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+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 06 : Crystal Imperfection"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1, Page No 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "t1 = 0       #temperature in kelvin\n",
+      "t2 = 300.0     #temperature in kelvin\n",
+      "t3 = 900.0    #temperature in kelvin\n",
+      "R = 8.314     #universal gas constant\n",
+      "del_hf_al = 68.0  #Enthalpy of formation of aluminium crystal in KJ\n",
+      "del_hf_ni = 168.0   #Enthalpy of formation of nickel crystal in KJ\n",
+      "print(\" Example 6.1\")\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"Equilibrium concentration of vacancies of aluminium at %.2f K is 0\" %t1)\n",
+      "n_N = math.exp(-del_hf_al*1e3/(R*t2))\n",
+      "print(\" Equilibrium concentration of vacancies of aluminium at %.2f K \" %t2)    # answer in book is 1.45e-12\n",
+      "print(\"is %.2e\" %(n_N))\n",
+      "n_N = math.exp(-del_hf_al*1e3/(R*t3))\n",
+      "print(\" Equilibrium concentration of vacancies of aluminium at %.2f K \" %t3)    # answer in book is 1.12e-4\n",
+      "print(\"is %.2e \" %(n_N))\n",
+      "\n",
+      "#Results\n",
+      "print(\"Equilibrium concentration of vacancies of Nickel at %.2f K is 0\" %t1)\n",
+      "n_N = math.exp(-del_hf_ni*1e3/(R*t2))\n",
+      "print(\" Equilibrium concentration of vacancies of Nickel at %.2fK \" %t2)    # answer in book is 1.45e-12\n",
+      "print(\"is %.2e\" %(n_N))\n",
+      "n_N = math.exp(-del_hf_ni*1e3/(R*t3))\n",
+      "print(\" Equilibrium concentration of vacancies of Nickel at %.2f K \" %t3)    # answer in book is 1.78e-10\n",
+      "print(\"is %.2e \" %(n_N))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Example 6.1\n",
+        "Equilibrium concentration of vacancies of aluminium at 0.00 K is 0\n",
+        " Equilibrium concentration of vacancies of aluminium at 300.00 K \n",
+        "is 1.44e-12\n",
+        " Equilibrium concentration of vacancies of aluminium at 900.00 K \n",
+        "is 1.13e-04 \n",
+        "Equilibrium concentration of vacancies of Nickel at 0.00 K is 0\n",
+        " Equilibrium concentration of vacancies of Nickel at 300.00K \n",
+        "is 5.59e-30\n",
+        " Equilibrium concentration of vacancies of Nickel at 900.00 K \n",
+        "is 1.77e-10 \n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2, Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a = 2.87    #lattice parameter in angstrom\n",
+      "b= 2.49   #magnitude of burgers vector in angstrom\n",
+      "G = 80.2   #shear modulus in GN\n",
+      "\n",
+      "#Calculations\n",
+      "E = G*1e9*(b*1e-10)**2*1.0/2 \n",
+      "\n",
+      "#Results\n",
+      "print(\"Line energy of dislocation is %.2e J m^-1\" %E)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Line energy of dislocation is 2.49e-09 J m^-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4, Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a = 1.0e10      # total number of edge dislocation \n",
+      "N = 6.023e23 \t# Avogadro number\n",
+      "R = 8.314 \t# Universal gas constant\n",
+      "t1 = 0 \t# initial temperature in K\n",
+      "t2 = 1000.0\t # Final temperature in K\n",
+      "del_hf = 100.0 \t # Enthalpy of vacancy formation in KJ\n",
+      "d = 2.0 # length of one step in angstrom\n",
+      "\n",
+      "#Calculations\n",
+      "v = 5.5/10**6  # volume of one mole crystal\n",
+      "n = N*math.exp(-(del_hf*1e03)/(R*(t2-t1)))/v\n",
+      "k = 1.0/(d*1e-10 ) # atoms required for 1 m climb\n",
+      "b = n/(k*a)# average amount of climb\n",
+      "c = b*d*1e-10 \n",
+      "\n",
+      "#Results\n",
+      "print(\"Average down climb of crystal is %.2e m\" %c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average down climb of crystal is 2.62e-06 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5 Page No 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "E = 56.4    #bond energy in KJ\n",
+      "N_a = 6.023e23 #Avogadro\u2019s number\n",
+      "n = 12.0       #number of bonds\n",
+      "m = 3.0    #number of broken bonds \n",
+      "N = 1.77e19   #number of atoms in copper crystal of type {111} per m^2\n",
+      "\n",
+      "#Calculations\n",
+      "b_e = 1.0/2*E*1e3*n/N_a    #bond energy per atom\n",
+      "e_b = b_e*m/n    #energy of broken bond at surface\n",
+      "s_e = e_b*N    #surface enthalpy of copper\n",
+      "\n",
+      "#Results\n",
+      "print(\"Surface enthalpy of copper is %0.2f J m^-2\" %s_e)\n",
+      "print(\"Surface enthalpy of copper is %0.2f erg cm^-2\" %(s_e*1e3))\n",
+      "# Answer in book is 2490 erg cm^-2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Surface enthalpy of copper is 2.49 J m^-2\n",
+        "Surface enthalpy of copper is 2486.17 erg cm^-2\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6, Page No 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "Gamma_gb = 1.0    #let, energy of grain boundary\n",
+      "\n",
+      "#Calculations\n",
+      "Gamma_s = 3.0*Gamma_gb     #energy of free surface\n",
+      "theta = 2*math.degrees(math.acos(1.0/2*Gamma_gb/Gamma_s))\n",
+      "\n",
+      "#Results\n",
+      "print(\"Angle at the bottom of groove of a boundary is %0.2f degrees.\" %math.ceil(theta))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angle at the bottom of groove of a boundary is 161.00 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file