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+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6 : Imperfections, Defects and Dislocations in Solids"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.1 pageno : 125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math \n",
+ "\n",
+ "# Variables\n",
+ "na = 6.023*10**23;\t\t\t#Avagadro's no.\n",
+ "p = 3170.;\t \t\t#density in kg/m**3\n",
+ "\n",
+ "# Calculations\n",
+ "mw = 7.9*10**-2;\t\t\t#molecular weight of CaF2\n",
+ "nl = na*p/mw;\t\t\t #calcium ions/cubic m\n",
+ "ni = 2.*nl;\t\t\t # /cubic m\n",
+ "t = 1300.;\t \t\t#in K\n",
+ "ef = 2.7*1.6*10**-19.;\t\t\t#energy of formation of one frenkel defect\n",
+ "k = 1.38*10**-23;\t\t\t#boltzmann constant\n",
+ "nf = math.sqrt(nl*ni)*math.exp(-ef/(2*k*t));\t\t\t#in /cubic m\n",
+ "\n",
+ "\n",
+ "# Results\n",
+ "print \"No. of Frenkel defect per unit volume of Calcium Fluoride (in /m**3) = %.2e /m**3\"%nf\n",
+ "\n",
+ "# note: Answers may vary because of rouding error."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No. of Frenkel defect per unit volume of Calcium Fluoride (in /m**3) = 2.02e+23 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.2 pageno : 127"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math \n",
+ "\n",
+ "# Variables\n",
+ "r = 8.314;\t\t\t# J/mol K\n",
+ "t1 = 300.;\t\t\t#in K\n",
+ "t2 = 1000.;\t\t\t#in K\n",
+ "\n",
+ "# Calculations\n",
+ "ent = 168*10**3; \t\t\t#enthalpy of formation of vacancy in J/mol\n",
+ "x1 = math.exp(-ent/(r*t1));\t\t\t#x1 = n/Na\n",
+ "x2 = math.exp(-ent/(r*t2));\t\t\t#x2 = n/Na\n",
+ "rt = x1/x2; \t\t\t#ratio \n",
+ "\n",
+ "# Results\n",
+ "print \"Ratio of no. of vacancies = %.2e\"%rt\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio of no. of vacancies = 3.34e-21\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.3 page no : 132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "# Calculations\n",
+ "# for 1/6[121] dislocation\n",
+ "a1 = 1./6 * 1 # x - component\n",
+ "b1 = 1./6 * 2 # y - component\n",
+ "c1 = 1./6 * 1 # z - component\n",
+ "\n",
+ "# for 1/6[211] dislocation\n",
+ "a2 = 1./6 * 2 # x - component\n",
+ "b2 = 1./6 * 1 # y - component\n",
+ "c2 = 1./6 * -1 # z - component\n",
+ "\n",
+ "a = a1 + a2 # x - component\n",
+ "b = b1 + b2 # y - component\n",
+ "c = c1 + c2 # z - component\n",
+ "\n",
+ "# Results\n",
+ "print \"X-component a = %.1f\"%a\n",
+ "print \"Y-component b = %.1f\"%b\n",
+ "print \"Z-component c = %.1f\"%c\n",
+ "\n",
+ "print \"Thus, [abc] = [%.1f %.1f %.1f] = 1/2[110]\"%(a,b,c)\n",
+ "print \"Hence, 1/6[121] + 1/6[211] -> 1/2[110] proved\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X-component a = 0.5\n",
+ "Y-component b = 0.5\n",
+ "Z-component c = 0.0\n",
+ "Thus, [abc] = [0.5 0.5 0.0] = 1/2[110]\n",
+ "Hence, 1/6[121] + 1/6[211] -> 1/2[110] proved\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.4 pageno : 136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math \n",
+ "\n",
+ "# Variables\n",
+ "v = 0.31; \t\t\t#poisson's ratio\n",
