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"source": [
"# Chapter 5 Imperfection in Solids"
]
},
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"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5_1 pgno:56"
]
},
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"name": "stdout",
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"text": [
"Example 5.1\n",
"\n",
"\n",
" Equilibrium number of vacancies/m**3 is for 1273K 2.18444488963e+25\n"
]
}
],
"source": [
"# given that\n",
"Na=6.023*10**23 #Avogadro No.\n",
"rho=8.4e6 #Density of Copper in g/m**3\n",
"A=63.5 #Atomic weight of Copper\n",
"Qv=0.9 #Activation energy in eV\n",
"k=8.62*10**-5 #Boltzmann Constant in eV/K\n",
"T=1000+273#Temperature in K\n",
"from math import exp\n",
"print\"Example 5.1\\n\"\n",
"N=Na*rho/A #No. of atomic site per cubic meter\n",
"Nv=N*exp(-Qv/(k*T))\n",
"print\"\\n Equilibrium number of vacancies/m**3 is for 1273K\",Nv\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5_3 pgno:57"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
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"outputs": [
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"name": "stdout",
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"text": [
" Example 5.3\n",
"\n",
"\n",
" Atomic of Al is 98.7039833218\n",
"\n",
" Atomic of Cu is 1.29601667817\n"
]
}
],
"source": [
"# given that\n",
"C_Al=97. #Aluminium wt%\n",
"C_Cu=3. #Copper wt%\n",
"A_Al=26.98 #Atomic wt of Aluminium\n",
"A_Cu=63.55 #Atomic wt of Copper\n",
"\n",
"print\" Example 5.3\\n\"\n",
"CAl=C_Al*A_Cu*100/((C_Al*A_Cu)+(C_Cu*A_Al))\n",
"CCu=C_Cu*A_Al*100/((C_Cu*A_Al)+(C_Al*A_Cu))\n",
"print\"\\n Atomic of Al is\",CAl\n",
"print\"\\n Atomic of Cu is\",CCu\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5_4 pgno:58"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 5.4\n",
"\n",
"\n",
" Number of Schottky defects are defects/m**3. 5.31422380078e+19\n"
]
}
],
"source": [
"# given that\n",
"Na=6.023*10**23 #Avogadro No.\n",
"rho=1.955 #Density of KCl in g/cm**3\n",
"A_k= 39.10 #Atomic weight of potassium in g/mol\n",
"A_cl= 35.45 #Atomic weight of Chlorine in g/mol\n",
"Qs=2.6 #Activation energy in eV\n",
"k=8.62*10**-5 #Boltzmann Constant in eV/K\n",
"T=500+273 #Temperature in K\n",
"from math import exp\n",
"\n",
"print\"Example 5.4\\n\"\n",
"A = A_k+A_cl # Molar mass of KCl in gram\n",
"N=Na*rho*1e6/A #No. of atomic site per cubic meter\n",
"Ns=N*exp(-Qs/(2*k*T))\n",
"print\"\\n Number of Schottky defects are defects/m**3.\",Ns\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5_6 pgno:58"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Example 5.6\n",
"\n",
"\n",
" Part A\n",
"\n",
" Grain size number is \n",
"6.49185309633\n",
"\n",
" Part B\n",
"\n",
" At magnification of 85x\n",
"\n",
" Number of grains per inch square are\n",
"62.2837370242\n"
]
}
],
"source": [
"# given that \n",
"N=45. #Number of grains per square inch\n",
"M=85. # magnification\n",
"from math import log\n",
"print\"Example 5.6\\n\"\n",
"print\"\\n Part A\"\n",
"n=(log(N)/log(2))+1 #calculation for grain size no. N=2**(n-1)\n",
"print\"\\n Grain size number is \\n\",n\n",
"print\"\\n Part B\"\n",
"Nm=(100/M)**2*2**(n-1)\n",
"print\"\\n At magnification of 85x\\n\"\n",
"print\" Number of grains per inch square are\\n\",Nm\n",
"# answer in book is 62.6. It is because of rounding off at intermediate stages\n"
]
},
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