{ "metadata": { "name": "Chapter 7" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": "Crystal Imperfections" }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": "Example number 7.1, Page number 207 " }, { "cell_type": "code", "collapsed": false, "input": "#To calculate the number of vacancies and vacancy fraction\n\n#importing modules\nimport math\n\n#Variable declaration\nk=1.38*10**-23;\nEv=0.98; #energy in eV/atom\nT1=900; #temperature in C\nT2=1000;\nA=6.022*10**26; #avagadro's constant\nw=196.9; #atomic weight in g/mol\nd=18.63; #density in g/cm^3\n\n#Calculation\nEv=Ev*1.6*10**-19; #converting eV to J\nd=d*10**3; #converting g/cm^3 into kg/m^3\nN=(A*d)/w;\nn=N*math.exp(-Ev/(k*T1));\n#let valency fraction n/N be V\nV=math.exp(-Ev/(k*T2));\n\n#Result\nprint(\"concentration of atoms per m^3 is\",N);\nprint(\"number of vacancies per m^3 is\",n);\nprint(\"valency fraction is\",V);\n", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "('concentration of atoms per m^3 is', 5.69780904012189e+28)\n('number of vacancies per m^3 is', 1.8742498047705634e+23)\n('valency fraction is', 1.1625392535344139e-05)\n" } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": "Example number 7.2, Page number 208 " }, { "cell_type": "code", "collapsed": false, "input": "#To calculate the energy for vacancy formation\n\n#importing modules\nimport math\n\n#Variable declaration\nk=1.38*10**-23;\nA=6.022*10**26; #avagadro's constant\nT=1073; #temperature in K\nn=3.6*10**23; #number of vacancies\nd=9.5; #density in g/cm^3\nw=107.9; #atomic weight in g/mol\n\n#Calculation\nd=d*10**3; #converting g/cm^3 into kg/m^3\nN=(A*d)/w; #concentration of atoms\nE=k*T*math.log((N/n), ); #energy in J\nEeV=E/(1.602176565*10**-19); #energy in eV\nEeV=math.ceil(EeV*10**2)/10**2; #rounding off to 2 decimals\n\n#Result\nprint(\"concentration of atoms per m^3 is\",N);\nprint(\"energy for vacancy formation in J\",E);\nprint(\"energy for vacancy formation in eV\",EeV);", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "('concentration of atoms per m^3 is', 5.3020389249304915e+28)\n('energy for vacancy formation in J', 1.762092900344914e-19)\n('energy for vacancy formation in eV', 1.1)\n" } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": "Example number 7.3, Page number 209 " }, { "cell_type": "code", "collapsed": false, "input": "#To calculate the number of Schotky defect\n\n#importing modules\nimport math\n\n#Variable declaration\nA=6.022*10**26; #avagadro's constant\nk=1.38*10**-23;\nw1=39.1; #atomic weight of K\nw2=35.45; #atomic weight of Cl\nEs=2.6; #energy formation in eV\nT=500; #temperature in C\nd=1.955; #density in g/cm^3\n\n#Calculation\nEs=Es*1.6*10**-19; #converting eV to J\nT=T+273; #temperature in K\nd=d*10**3; #converting g/cm^3 into kg/m^3\nN=(A*d)/(w1+w2);\nn=N*math.exp(-Es/(2*k*T));\n\n#Result\nprint(\"number of Schotky defect per m^3 is\",n);\n\n#answer given in the book is wrong by 3rd decimal point", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "('number of Schotky defect per m^3 is', 5.373777171020081e+19)\n" } ], "prompt_number": 7 }, { "cell_type": "code", "collapsed": false, "input": "", "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }