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diff --git a/Solid_state_physics_by_P._K._Palanisamy/Chapter_2.ipynb b/Solid_state_physics_by_P._K._Palanisamy/Chapter_2.ipynb new file mode 100755 index 00000000..99284878 --- /dev/null +++ b/Solid_state_physics_by_P._K._Palanisamy/Chapter_2.ipynb @@ -0,0 +1,331 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:64e1fbee0e1d9b8157cae12a98a7773b847a3c5e842a9d3f124a5485fe931875" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2: Crystal Structure" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1, Page number 2.16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import sqrt \n", + "\n", + "#Variable declaration\n", + "#Assuming r=1 for simpliciy in calculations\n", + "r = 1\n", + "\n", + "#Calculations\n", + "a = (4*r)/sqrt(3)\n", + "#Let R be the radius of interstitial sphere that can fit into the void,therefore,\n", + "R = (a-2*r)/2 \n", + "\n", + "#Result\n", + "print \"The maximum radius of interstitial sphere is\",round(R,3),\"r\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum radius of interstitial sphere is 0.155 r\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2, Page number 2.17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "r1 = 1.258*10**-10 #atomic radius(m)\n", + "r2 = 1.292*10**-10 #atomic radius(m)\n", + "\n", + "#Calculations\n", + "#In BCC\n", + "a_bcc = (4*r1)/sqrt(3)\n", + "v_bcc = a_bcc**3 #volume of unit cell(m^3)\n", + "n1 = ((1./8.)*8.)+1\n", + "V1 = v_bcc/n1 #volume occupied by 1 atom(m^3)\n", + "\n", + "#In FCC\n", + "a_fcc = 2*sqrt(2)*r2\n", + "v_fcc = a_fcc**3 #volume of unit cell(m^3)\n", + "n2 = ((1./2.)*6.)+((1./8.)*8.)\n", + "V2 = v_fcc/n2 #volume occupied by 1 atom(m^3)\n", + "\n", + "del_v = ((V1-V2)/V1)*100 #change in volume\n", + "\n", + "#Result\n", + "print \"During the conversion of iron from BCC to FCC, the decrease in volume is\",round(del_v,1),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "During the conversion of iron from BCC to FCC, the decrease in volume is 0.5 %\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3, Page number 2.17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "a = 0.27*10**-9 #nearest neighbour distance(m)\n", + "c = 0.494*10**-9 #height of unit cell(m)\n", + "M = 65.37 #atomic weight of zinc\n", + "N = 6.023*10**26 #Avogadro's number(k/mol)\n", + "\n", + "#Calculations\n", + "V = (3*sqrt(3)*a**2*c)/2 #volume of unit cell\n", + "rho = (6*M)/(N*V) #density of crystal\n", + "\n", + "#Results\n", + "print \"Volume of unit cell =\",round((V/1E-29),2),\"*10^29 m^3\"\n", + "print \"Density of zinc =\",round(rho),\"kg/m^3\"\n", + "print \"\\nThe solution differs because of rounding-off of the digits in the textbook\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volume of unit cell = 9.36 *10^29 m^3\n", + "Density of zinc = 6960.0 kg/m^3\n", + "\n", + "The solution differs because of rounding-off of the digits in the textbook\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4, Page number 2.18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import sqrt\n", + "\n", + "#Varaible declaration\n", + "#Let r be the radius of atom and R be the radius of sphere\n", + "#For simplicity in calculations, let us assume r =1\n", + "r = 1\n", + "\n", + "#Calculations\n", + "#For FCC structure\n", + "a = (4*r)/sqrt(2)\n", + "R = (a/2)-r\n", + "\n", + "print \"Maximum radius of sphere =\",round(R,3),\"r\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum radius of sphere = 0.414 r\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5, Page number 2.19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "a = 0.356*10**-9 #cube edge(m)\n", + "M = 12.01 #atomic weight of carbon\n", + "N = 6.023*10**26 #Avogadro's number(k/mol)\n", + "na = 1.77*10**29 #no. of atoms per meter cube\n", + "\n", + "#Calculations\n", + "#Diamond has 2 interpenetrating FCC lattices. Since each FCC unit cell has 4 atoms, the total no. of atoms per unit cell is 8\n", + "n = 8/a**3\n", + "m = M/N\n", + "rho = m*na\n", + "\n", + "#Result\n", + "print \"Number of atoms =\",round((n/1E+29),3),\"*10^29\"\n", + "print \"The density of diamond is\",round(rho,1),\"kg/m^3(Calculation mistake in the textbook)\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of atoms = 1.773 *10^29\n", + "The density of diamond is 3529.4 kg/m^3(Calculation mistake in the textbook)\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.6, Page number 2.19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "rho = 2.18 #density of NaCl(gm/cm^3)\n", + "N = 6.023*10**23 #Avogadro's number(/mol)\n", + "\n", + "#Caculations\n", + "w = 23+35.5 #molecular weight of NaCl\n", + "m = w/N #mass of NaCl molecule\n", + "nm = rho/m #no. of molecules per unit volume(molecule/cm^3)\n", + "#Since NaCl is diatomic\n", + "n = 2*nm\n", + "#Let a be the distance between adjacent atoms in NaCl and\n", + "# n be the no. of atoms along the edge of the cube\n", + "#length of an edge = na\n", + "#volume of unit cube = n^3*a^3\n", + "a = (1/n)**(1./3.)\n", + "\n", + "#Result\n", + "print \"The distance between two adjacent atoms is\",round((a/1E-8),2),\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The distance between two adjacent atoms is 2.81 A\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7, Page number 2.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "w = 63.5 #atomic weight of copper\n", + "r = 1.278*10**-8 #atomic rdius(m)\n", + "N = 6.023*10**23 #Avogadro's number(/mol)\n", + "\n", + "#Calculations\n", + "m = w/N #mass of each copper atom(gm)\n", + "#Since copper has FCC structure lattice constant\n", + "a = (4*r)/sqrt(2) \n", + "n = 4 #no. of atoms in unit cell of FCC structure\n", + "M = n*m #mass of unit cell\n", + "rho = M/a**3 #density\n", + "\n", + "#Result\n", + "print \"Density of copper =\",round(rho,2),\"gm/cm^3\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Density of copper = 8.93 gm/cm^3\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
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