{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#3(A):Defects In Solids "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##Example number 3.1, Page number 3.17"
   ]
  },
  {
   "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 3.2, Page number 3.16"
   ]
  },
  {
   "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 3.3, Page number 3.17"
   ]
  },
  {
   "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 3.4, Page number 3.18"
   ]
  },
  {
   "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\""
   ]
  }
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