is claimed that vacancies are not true "point" defects (zero-dimensional) because they exert a tensile stress on the atoms surrounding them, extending over finite dimensions. Consider the illustration of a single close-packed plane containing a vacant site shown to the right. Is there a tensile stress in the vicinity of the vacancy or not? Explain . There appears to be no room to "collapse" into the vacant site when all atoms are touching one another in this plane, but the "hard sphere" model is inaccurate here. There is a collapse into the gap by all contiguous atoms, which in turn stretch their bonds to their nearest neighbors, the equivalent response to a tensile load. So the strain field is "tensile." b . Vacancies are created at high temperatures in concentrations that obey an Arrhenius-type relation. Also at temperatures sufficient to cause diffusion, vacancies can "condense" into dislocation "loops," as shown in an edge-on projection here, with time increasing from left to right. Label both the line direction vector ( ξ ) and the Burgers vector ( b ) of the dislocation loop in the final frame. What type of dislocation (edge or screw) is it? The line direction vector ( ξ ) runs both into and out of the plane of the paper, because it follows the loop, either in at the top, out at the bottom, or the reverse (it may arbitrarily assigned). The Burgers vector ( b ) however is fixed by the FSRH convention to point to the right with a magnitude of one interatomic spacing . Because ξ b , the dislocation loop is pure edge . ! !! !!! !" E45 Fall 10 Midterm 01 Solutions Professor R. Gronsky page10 of 11 b ξ (in) ξ (out)
5. Crystal Defects c . In BCC crystals, active slip systems consist of close-packed < 111 > directions and the { 1 ¯ 10 } planes that contain those directions. Consider an edge dislocation with a Burgers vector Specify the possible slip planes for this edge dislocation. By definition, the slip plane must contain the Burgers vector, and in this instance there are three possible candidate slip planes. Check: Dot product between b and all three plane normals = 0 d . Nearly all engineering materials are polycrystalline, with individual grains meeting at internal interfaces known as grain boundaries to establish a very important characteristic "grain size." A universal tenet of engineering materials is "fine-grained alloys are stronger than coarse-grained alloys." Explain .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business â€˜17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. Itâ€™s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania â€˜17, Course Hero Intern

• The ability to access any universityâ€™s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLAâ€™s materials to help me move forward and get everything together on time.

Jill Tulane University â€˜16, Course Hero Intern