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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#8:Defects In Solids "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 8.1, Page number 8.16"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "at 0K, The number of vacancies per kilomole of copper is 0\n",
+      "at 300K, The number of vacancies per kilomole of copper is 7.577 *10**5\n",
+      "at 900K, The numb ber of vacancies per kilomole of copper is 6.502 *10**19\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "N=6.023*10**26\n",
+    "deltaHv=120\n",
+    "B=1.38*10**-23\n",
+    "k=6.023*10**23\n",
+    "\n",
+    "#Calculations\n",
+    "n0=0                                          # 0 in denominator\n",
+    "n300=N*math.exp(-deltaHv*10**3/(k*B*300))     #The number of vacancies per kilomole of copper\n",
+    "n900=N*math.exp(-(deltaHv*10**3)/(k*B*900))\n",
+    "\n",
+    "#Results\n",
+    "print\"at 0K, The number of vacancies per kilomole of copper is\",n0\n",
+    "print\"at 300K, The number of vacancies per kilomole of copper is\",round(n300/10**5,3),\"*10**5\"\n",
+    "print\"at 900K, The numb ber of vacancies per kilomole of copper is\",round(n900/10**19,3),\"*10**19\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 8.2, Page number 8.17"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Fraction of vacancies at 1000 degrees C = 8.5 *10**-7\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "from sympy import Symbol\n",
+    "\n",
+    "#Variable declaration\n",
+    "F_500=1*10**-10\n",
+    "delta_Hv=Symbol('delta_Hv')\n",
+    "k=Symbol('k')\n",
+    "T1=500+273\n",
+    "T2=1000+273\n",
+    "\n",
+    "\n",
+    "#Calculations\n",
+    "lnx=math.log(F_500)*T1/T2;\n",
+    "x=math.exp(round(lnx,2))\n",
+    "\n",
+    "print\"Fraction of vacancies at 1000 degrees C =\",round(x*10**7,1),\"*10**-7\" "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 8.3, Page number 8.17"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Volume of unit cell of NaCl = 1.794 *10**-28 m**3\n",
+      "Total number of ion pairs 'N' =' 2.23 *10**28\n",
+      "The concentration of Schottky defects per m**3 at 300K = 6.42 *10**11\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=(2*2.82*10**-10)\n",
+    "delta_Hs=1.971*1.6*10**-19\n",
+    "k=1.38*10**-23\n",
+    "T=300\n",
+    "\n",
+    "#Calculations\n",
+    "V=a**3                           #Volume of unit cell of NaCl\n",
+    "N=4/V                            #Total number of ion pairs\n",
+    "n=N*math.e**-(delta_Hs/(2*k*T))  \n",
+    "\n",
+    "#Result\n",
+    "print\"Volume of unit cell of NaCl =\",round(V*10**28,3),\"*10**-28 m**3\"\n",
+    "print\"Total number of ion pairs 'N' ='\",round(N/10**28,2),\"*10**28\"\n",
+    "print\"The concentration of Schottky defects per m**3 at 300K =\",round(n/10**11,2),\"*10**11\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 8.4, Page number 8.18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 36,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The number that must be created on heating from 0 to 500K is n= 9.22 *10**12 per cm**3\n",
+      "As one step is 2 Angstorms, 5*10**7 vacancies are required for 1cm\n",
+      "The amount of climb down by the dislocation is 0.369 cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "N=6.023*10**23\n",
+    "delta_Hv=1.6*10**-19\n",
+    "k=1.38*10**-23\n",
+    "T=500\n",
+    "mv=5.55;     #molar volume\n",
+    "x=2*10**-8;     #numbber of cm in 1 angstrom\n",
+    "\n",
+    "#Calculations\n",
+    "n=N*math.exp(-delta_Hv/(k*T))/mv\n",
+    "a=round(n/(5*10**7*10**6),4)*x;\n",
+    "\n",
+    "#Result\n",
+    "print\"The number that must be created on heating from 0 to 500K is n=\",round(n/10**12,2),\"*10**12 per cm**3\" #into cm**3\n",
+    "print\"As one step is 2 Angstorms, 5*10**7 vacancies are required for 1cm\"\n",
+    "print\"The amount of climb down by the dislocation is\",a*10**8,\"cm\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}