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-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c896524cb6d8dfdd75df5649979d411984ee380fa5c3cbff49f27851e63b1fb4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4:Defects in Solids"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.1, Page number 4.6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "k = 1.38*10**-23   #Boltzmann constant(eV/K)\n",
-      "e = 1.6*10**-19    #Electronic charge(C)\n",
-      "T1 = 500           #First temperature for metal(K)\n",
-      "T2 = 1000          #Second temperature for metal(K)\n",
-      "Ev = 1             #Average energy required to create a vacancy in metal(eV)\n",
-      "\n",
-      "#Calculations\n",
-      "x = k/e\n",
-      "#n_500 = N*exp(-Ev/T1*k)  ---(1)\n",
-      "#n_1000 = N*exp(-Ev/T2*k) ---(2)\n",
-      "#Dividing (1) by (2), we get the following expression\n",
-      "n = math.exp(Ev/(T2*x))\n",
-      "\n",
-      "#Result\n",
-      "print \"Ratio of vacancies=\",round((n/1E+5),3),\"*10^5\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Ratio of vacancies= 1.085 *10^5\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.2, Page number 4.7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1_by_N = 1.*10**-10  #frequency of vacancy sites at 500 C\n",
-      "T1 = 500.+273.        #K\n",
-      "T2 = 1000.+273.       #K\n",
-      "\n",
-      "#Calculations\n",
-      "x = math.exp((T1/T2)*math.log(n1_by_N))\n",
-      "\n",
-      "#Result\n",
-      "print \"Frequency of vacancy sites at 1000 C =\",round((x/1E-7),4),\"*10^-7\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Frequency of vacancy sites at 1000 C = 8.467 *10^-7\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.3, Page number 4.9"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "r = 2.82*10**-10  #interionic distance(m)\n",
-      "n = 5*10**11      #density of Schottky defect(per m^3)\n",
-      "T = 25+273        #temperature(K)\n",
-      "k = 8.625*10**-5  #Boltzmann constant(/K)\n",
-      "\n",
-      "#Calculations\n",
-      "v = (2*r)**3       #volume of one unit cell(m^3)\n",
-      "N = 4/v           #density of ion pairs\n",
-      "Es = 2*k*T*2.303*math.log10(N/n)\n",
-      "\n",
-      "#Result\n",
-      "print \"The average energy required for creation of one Schottky defect is\",round(Es,3),\"eV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average energy required for creation of one Schottky defect is 1.971 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4, Page number 4.11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "T1 = 20+273  #K\n",
-      "T2 = 300+273 #K\n",
-      "Ef = 1.4     #average energy for creating a Freknel defect(eV)\n",
-      "k = 8.625*10**-5  #Boltzmann constant(J/K)\n",
-      "N = 1             #For simplicity assume total number of metal ions to be unity\n",
-      "Ni = 1            #For simplicity assume total number of metal ions to be unity\n",
-      "\n",
-      "#Calculations\n",
-      "n1 = (N*Ni)**0.5*math.exp(-Ef/(2*k*T1)) \n",
-      "n2 = (N*Ni)**0.5*math.exp(-Ef/(2*k*T2))      \n",
-      "x = n1/n2\n",
-      "\n",
-      "#Result\n",
-      "print \"The ratio of the number of Frenkel defects is\",round((x/1E-6),2),\"*10^-6 or\",round(((1/x)/1E+5),2),\"*10^5\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ratio of the number of Frenkel defects is 1.32 *10^-6 or 7.56 *10^5\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file