# Figure 49 of the text p 109 gives a pictorial

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Unformatted text preview: on site coupled to a vacant cation site, which together preserve charge neutrality. Figure 4.9 of the text (p. 109) gives a pictorial illustration. In MgO, such a pair consists of one Mg2+ cation vacancy and one O2– anion vacancy. To begin this calculation, note that by definition, the number of Mg2+- O2– “pairs” in one mole of MgO is Avogadro's number. From the density of MgO given in the problem statement and the molar mass of MgO obtained from data on atomic masses found in Appendix 1, the number of pairs per unit volume can be determined. The problem statement demands that the volume have units of m–3, so attention to units is required, and the calculation is 3.60 Mg/m3 × 106 g/Mg × 0.602 × 1024 ”pairs”/mol = 5.38 × 1028 pairs/m3 [24.31 + 16.00] g/mol Now using the fraction of vacant sites given in the problem statement, the density of Schottky pairs can be directly calculated. It is important to note that the problem statement gives the fraction of vacant lattice sites, (not vacant ionic sites). In MgO, each lattice site is adorned by twoion motif, one cation and one anion (see Figure 3.9 for review of this concept). The calculation of the Schottky pair density therefore proceeds as follows 5 × 10−6 Schottky pairs lattice site pairs ×1 × 5.38 × 1028 lattice site pair m3 to yield the density of Schottky pairs in this sample in the required units = 0.27 x 1024 m–3 Problem 4.19 Solution Professor R. Gronsky page 1 of 1 Problem 4.23 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 4.23 The comments in Problem 4.21 also apply for the HCP metal structure. Calculate (relative to Eb = [11¯ ) the dislocation energies for (a) Eb = [1¯ and (b) Eb = [0001]. 20] 100] SOLUTION The statement of Problem 4.21 reveals an important aspect of the “energy” associated with the class of lattice defects known as dislocations. It states that the energy necessary to generate a dislocation is proportional to the square of the length of its Burgers vector |b|2. Consequently, to determine the most stable (lowest energy...
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## This note was uploaded on 01/21/2011 for the course E 45 taught by Professor Gronsky during the Fall '08 term at University of California, Berkeley.

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