# Ciencia dos materiais

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12.5 On the basis of ionic charge and ionic radii given in Table 12.3, predict crystal structures for the following materials: (a) CsI, (b) NiO, (c) KI, and (d) NiS. Justify your selections. Solution This problem calls for us to predict crystal structures for several ceramic materials on the basis of ionic charge and ionic radii. (a) For CsI, using data from Table 12.3

rCs + rI−

=

0.170nm = 0.773 0.220 nm

Now, from Table 12.2, the coordination number for each cation (Cs+) is eight, and, using Table 12.4, the predicted crystal structure is cesium chloride. (b) For NiO, using data from Table 12.3

rNi2+ rO 2−

=

0.069 nm = 0.493 0.140 nm

The coordination number is six (Table 12.2), and the predicted crystal structure is sodium chloride (Table 12.4). (c) For KI, usingdata from Table 12.3

rK + rI−

=

0.138 nm = 0.627 0.220 nm

The coordination number is six (Table 12.2), and the predicted crystal structure is sodium chloride (Table 12.4). (d) For NiS, using data from Table 12.3

rNi2+ rS 2−

=

0.069 nm = 0.375 0.184 nm

The coordination number is four (Table 12.2), and the predicted crystal structure is zinc blende (Table 12.4).
Excerptsfrom this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 12.13 Calculate the density of FeO, given that it has the rock salt crystal structure. Solution We are asked to calculate the theoretical density of FeO. This density may be computed using Equation (12.1) as

ρ =

n′( AFe + AO ) VC N A

Since the crystal structure is rock salt, n' = 4 formula units per unit cell. Using the ionic radii for Fe2+ and O2- from Table 12.3, the unit cell volumeis computed as follows:

VC = a 3 = 2rFe 2+ + 2rO 2-

(

)

3

= [ 2 (0.077 nm) + 2 (0.140 nm)]

3

= 0.0817

cm3 nm3 = 8.17 × 10 -23 unit cell unit cell

Thus,

ρ =

(4 formula units/unit cell)(55.85 g/mol + 16.00 g/mol)

(8.17 × 10-23 cm3/unit cell) (6.022 × 1023 formula units/mol)
= 5.84 g/cm3

Excerpts from this work may be reproduced by instructors for distributionon a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

12.29 Calculate the fraction of lattice sites that areSchottky defects for sodium chloride at its melting temperature (801°C). Assume an energy for defect formation of 2.3 eV. Solution We are asked in this problem to calculate the fraction of lattice sites that are Schottky defects for NaCl at its melting temperature (801°C), assuming that the energy for defect formation is 2.3 eV. In order to solve this problem it is necessary to use Equation 12.3 andsolve for the Ns/N ratio. Rearrangement of this expression and substituting values for the several parameters leads to

 Q  Ns = exp − s   2kT  N
  2.3 eV = exp −   (2)(8.62 × 10 -5 eV/K)(801 + 273 K) 
= 4.03 × 10-6

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12.34 (a) Suppose that Li2O is added as an impurity to CaO. If the Li+ substitutes for Ca2+, what kind of vacancies would you expect to form? How...