From 4a1f703f1c1808d390ebf80e80659fe161f69fab Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 28 Aug 2015 16:53:23 +0530 Subject: add books --- Applied_Physics-I_by_I_A_Shaikh/Chapter1.ipynb | 2515 ++++++++++++++++++++++++ 1 file changed, 2515 insertions(+) create mode 100755 Applied_Physics-I_by_I_A_Shaikh/Chapter1.ipynb (limited to 'Applied_Physics-I_by_I_A_Shaikh/Chapter1.ipynb') diff --git a/Applied_Physics-I_by_I_A_Shaikh/Chapter1.ipynb b/Applied_Physics-I_by_I_A_Shaikh/Chapter1.ipynb new file mode 100755 index 00000000..da15fd14 --- /dev/null +++ b/Applied_Physics-I_by_I_A_Shaikh/Chapter1.ipynb @@ -0,0 +1,2515 @@ +{ + "metadata": { + "celltoolbar": "Raw Cell Format", + "name": "", + "signature": "sha256:c884249c28dc1486b445f3d4013b1d4277c7f2f132398c648a2fa82d33db0def" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1: Crystallography" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3.1,Page number 1-14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "A=26.98 #atomic weight of Al\n", + "N=6.023*10**26 #Avogadro's number\n", + "p=2700 #Density\n", + "n=4 #FCC structure\n", + "\n", + "a=(n*A/(N*p))**(1./3)\n", + "\n", + "print\"Unit cell dimension of Al=\",\"{0:.3e}\".format(a),\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Unit cell dimension of Al= 4.049e-10 m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3.2,Page number 1-15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "As=28.1 #atomic weight of Si\n", + "Ag=69.7 #atomic weight of Ga\n", + "Aa=74.9 #atomic weight of As\n", + "a_s=5.43*10**-8 #lattice constant of Si\n", + "aga=5.65*10**-8 #lattice constant of GaAs\n", + "ns=8 #no of atoms/unit cell in Si\n", + "nga=4 #no of atoms/unit cell in GaAs\n", + "N=6.023*10**23 #Avogadro's number\n", + "\n", + "#p=(n*A)/(N*a**3) this is formula for density\n", + "\n", + "#for Si\n", + "\n", + "ps=(ns*As)/(N*a_s**3)\n", + "\n", + "print\"1) Density of Si=\",round(ps,4),\"gm/cm^3\"\n", + "\n", + "#for GaAs\n", + "\n", + "Aga=Ag+Aa #molecular wt of GaAs\n", + "\n", + "pga=(nga*Aga)/(N*aga**3)\n", + "\n", + "print\"2) Density of GaAs=\",round(pga,4),\"gm/cm^3\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1) Density of Si= 2.3312 gm/cm^3\n", + "2) Density of GaAs= 5.3244 gm/cm^3\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3.3,Page number 1-16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "A=63.5 #atomic weight of Cu\n", + "N=6.023*10**23 #Avogadro's number\n", + "n=4 #FCC structure\n", + "r=1.28*10**-8 #atomic radius of Cu\n", + "\n", + "#for FCC\n", + "\n", + "a=4*r/(sqrt(2)) #lattice constant\n", + "p=(n*A)/(N*a**3)\n", + "\n", + "print\"Density of Cu=\",round(p,4),\"gm/cm^3\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Density of Cu= 8.887 gm/cm^3\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3.4,Page number 1-17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "A=50 #atomic weight of chromium\n", + "N=6.023*10**23 #Avogadro's number\n", + "p=5.96 #Density\n", + "n=2 #BCC structure\n", + "\n", + "#step 1 : claculation for lattice constant (a)\n", + "\n", + "a=(n*A/(N*p))**(1./3)\n", + "\n", + "#step 2 : radius of an atom in BCC\n", + "\n", + "r=sqrt(3)*a/4\n", + "\n", + "#step 3 : Atomic packing factor (APF)\n", + "\n", + "APF=n*((4./3)*math.pi*r**3)/a**3\n", + "\n", + "print\"Atomic packing factor (APF)=\",round(APF,4)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Atomic packing factor (APF)= 0.6802\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3.5,Page number 1-17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "A=120 #atomic weight of chromium\n", + "N=6.023*10**23 #Avogadro's number\n", + "p=5.2 #Density\n", + "n=2 #BCC structure\n", + "m=20 #mass\n", + "\n", + "#step 1 : claculation for volume of unit cell(a**3)\n", + "\n", + "a=(n*A/(N*p))\n", + "\n", + "#step 2 : volume of 20 gm of the element\n", + "\n", + "v=m/p\n", + "\n", + "#step 3 :no of unit cell\n", + "\n", + "x=v/a\n", + "\n", + "print\"no of unit cell=\",\"{0:.3e}\".format(x)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "no of unit cell= 5.019e+22\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3.6,Page number 1-18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "A=132.91 #atomic weight of chromium\n", + "N=6.023*10**26 #Avogadro's number\n", + "p=1900 #Density\n", + "a=6.14*10**-10 #lattice constant\n", + "\n", + "#step 1 : type of structure\n", + "\n", + "n=(p*N*a**3)/A\n", + "\n", + "print\"n =\",round(n)\n", + "\n", + "print\"BCC structure\"\n", + "\n", + "#step 2: no of atoms/m**3\n", + "\n", + "x=n/a**3\n", + "\n", + "print\"no of atoms/m^3=\",\"{0:.3e}\".format(x)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "n = 2.0\n", + "BCC structure\n", + "no of atoms/m^3= 8.610e+27\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3.7,Page number 1-18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "a=0.4049*10**-9 #lattice constant\n", + "t=0.006*10**-2 #thickness of Al foil\n", + "A=50*10**-4 #Area of foil\n", + "\n", + "V1=a**3 #volume of unit cell\n", + "\n", + "V=A*t #volume of the foil\n", + "\n", + "N=V/V1 #no of unit cell in the foil\n", + "\n", + "print\"no of unit cell in the foil=\",\"{0:.3e}\".format(N)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "no of unit cell in the foil= 4.519e+21\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5.1,Page number 1-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#refer diagram from textbook\n", + "\n", + "#on joining centre of 3 anions,an equilateral triangle is formed and on joining centres of any anion and cation a right angle triangle ABC os formed\n", + "\n", + "#where AC=rc+ra\n", + "\n", + "#and BC=ra\n", + "\n", + "#m(angle (ACB))=30 degree\n", + "\n", + "#therefore cos (30)=ra/(rc+ra)\n", + "\n", + "#assume rc/ra=r\n", + "\n", + "r=(1.0-math.cos(30.0*math.pi/180))/math.cos(math.pi*30/180) #by arrangimg terms we get value of r\n", + "\n", + "print\"critical radius ratio of ligancy 3=\",round(r,4)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "critical radius ratio of ligancy 3= 0.1547\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5.2,Page number 1-30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#refer diagram from textbook\n", + "\n", + "#in the said arrangement a cation is squeezed into 4 anions in a plane and 5th anion is in upper layer and 6th in bottom layer \n", + "\n", + "#join cation anion centres E and B and complete the triangle EBF\n", + "\n", + "#in triangle EBF m(angle F)=90 and EF=BF\n", + "\n", + "#m(angle B)=m(angle E)=45\n", + "\n", + "#and EB=rc+ra and BF=ra\n", + "\n", + "#cos(45)=ra/(rc+ra)\n", + "\n", + "#assume rc/ra=r\n", + "\n", + "r=(1-math.cos(45*math.pi/180))/math.cos(45*math.pi/180) #by arrangimg terms we get value of r\n", + "\n", + "print\"critical radius ratio for ligancy 6 =\",round(r,4)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5.3,Page number 1-30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#refer diagram from textbook\n", + "\n", + "#since plane is square hence it is same as ligancy 6\n", + "\n", + "#in the said arrangement a cation is squeezed into 4 anions in a plane and 5th anion is in upper layer and 6th in bottom layer \n", + "\n", + "#join cation anion centres E and B and complete the triangle EBF\n", + "\n", + "#in triangle EBF m(angle F)=90 and EF=BF\n", + "\n", + "#m(angle B)=m(angle E)=45\n", + "\n", + "#and EB=rc+ra and BF=ra\n", + "\n", + "#cos(45)=ra/(rc+ra)\n", + "\n", + "#assume rc/ra=r\n", + "\n", + "r=(1-math.cos(45*math.pi/180))/math.cos(45*math.pi/180) #by arrangimg terms we get value of r\n", + "\n", + "print\"critical radius ratio for ligancy 8 =\",round(r,4)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "critical radius ratio for ligancy 8 = 0.4142\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5.4,Page number 1-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#a tetrahedron CAEH can be considered with C as the apex of the tetrahedron.\n", + "\n", + "#the edges AE,AH and EH of the tetrahedron will then be the face of the cube faces ABEF,ADHF,EFHG resp.\n", + "\n", + "#from fig\n", + "\n", + "#AO=ra+rc and AJ=ra\n", + "\n", + "#AE=root(2)*a and AG=root(3)*a\n", + "\n", + "#AO/AJ=AG/AE=(ra+rc)/ra=root(3)*a/root(2)*a\n", + "\n", + "#assume rc/ra=r\n", + "r=(math.sqrt(3)-math.sqrt(2))/math.sqrt(2)\n", + "\n", + "print\"critical radius ratio for ligancy 4 = \",round(r,4)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "critical radius ratio for ligancy 4 = 0.2247\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5.5,Page number 1-32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#ligancy 8 represents cubic arrangment .8 anions are at the corners and touch along cube edgs.Along the body diagonal the central cation and the corner anion are in contact.