University:
Massachusetts Institute of Technology
Course:
3.091SC | Introduction to Solid State Chemistry
Academic year:

2020
Views:

185

Pages:

3
Author:

DemCedgevexynw

Session #19: Homework Solutions Problem #1 Calculate the vacancy fraction in copper (Cu) at (a) 20°C and (b) the melting point, 1085°C. Measurements have determined the values of the enthalpy of vacancy formation, ΔHv , to be 1.03 eV and the entropic prefactor, A, to be 1.1. Solution number of sites / unit volume (also known as site density) is given by: NA Vmolar ∴ site density = 6.02 x 1023 / 7.11 cm3 = 8.47 x 1022 → vacancy density = fv x site density - (a) fV = Ae ΔHV kB T - = 1.1 × e 1.03×1.6 ×10-19 1.38×10−22 ×(20 + 273) = 2.19 × 10-18 vacancy density at 20°C = 1.85 x 105 cm-3 - (b) fv = Ae ΔHv kB T - = 1.1 × e 1.03×1.6 ×10-19 1.38×10−22 ×(20 +1085) = 1.67 × 10-4 vacancy density at 1085°C = 1.41 x 1019 cm-3 Note that the ratio of fv(1085°C) / fv(20°C) = 7.62 x 1013 ! Problem #2 In iridium (Ir), the vacancy fraction, nv/N, is 3.091x10–5 at 1234ºC and 5.26x10–3 at the melting point. Calculate the enthalpy of vacancy formation, ΔHv . Solution All we need to know is the temperature dependence of the vacancy density: − nv = Ae N ΔHv RT -5 3.091 × 10 -3 5.26 × 10 Taking the ratio: , where T is in Kelvins and the melting point of Ir is 2446°C - = Ae - = Ae ΔHv RT1 ΔHv RT2 , where T1 = 1234oC = 1507 K , where T2 = 2446oC = 2719 K -3 5.26 × 10 3.091 × 10-5 = ΔHv RT1 - ΔHv RT2 Ae Ae ∴ ln 170.2 = - ∴ ΔHv = - - =e ΔHv ⎛ 1 1 ⎞ − ⎜ ⎟ R ⎜⎝ T1 T2 ⎟⎠ ΔHv ⎛ 1 1 ⎞ − ⎜ ⎟ R ⎝ T1 T2 ⎠ R × ln 170.2 8.314 × ln 170.2 == 1.44 × 105 J/mole ⋅ vac 1 1 1 1 − − 1507 2719 1507 2719 1.44 × 105 ∴ ΔHv = - 23 6.02 × 10 = 2.40 × 10-19 J/vac = 1.5 eV/vac Problem #3 A formation energy of 2.0 eV is required to create a vacancy in a particular metal. At 800ºC there is one vacancy for every 10,000 atoms. (a) At what temperature will there be one vacancy for every 1,000 atoms? (b) Repeat the calculation, but this time with an activation energy of 1.0 eV. Note the big change in the temperature interval necessary to obtain the same change in vacancy concentration as was the case with an activation energy of 2.0 eV. Solution (a) We need to know the temperature dependence of the vacancy density: 1 104 = Ae − ΔHv kT1 and 1 103 1 From the ratio: ∴ = Ae − ΔHv kTx − ΔHv/kT 1 3 104 = 10 = Ae − 1 104 Ae 3 10 ΔHv/kT x we get -ln 10 = − ΔHv ⎛ 1 1 ⎞ − ⎜ ⎟ k ⎝ T1 Tx ⎠ ⎛ 1 1 ⎞ k ln 10 − ⎜ ⎟= ΔHv ⎝ T1 Tx ⎠ 1 1 k ln 10 1 1.38 × 10-23 × ln 10 = = = 8.33 × 10−4 −19 Tx T1 ΔHv 1073 2 × 1.6 × 10 Tx = 1200 K = 928oC (b) Repeat the calculation following the method given above but with ΔHv = 1.0 eV to find that Tx = 1364 K = 1091oC Note: the change in ΔHv from 2.0 eV to 1.0 eV resulted in a change from 128 K to 291 K in ΔT. Problem #4 On appropriate schematic drawings show the generation and characteristics of Schottky defects in (a) a closed-packed metal, (b) an ionic crystal and (c) a semiconductor. ÉÉ ÉÉ ÉÉ Solution (a) ÉÉÉ ÉÉÉ ÉÉÉ → (b) (c) + + – – + + – + – + – + – – + – + – + – VC – + – + – + – + – + VA + – – + – + – + – + – + + – – + – +