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diff --git a/Materials_Science_and_Engineering/Chapter5.ipynb b/Materials_Science_and_Engineering/Chapter5.ipynb new file mode 100755 index 00000000..446f509f --- /dev/null +++ b/Materials_Science_and_Engineering/Chapter5.ipynb @@ -0,0 +1,227 @@ +{
+ "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": {}
+ }
+ ]
+}
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