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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 05 : The structure of Solids"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.1, Page No 86"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"n_c = 1.0/8 \t# sharing of corner atom in a unit cell\n",
"N_c = 8.0 \t\t# Number of corner atoms in unit cell\n",
"n_b = 1.0 \t\t# sharing of body centered atom in a unit cell\n",
"N_b = 4.0 \t\t# Number of body centered atoms in unit cell\n",
"n_f = 0.5\t\t# sharing of face centered atom in a unit cell\n",
"N_f = 6.0\t\t# Number of face centered atoms in unit cell\n",
"a = 1.0\t\t\t # let lattice parameter\n",
"m = 12.0 \t\t# mass of carbon\n",
"\n",
"#Calculations\n",
"N = n_c*N_c+n_b*N_b+n_f*N_f # effective number of atoms\n",
"r = a*math.sqrt(3.0)/8\n",
"p_e = N*4/3*math.pi*r**3/a**3 \t\t# packing efficiency\n",
"print(\"Packing efficiency of diamond is %.2f\" %p_e)\n",
"a = 3.57 # lattice parameter in angstrom\n",
"d = m*1.66e-27*N/(a*1e-10)**3\n",
"\n",
"#Results\n",
"print(\"Density of diamond is %d Kg/m^3\" %d)\t\t\t# numerical answer in book is 3500\n",
"print(\"Density of diamond is %.1f g/cm^3\" %(d/1000))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Packing efficiency of diamond is 0.34\n",
"Density of diamond is 3502 Kg/m^3\n",
"Density of diamond is 3.5 g/cm^3\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.3, Page No 92"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"a = 1.0 # let\n",
"PR = a\n",
"\n",
"#Calculations\n",
"RT = a/math.sqrt(3)\n",
"PT = math.sqrt(PR**2-RT**2)\n",
"c_a = 2*PT/PR\n",
"\n",
"#Results\n",
"# Calculations are made on the crystal structure drawn in book\n",
"print(\"c/a ratio for an ideally close packed HCP crystal is %0.3f \" %c_a)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"c/a ratio for an ideally close packed HCP crystal is 1.633 \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.4, Page No 94"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"r = 1.0 \t\t# let\n",
"a = 3.0/4.0\n",
"\n",
"#Calculations\n",
"pt = 2*math.sqrt(2/3)*r\n",
"s = a*pt-r \t\t\t# size of sphere\n",
"\n",
"#Results\n",
"print(\"Size of largest sphere that can fit into a tetrahedral void is %.3f r\" %s)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Size of largest sphere that can fit into a tetrahedral void is -1.000 r\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.5 Page No 99"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"theta = 60.0\t # angle in degree\n",
"\n",
"#Calculations\n",
"r_c_a = (2.0/3*2*math.sin(theta*math.pi/180))-1 # ratio calculation\n",
"\n",
"#Results\n",
"print(\"Critical radius ratio for triangular coordination is %0.3f \" %r_c_a)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Critical radius ratio for triangular coordination is 0.155 \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.6, Page No 101"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"r_mg = 0.78 \t# radius of magnesium cation in angstrom\n",
"r_o = 1.32 \t\t# radius of oxygen anion in angstrom\n",
"n = 4.0 \t # effective number of unit cell\n",
"m_o = 16.0\t # mass of oxygen\n",
"m_mg = 24.3 \t# mass of magnesium\n",
"\n",
"#Calculations\n",
"a = 2*(r_mg+r_o)\t\t\t\t\t\t\t# lattice parameter\n",
"d = (m_o+m_mg)*1.66e-27*n/(a*1e-10)**3\t\t# density \n",
"\n",
"#Results\n",
"print(\"Density of MgO is %d Kg/m^3\" %d) \t# answer is 3610 kg/m^3\n",
"print(\"Density of MgO is %0.2f g/cm^3\" %(d/1000))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Density of MgO is 3611 Kg/m^3\n",
"Density of MgO is 3.61 g/cm^3\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}
|
