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-{
- "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": 16,
- "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 *\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\""
- ]
- }
- ],
- "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
-}