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{
"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": {}
}
]
}
|