\n", + "\n", + "#cube edge=2*ra\n", + "\n", + "#refer diagram from textbook\n", + "\n", + "#and body diagonal=root(3)*cube edge=root(3)[2*(rc+ra)]\n", + "\n", + "#assume rc/ra=r\n", + "\n", + "r=math.sqrt(3)-1.0\n", + "\n", + "print\"critical radius ratio of ligancy 8=\",round(r,4)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "critical radius ratio of ligancy 8= 0.7321\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5.6,Page number 1-32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#for an ionic crystal exibiting HCP structure the arrangment of ions refere from textbook\n", + "\n", + "#at centre we have a cation with radius rc=OA\n", + "\n", + "#it is an touch with 6 anions with radius ra=AB\n", + "\n", + "#OB=OC=ra+rc\n", + "\n", + "#intrangle ODB ,m(angle (OBC))=60 degree ,m(angle (ODB))=90 degree\n", + "\n", + "#therefore cos(60)=BD/OB=AB/(OA+OB)=ra/(rc+ra)\n", + "\n", + "#assume rc/ra=r\n", + "\n", + "r=(1.-math.cos(60*math.pi/180))/math.cos(60*math.pi/180) #by arrangimg terms we get value of r\n", + "\n", + "print\"critical radius ratio 0f HCP structure=\",round(r,4)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "critical radius ratio 0f HCP structure= 1.0\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6.2,Page number 1-35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#intercept of planeare in proportion a,b/3,2*c\n", + "\n", + "#as a,b and c are basic vectors the proportin of intercepts 1:1/3:2\n", + "\n", + "#therefore reciprocal\n", + "\n", + "r1=1\n", + "\n", + "r2=3\n", + "\n", + "r3=1./2\n", + "\n", + "#taking LCM of 2 and 1 is 2\n", + "\n", + "l=2\n", + "\n", + "m1=(l*r1)\n", + "\n", + "m2=(l*r2)\n", + "\n", + "m3=(l*r3)\n", + "\n", + "print\"miler indices=\",m3,m2,m1\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "miler indices= 1.0 6 2\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6.4,Page number 1-38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "r=1.414 #atomic radius in amstrong unit\n", + "\n", + "#for FCC structure\n", + "\n", + "a=4*r/math.sqrt(2)\n", + "\n", + "#part 1: plane(2,0,0)\n", + "\n", + "#the interplanar spacing of plane\n", + "\n", + "h1=2\n", + "k1=0\n", + "l1=0\n", + "\n", + "#we know that d=a/sqrt(h**2+k**2+l**2)\n", + "\n", + "d1=a/sqrt(h1**2+k1**2+l1**2)\n", + "\n", + "print\"1)interplanar spacing for (2,0,0) plane=\",round(d1,4),\"amstrong\"\n", + "\n", + "#part 2: plane(1,1,1)\n", + "\n", + "#the interplanar spacing of plane\n", + "\n", + "h2=1\n", + "k2=1\n", + "l2=1\n", + "\n", + "#we know that d=a/sqrt(h**2+k**2+l**2)\n", + "\n", + "d2=a/sqrt(h2**2+k2**2+l2**2)\n", + "\n", + "print\"2)interplanar spacing for(1,1,1) plane=\",round(d2,4),\"amstrong\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1)interplanar spacing for (2,0,0) plane= 1.9997 amstrong\n", + "2)interplanar spacing for(1,1,1) plane= 2.3091 amstrong\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.1,Page number 1-58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=4 #FCC structure\n", + "ro=2180 #density of NaCl\n", + "M=23+35.5 #molecular weight of NaCl\n", + "N=6.023*10**26 #Avogadro's number\n", + "\n", + "a=((n*M)/(N*ro))**(1.0/3)\n", + "\n", + "print\"Lattice constant=\",\"{0:.3e}\".format(a),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lattice constant= 5.627e-10 m\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.2,Page number 1-58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=4 #FCC structure\n", + "ro=8.9 #density of Cu atom\n", + "A=63.55 #atomic weight of Cu atom\n", + "N=6.023*10**23 #Avogadro's number\n", + "\n", + "a=((n*A)/(N*ro))**(1./3)\n", + "\n", + "print\"1) Lattice constant=\",\"{0:.3e}\".format(a),\"cm\"\n", + "\n", + "r=math.sqrt(2)*a/4 #radius of Cu atom\n", + "\n", + "d=2*r #diameter of Cu atom\n", + "\n", + "print\"2) Diameter of Cu atom=\",\"{0:.3e}\".format(d),\"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1) Lattice constant= 3.620e-08 cm\n", + "2) Diameter of Cu atom= 2.559e-08 cm\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.3,Page