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CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS 4.1 In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation (4.1). As stated in the problem, Qv= 0.55 eV/atom. Thus, NVN= exp −QVkT⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = exp −0.55 eV /atom8.62 x 10−5eV /atom -K()(600 K)⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = 2.41 x 10-54.2 Determination of the number of vacancies per cubic meter in gold at 900°C (1173 K) requires the utilization of Equations (4.1) and (4.2) as follows: NV= N exp −QVkT⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = NAρAuAAuexp −QVkT⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 6.023 x 1023atoms/mol()19.32 g/ cm3()196.9 g/molexp −0.98 eV / atom8.62 x 10−5eV /atom−K()(1173 K)⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = 3.65 x 1018cm-3= 3.65 x 1024m-34.3 This problem calls for the computation of the energy for vacancy formation in silver. Upon examination of Equation (4.1), all parameters besides Qvare given except N, the total number of atomic sites. However, Nis related to the density, (ρ), Avogadro's number (NA), and the atomic weight (A) according to Equation (4.2) as N = NAρAgAAg= 6.023 x 1023atoms /mol()10.49 g/cm3()107.87 g/mol1
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