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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#8:Defects In Solids "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##Example number 8.1, Page number 8.16"
]
},
{
"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 8.2, Page number 8.17"
]
},
{
"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 8.3, Page number 8.17"
]
},
{
"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 8.4, Page number 8.18"
]
},
{
"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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|
