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{
"metadata": {
"name": "",
"signature": "sha256:f1c728f94d30127360e83c10a55164d3b256772685ed9e2e28ae6d78b3f4a0ae"
},
"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": [
"\n",
"\n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable declaration\n",
"k=1.38*10**-23;\n",
"Ev=0.98; #energy in eV/atom\n",
"T1=900; #temperature in C\n",
"T2=1000;\n",
"A=6.022*10**26; #avagadro's constant\n",
"w=196.9; #atomic weight in g/mol\n",
"d=18.63; #density in g/cm^3\n",
"\n",
"#Calculation\n",
"Ev=Ev*1.6*10**-19; #converting eV to J\n",
"d=d*10**3; #converting g/cm^3 into kg/m^3\n",
"N=(A*d)/w;\n",
"n=N*math.exp(-Ev/(k*T1));\n",
"#let valency fraction n/N be V\n",
"V=math.exp(-Ev/(k*T2));\n",
"\n",
"#Result\n",
"print(\"concentration of atoms per m^3 is\",N);\n",
"print(\"number of vacancies per m^3 is\",n);\n",
"print(\"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": [
"\n",
"\n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable declaration\n",
"k=1.38*10**-23;\n",
"A=6.022*10**26; #avagadro's constant\n",
"T=1073; #temperature in K\n",
"n=3.6*10**23; #number of vacancies\n",
"d=9.5; #density in g/cm^3\n",
"w=107.9; #atomic weight in g/mol\n",
"\n",
"#Calculation\n",
"d=d*10**3; #converting g/cm^3 into kg/m^3\n",
"N=(A*d)/w; #concentration of atoms\n",
"E=k*T*math.log((N/n), ); #energy in J\n",
"EeV=E/(1.602176565*10**-19); #energy in eV\n",
"EeV=math.ceil(EeV*10**2)/10**2; #rounding off to 2 decimals\n",
"\n",
"#Result\n",
"print(\"concentration of atoms per m^3 is\",N);\n",
"print(\"energy for vacancy formation in J\",E);\n",
"print(\"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": [
"\n",
"#importing modules\n",
"import math\n",
"\n",
"#Variable declaration\n",
"A=6.022*10**26; #avagadro's constant\n",
"k=1.38*10**-23;\n",
"w1=39.1; #atomic weight of K\n",
"w2=35.45; #atomic weight of Cl\n",
"Es=2.6; #energy formation in eV\n",
"T=500; #temperature in C\n",
"d=1.955; #density in g/cm^3\n",
"\n",
"#Calculation\n",
"Es=Es*1.6*10**-19; #converting eV to J\n",
"T=T+273; #temperature in K\n",
"d=d*10**3; #converting g/cm^3 into kg/m^3\n",
"N=(A*d)/(w1+w2);\n",
"n=N*math.exp(-Es/(2*k*T));\n",
"\n",
"#Result\n",
"print(\"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": {}
}
]
}
|