number 1-59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=8 #diamond structure\n", + "A=12.01 #atomic wt\n", + "N=6.023*10**23 #Avogadro's number\n", + "a=3.75*10**-8 #lattice constant of diamond\n", + "\n", + "ro=(n*A)/(N*(a**3))\n", + "\n", + "print\"Density of diamond=\",round(ro,4),\"gm/cc\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Density of diamond= 3.025 gm/cc\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.4,Page number 1-59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#intercept of planeare in proportion 3a:4b:infinity (plane parallel to z axis)\n", + "\n", + "#as a,b and c are basic vectors the proportin of intercepts 3:4:infinity\n", + "\n", + "#therefore reciprocal\n", + "\n", + "r1=1./3\n", + "r2=1./4\n", + "r3=0\n", + "\n", + "#taking LCM of 3 and 4 i.e. 12\n", + "\n", + "l=12\n", + "\n", + "m1=(l*r1)\n", + "\n", + "m2=(l*r2)\n", + "\n", + "m3=(l*r3)\n", + "\n", + "print\"miler indices=\",(m3,m2,m1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "miler indices= (0, 3.0, 4.0)\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.5,Page number 1-59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#intercept of planeare in proportion 3a:-2b:3/2c\n", + "\n", + "#as a,b and c are basic vectors the proportin of intercepts 3:-2:3/2\n", + "\n", + "#therefore reciprocal\n", + "\n", + "r1=1./3\n", + "r2=-1./2\n", + "r3=2./3\n", + "\n", + "#taking LCM of 3, 2 and 3/2 is 6\n", + "\n", + "l=6\n", + "\n", + "m1=(l*r1)\n", + "\n", + "m2=(l*r2)\n", + "\n", + "m3=(l*r3)\n", + "\n", + "print\"miler indices=\",(m3,m2,m1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "miler indices= (4.0, -3.0, 2.0)\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.6,Page number 1-59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#if a plane cut at length m,n,p on the three crystal axes,then\n", + "\n", + "#m:n:p=xa:yb:zc\n", + "\n", + "#when primitive vectors of unit cell and numbers x,y,z,are related to miller indices (h,k,l)of the plane by relation\n", + "\n", + "#1/x:1/y:1/z=h:k:l\n", + "\n", + "#since a=b=c (crystal is simple cubic)\n", + "\n", + "#and (h,k,l)=(1,2,3)\n", + "\n", + "#therefore reciprocal\n", + "\n", + "r1=1./1\n", + "r2=1./2\n", + "r3=1./3\n", + "\n", + "#taking LCM of 1 ,2 and 3 is 6\n", + "\n", + "l=6\n", + "\n", + "m=(l*r1)\n", + "\n", + "n=(l*r2)\n", + "\n", + "p=(l*r3)\n", + "\n", + "print\"ratio of intercepts=\",(m,n,p)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio of intercepts= (6.0, 3.0, 2.0)\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.7,Page number 1-60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#primitive vectors\n", + "\n", + "a=1.2 #in amstrong unit\n", + "b=1.8 #in amstrong unit\n", + "c=2 #in amstrong unit\n", + "\n", + "#miller indices of the plane\n", + "\n", + "h=2\n", + "k=3\n", + "l=1\n", + "\n", + "#therefore intercepts are a/h,b/k,c/l\n", + "\n", + "x=a/h\n", + "y=b/k\n", + "z=c/l\n", + "\n", + "#this gives intercepts along x axis as x amstrong but it is given tthat plane cut x axis at 1.2 amstrong .\n", + "\n", + "t=1.2/x\n", + "\n", + "#this shows that the plane under consideration is another plane which is parallel to it(to keep miller indices same)\n", + "\n", + "n=t*y #Y intercept\n", + "\n", + "p=t*z #Z intercept\n", + "\n", + "print\"1) Y intercept=\",n,\"amstrong\"\n", + "\n", + "print\"2)Z intercept=\",p,\"amstrong\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1) Y intercept= 1.2 amstrong\n", + "2)Z intercept= 4.0 amstrong\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.8,Page number 1-61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#the interplanar spacing of plane\n", + "\n", + "h=1\n", + "k=1\n", + "l=0\n", + "d=2 #interpanar spacing in amstrong unit\n", + "\n", + "#we know that d=a/sqrt(h**2+k**2+l**2) therefore\n", + "\n", + "a=d*math.sqrt(h**2+k**2+l**2)\n", + "\n", + "#for FCC structure\n", + "\n", + "r=math.sqrt(2)*a/4\n", + "\n", + "print\"radius r=\",(r),\"amstrong\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "radius r= 1.0 amstrong\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.9,Page number 1-61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=4 #for FCC structure\n", + "\n", + "#the interplanar spacing of