+ "bv = .25*10**-9;\t\t\t#burger's vector in m\n",
+ "ri = 1.1*10**-9;\t\t\t#in m\n",
+ "\n",
+ "# Calculations\n",
+ "r0 = 10.**5*bv;\t \t\t#in m\n",
+ "sm = 45.*10**9;\t\t \t#shear modulous in n/sqm\n",
+ "gb_2 = sm*bv**2;\n",
+ "u_ed = (gb_2/(4*3.14*(1-v)))*math.log(r0/ri);\n",
+ "u_sd = (gb_2/(4*3.14))*math.log(r0/ri);\n",
+ "r = u_ed/u_sd;\t\t\t#ratio\n",
+ "\n",
+ "# Results\n",
+ "print \"Elastic Strain Energy of Edge dislocation (in J/m) = %.2e J/m\"%u_ed\n",
+ "print \"Elastic Strain Energy of Screw dislocation (in J/m) = %.2e J/m\"%u_sd\n",
+ "print \"Ratio of energies of edge dislocation over screw dislocation = %.2f\"%r\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Elastic Strain Energy of Edge dislocation (in J/m) = 3.26e-09 J/m\n",
+ "Elastic Strain Energy of Screw dislocation (in J/m) = 2.25e-09 J/m\n",
+ "Ratio of energies of edge dislocation over screw dislocation = 1.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.5 pageno : 139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "# Variables\n",
+ "r = 1.7*10**-10;\t\t\t#atomic radius in m\n",
+ "n1 = 10.**-3; \t\t\t#1mm = 10**-3m\n",
+ "\n",
+ "# Calculations\n",
+ "a = 2*r;\t\t \t#in m\n",
+ "n = n1/a;\n",
+ "ed = 2.*10**-6; \t\t\t#edge dislocation in m\n",
+ "ns = ed/a;\n",
+ "nv = n*ns;\n",
+ "\n",
+ "# Results\n",
+ "print \"Total no. of created vacancies = %.2e\"%nv\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total no. of created vacancies = 1.73e+10\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.6 pageno : 142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Variables\n",
+ "c1 = 3.61 # lattice parameter of copper\n",
+ "c2 = 44 # shear modulus of copper\n",
+ "\n",
+ "# Calculations\n",
+ "b = c1*math.sqrt(1**2 + 1**2 + 0**2)/2.\n",
+ "U = c2*b**2 * 10**9 * 10**-20/2.\n",
+ "\n",
+ "# Results\n",
+ "print \"The magnitude of the Burgers vector : %.3f A\"%b\n",
+ "print \"LIne energy of dislocation U : %.3e J/m\"%U"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of the Burgers vector : 2.553 A\n",
+ "LIne energy of dislocation U : 1.434e-09 J/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.7 pageno : 144"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Variables\n",
+ "a = 3.84 # lattice constant\n",
+ "h = 1\n",
+ "k = 1\n",
+ "l = 0\n",
+ "\n",
+ "# calculations\n",
+ "b = a*math.sqrt(h**2 + k**2 + l**2)/2.\n",
+ "h1 = b/math.tan(math.radians(1))\n",
+ "h3 = b/math.tan(math.radians(3))\n",
+ "\n",
+ "# Results\n",
+ "print \"b = %.3f A\"%b\n",
+ "print \"Spacing between dislocaitons in a low angle tilt boundary in iridium\"\n",
+ "print \"When,\"\n",
+ "print \"Angle of tilts 1 = %.2f A\"%h1\n",
+ "print \"Angle of tilts 3 = %.2f A\"%h3"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "b = 2.715 A\n",
+ "Spacing between dislocaitons in a low angle tilt boundary in iridium\n",
+ "When,\n",
+ "Angle of tilts 1 = 155.56 A\n",
+ "Angle of tilts 3 = 51.81 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
generated by cgit (git 2.34.1) at 2024-12-20 16:52:22 +0000