plane\n", + "\n", + "h=1\n", + "k=1\n", + "l=1\n", + "d=2.08*10**-10 #distance\n", + "A=63.54 #atomic weight of Cu\n", + "N=6.023*10**26 #amstrong no\n", + "\n", + "#we know that d=a/sqrt(h**2+k**2+l**2) therefore\n", + "\n", + "a=d*math.sqrt(h**2+k**2+l**2)\n", + "\n", + "#also (a**3*q)=n*A/N\n", + "\n", + "q=n*A/(N*a**3)\n", + "\n", + "print\"1)density=\",round(q,4),\"kg/m^3\"\n", + "\n", + "#for FCC structure\n", + "\n", + "r=math.sqrt(2)*a/4\n", + "\n", + "d=r*2\n", + "\n", + "print\"2)radius r=\",\"{0:.3e}\".format(r),\"m\"\n", + "\n", + "print\"3)diameter d=\",\"{0:.3e}\".format(d),\"m\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1)density= 9024.4855 kg/m^3\n", + "2)radius r= 1.274e-10 m\n", + "3)diameter d= 2.547e-10 m\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.10,Page number 1-62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "A=63.546 #atomic weight of Cu\n", + "N=6.023*10**26 #Avogadro's number\n", + "p=8930 #Density\n", + "n=1.23 #no.of electron per atom\n", + "\n", + "#density=mass/volume\n", + "\n", + "#therfore 1/volume=density/mass\n", + "\n", + "#since electron concentration is needed, let us find out no of atoms/volume(x)\n", + "\n", + "x=N*p/A\n", + "\n", + "#now one atom contribute n=1.23 electron\n", + "\n", + "#therefore x atoms contribute y no of free electron\n", + "\n", + "y=x*n\n", + "\n", + "print\"free electron concentration=\",\"{0:.3e}\".format(y),\"electron/m^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "free electron concentration= 1.041e+29 electron/m^3\n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.11,Page number 1-62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#primitive vectors\n", + "\n", + "a=1.5 #in amstrong unit\n", + "b=2 #in amstrong unit\n", + "c=4. #in amstrong unit\n", + "\n", + "#miller indices of the plane\n", + "\n", + "h=3\n", + "k=2\n", + "l=6\n", + "\n", + "#therefore intercepts are a/h,b/k,c/l\n", + "\n", + "x=a/h\n", + "y=b/k\n", + "z=c/l\n", + "\n", + "#this gives intercepts along x axis as x amstrong but it is given that plane cut x axis at 1.2 amstrong .\n", + "\n", + "t=1.5/x\n", + "\n", + "#this shows that the plane under consideration is another plane which is parallel to it(to keep miller indices same)\n", + "\n", + "n=t*y #Y intercept\n", + "\n", + "p=t*z #Z intercept\n", + "\n", + "print\"1) Y intercept=\",(n),\"amstrong\"\n", + "\n", + "print\"2)Z intercept=\",(p),\"amstrong\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1) Y intercept= 3.0 amstrong\n", + "2)Z intercept= 2.0 amstrong\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.12,Page number 1-63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "ro=7.87 #density of metal\n", + "A=55.85 #atomic wt of metal\n", + "N=6.023*10**23 #Avogadro's number\n", + "a=2.9*10**-8 #lattice constant of metal\n", + "\n", + "n=(N*(a**3)*ro)/A\n", + "\n", + "print\"Number of atom per unit cell of a metal=\",round(n,0)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of atom per unit cell of a metal= 2.0\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.13,Page number 1-63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=2 #BCC structure\n", + "ro=9.6*10**2 #density of sodium crystal\n", + "A=23 #atomic weight of sodium crystal\n", + "N=6.023*10**26 #Avogadro's number\n", + "\n", + "a=((n*A)/(N*ro))**(1./3)\n", + "\n", + "print\"Lattice constant=\",\"{0:.3e}\".format(a),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lattice constant= 4.301e-10 m\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.15,Page number 1-64" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "ro=2.7*10**3 #density of metal\n", + "A=27 #atomic wt of metal\n", + "N=6.023*10**26 #Avogadro's number\n", + "a=4.05*10**-10 #lattice constant of metal\n", + "\n", + "n=(N*(a**3)*ro)/A\n", + "\n", + "print\"1) Number of atom per unit cell of a metal=\",round(n,0)\n", + "\n", + "r=math.sqrt(2)*a/4 #radius of metal\n", + "\n", + "print\"2) atomic radius of a metal=\",\"{0:.3e}\".format(r),\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1) Number of atom per unit cell of a metal= 4.0\n", + "2) atomic radius of a metal= 1.432e-10 m\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.16,Page number 1-64" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=2 #BCC structure\n", + "ro=5.98*10**3 #density of chromium\n", + "A=50 #atomic wt of chromium\n", + "N=6.023*10**26 #Avogadro's number\n", + "\n", + "a=((n*A)/(N*ro))**(1./3)\n", + "\n", + "print\"1) Lattice constant=\",\"{0:.3e}\".format(a),\"m\"\n", + "\n", + "#for BCC\n", + "\n", + "r=math.sqrt(3)*a/4 #radius of chromium\n", + "\n", + "APF=(n*(4./3)*math.pi*(r**3))/(a**3)\n", + "\n", + "print\"2) A.P.F. for chromium=\",round(APF,4)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1) Lattice constant= 3.028e-10 m\n", + "2) A.P.F. for chromium= 0.6802\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.17,Page number 1-65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=4 #FCC structure\n", + "ro=6250 #density\n", + "M=60.2 #molecular weight\n", + "N=6.023*10**26 #Avogadro's number\n", + "\n", + "a=((n*M)/(N*ro))**(1./3)\n", + "\n", + "print\"Lattice constant=\",\"{0:.3e}\".format(a),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lattice constant= 3.999e-10 m\n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.19,Page number 1-66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "a=2.82*10**-9 #lattice constant\n", + "n=2 #FCC crystal\n", + "t=17.167 #glancing angle in degree\n", + "q=math.pi/180*t #glancing angle in radians\n", + "\n", + "#assuming reflection in (1,0,0) plane\n", + "\n", + "h=1\n", + "k=0\n", + "l=0\n", + "\n", + "d=a/math.sqrt(h**2+k**2+l**2)\n", + "\n", + "#using Bragg's law , 2*d*sin(q)=n*la\n", + "\n", + "la=2*d*sin(q)/n\n", + "\n", + "print\"wavlength of X-ray=\",\"{0:.3e}\".format(la),\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavlength of X-ray= 8.323e-10 m\n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.20,Page number 1-66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=8 #Diamond structure\n", + "ro=2.33*10**3 #density of diamond\n", + "M=28.9 #atomic weight of diamond\n", + "N=6.023*10**26 #Avogadro's number\n", + "\n", + "a=((n*M)/(N*ro))**(1./3)\n", + "\n", + "print\"1) Lattice constant=\",\"{0:.3e}\".format(a),\"m\"\n", + "\n", + "r=math.sqrt(3)*a/8 #radius of diamond structure\n", + "\n", + "print\"2) atomic radius of a metal=\",\"{0:.3e}\".format(r),\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1) Lattice constant= 5.482e-10 m\n", + "2) atomic radius of a metal= 1.187e-10 m\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14.21,Page number 1-66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=2 #BCC structure\n", + "ro=8.57*10**3 #density of chromium\n", + "d=2.86*10**-10 #nearest atoms distance\n", + "\n", + "#d=sqrt(3)/2*a\n", + "\n", + "a=2*d/math.sqrt(3)\n", + "\n", + "#now use formulae a**3*ro=n*A/N\n", + "\n", + "#therefore a**3*ro/n=mass of unit cell/(no of atoms pre unit cell)=mass of one atom\n", + "\n", + "m=a**3*ro/n\n", + "\n", + "print\"mass of one atom=\",\"{0:.3e}\".format(m),\"kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "mass of one atom= 1.543e-25 kg\n" + ] + } + ], + "prompt_number": 68 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.1,Page number 1-68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "d=4.255*10**-10 #interplaner spacing\n", + "l=1.549*10**-10 #wavelength of x ray\n", + "\n", + "#part 1: for smallest glancing angle(n=1)\n", + "\n", + "n1=1\n", + "\n", + "#using Bragg's law n*l=2*d*sin(q)\n", + "\n", + "q=math.degrees(math.asin(n1*l/(2*d)))\n", + "\n", + "print\"1)glancing angle=\",round(q,4),\"degree\"\n", + "\n", + "#part 2: for highst order\n", + "\n", + "#for highest order sin(q) not exceed one i.e maximum value is one\n", + "\n", + "#using Bragg's law n*l=2*d*sin(q)\n", + "\n", + "n2=2*d/l #since sin(q)is one\n", + "\n", + "print\"2)highest order possible =\",math.floor(n2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1)glancing angle= 10.4875 degree\n", + "2)highest order possible = 5.0\n" + ] + } + ], + "prompt_number": 70 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.2,Page number 1-69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "a=2.125*10**-10 #lattice constant\n", + "d=a/2 #interplaner spacing\n", + "n=2 #second order maximum\n", + "l=0.592*10**-10 #wavelength of rock salt crystal\n", + "\n", + "#using Bragg's law\n", + "\n", + "q=math.degrees(math.asin((n*l)/(2*d))) #glancing angle\n", + "\n", + "print\"glancing angle=\",round(q,4),\"degree\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "glancing angle= 33.8608 degree\n" + ] + } + ], + "prompt_number": 72 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.3,Page number 1-69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n1=1 #for 1st order\n", + "n2=2 #for 2nd order\n", + "t=3.4 #angle where 1st order reflection done\n", + "t1=t*math.pi/180 #convert degree to radian\n", + "\n", + "m=math.sin(t1)\n", + "\n", + "#but from Bragg's law\n", + "\n", + "#n*l=2*d*sin(t)\n", + "\n", + "#for for constant distance(d) and wavelength(l) \n", + "\n", + "#order(n) is directly proportionl to sine of angle i.e (sin(t))\n", + "\n", + "#n1/n2=sin(t1)/sin(t2)\n", + "\n", + "#assume sin(t2)=a\n", + "\n", + "a=n2/n1*m\n", + "\n", + "t2=math.degrees(math.asin(a)) #taking sin inverese in degree\n", + "\n", + "print\"second order reflection take place at an angle=\",round(t2,4),\"degree\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "second order reflection take place at an angle= 6.812 degree\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.4,Page number 1-70" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "V=50*10**3 #operating voltage of x-ray\n", + "M=74.6 #molecular weight\n", + "p=1.99*10**3 #density\n", + "n=4 #no of atoms per unit cell(for FCC structure)\n", + "h=6.63*10**-34 #plank's constant\n", + "c=3*10**8 #velocity \n", + "e=1.6*10**-19 #charge on electron\n", + "N=6.023*10**26 #Avogadro's number\n", + "\n", + "#step 1:clculating shortest wavelength\n", + "\n", + "l=h*c/(e*V)\n", + "\n", + "print\"1)shortest wavelength=\",(l),\"m\"\n", + "\n", + "#step:2 calculating distance(d)\n", + "\n", + "#now a**3*p=n*M/N therefore,\n", + "\n", + "a=(n*M/(N*p))**(1./3)\n", + "\n", + "#since KCl is ionic crystal herefore,\n", + "\n", + "d=a/2\n", + "\n", + "#step 3: calculaing glancing angle\n", + "\n", + "#using Bragg's law\n", + "\n", + "#n*l=2*d*sin(t)\n", + "\n", + "#assume sin(t)=a, wavelength is minimum i.e l and n=1\n", + "\n", + "n=1\n", + "\n", + "a=n*l/(2*d)\n", + "\n", + "t=math.degrees(math.asin(a)) #taking sin inverese in degree\n", + "\n", + "print\"2) glancing angle=\",round(t,4),\"degree\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1)shortest wavelength= 2.48625e-11 m\n", + "2) glancing angle= 2.265 degree\n" + ] + } + ], + "prompt_number": 77 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.5,Page number 1-70" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=1.0 #first order maximum\n", + "l=0.82*10**-10 #wavelength of X ray\n", + "qd=7.0 #glancing angle in degree\n", + "qm=51./60 #glancing angle in minute\n", + "qs=48./3600 #glancing angle in second\n", + "\n", + "q=qd+qm+qs #total glancin angle in degree\n", + "\n", + "#using Bragg's law n*l=2*d*sin(q)\n", + "\n", + "d=n*l/(2*math.sin(q*math.pi/180))\n", + "\n", + "a=3*10**-10 #lattice constant\n", + "\n", + "#we know that d=a/root(h**2+k**2+l**2)\n", + "\n", + "#assume root(h**2+k**2+l**2) =m\n", + "\n", + "#arranging terms we get\n", + "\n", + "m=a/d\n", + "\n", + "print\"square root(h**2+k**2+l**2)=\",round(m,0)\n", + "\n", + "print\"hence possible solutions are (100),(010),(001)\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "square root(h**2+k**2+l**2)= 1.0\n", + "hence possible solutions are (100),(010),(001)\n" + ] + } + ], + "prompt_number": 90 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.6,Page number 1-71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=1 #first order maximum\n", + "l=1j #wavelength of X ray\n", + "\n", + "#part 1:for(100)\n", + "\n", + "#using Bragg's law n*l=2*d*sin(q)\n", + "\n", + "q1=5.4 #glancing angle in degree\n", + "\n", + "dl1=n*l/(2*math.sin(q1*math.pi/180))\n", + "\n", + "#part 2:for(110)\n", + "\n", + "#using Bragg's law n*l=2*d*sin(q)\n", + "\n", + "q2=7.6 #glancing angle in degree\n", + "\n", + "dl2=n*l/(2*math.sin(q2*math.pi/180))\n", + "\n", + "#part 3:for(111)\n", + "\n", + "#using Bragg's law n*l=2*d*sin(q)\n", + "\n", + "q3=9.4 #glancing angle in degree\n", + "\n", + "dl3=n*l/(2*math.sin(q3*math.pi/180))\n", + "\n", + "#for taking ratio divide all dl by dl1\n", + "\n", + "d1=dl1/dl1\n", + "\n", + "d2=dl2/dl1\n", + "\n", + "d3=dl3/dl1\n", + "\n", + "print\"cubic lattice structure is=\",d1,d2,d3" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " cubic lattice structure is= (1+0j) (0.711559669333+0j) (0.576199350225+0j)\n" + ] + } + ], + "prompt_number": 94 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.7,Page number 1-71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=1 #first order maximum\n", + "l=1.54*10**-10 #wavelength of rock salt crystal\n", + "q=21.7 #glancing angle in degree\n", + "\n", + "#using Bragg's law n*l=2*d*sin(q)\n", + "\n", + "d=n*l/(2*math.sin(q*math.pi/180))\n", + "\n", + "print\"lattice constant of crystal=\",\"{0:.3e}\".format(d),\"meter\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "lattice constant of crystal= 2.083e-10 meter\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.8,Page number 1-72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "a=2.814*10**-10 #lattice constant\n", + "\n", + "#the interplanar spacing of plane\n", + "\n", + "h=1\n", + "k=0\n", + "l=0\n", + "\n", + "d=a/math.sqrt(h**2+k**2+l**2)\n", + "\n", + "n=2 #first order maximum\n", + "\n", + "l=0.714*10**-10 #wavelength of X-ray crystal\n", + "\n", + "#using Bragg's law\n", + "\n", + "q=math.degrees(math.asin((n*l)/(2*d))) #glancing angle\n", + "\n", + "print\"glancing angle=\",round(q,4),\"degree\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "glancing angle= 14.6984 degree\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.9,Page number 1-72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "d=2.82*10**-10 #interplaner spacing\n", + "t=10 #glancing angle\n", + "\n", + "#for part 1\n", + "\n", + "n=1 #first order maximum\n", + "\n", + "#using Bragg's law n*l=2*d*sin(t)\n", + "\n", + "l=2*d*math.sin(math.pi*t/180)/n\n", + "\n", + "print\"1)wavelength=\",\"{0:.3e}\".format(l),\"meter\"\n", + "\n", + "#for part 2\n", + "\n", + "n1=2\n", + "\n", + "#using Bragg's law n*l=2*d*sin(q)\n", + "\n", + "q=math.degrees(math.asin(n1*l/(2*d)))\n", + "\n", + "print\"2)glancing angle=\",round(q,4),\"degree\"\n", + "\n", + "#for part 3\n", + "\n", + "#for highest order sin(q) not exceed one i.e maximum value is one\n", + "\n", + "#using Bragg's law n*l=2*d*sin(q)\n", + "\n", + "n2=2*d/l #since sin(q)is one\n", + "\n", + "print\"3)highest order possible =\",(floor(n2))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1)wavelength= 9.794e-11 meter\n", + "2)glancing angle= 20.322 degree\n", + "3)highest order possible = 5.0\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.10,Page number 1-73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "#for line -A\n", + "\n", + "n1=1 #1st order maximum\n", + "q1=30 #glancing angle in degree\n", + "\n", + "#using Bragg's law for line A n1*l1=2*d1*sin(q1)\n", + "\n", + "#d1=n1*l1/(2*sin(q1))\n", + "\n", + "#for line B\n", + "\n", + "l2=0.97 #wavelength in amstrong unit\n", + "n2=3 #1st order maximum\n", + "q2=60 #glancing angle in degree\n", + "\n", + "#using Bragg's law for line B n2*l2=2*d2*sin(q2)\n", + "\n", + "#since for both lines A and B we use same plane of same crystal,therefore\n", + "\n", + "#d1=d2\n", + "\n", + "#therefore equution became n2*l2=2*n1*l1/(2*sin(q1))*sin(q2)\n", + "\n", + "#by arranging terms we get\n", + "\n", + "\n", + "l1=n2*l2*2*math.sin(q1*math.pi/180)/(2*n1*math.sin(q2*math.pi/180))\n", + "\n", + "print\"wavelength of the line A=\",round(l1,4),\"amstrong\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength of the line A= 1.6801 amstrong\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.11,Page number 1-74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "n=1.0 #first order minimum\n", + "d=5.5*10**-11 #atomic spacing\n", + "e=1.6*10**-19 #charge on one electron\n", + "Ee=10*10**3 #energy in eV\n", + "E=e*Ee #energy in Joule\n", + "m=9.1*10**-31 #mass of elelctron\n", + "h=6.63*10**-34 #plank's constant\n", + "\n", + "l=h/math.sqrt(2*m*E) #wavelength\n", + "\n", + "#using Bragg's law\n", + "\n", + "q=math.degrees(math.asin((n*l)/(2*d))) #glancing angle\n", + "\n", + "print\"glancing angle=\",round(q,4),\"degree\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "glancing angle= 6.4129 degree\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15.12,Page number 1-74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "a=2.814*10**-10 #lattice constant\n", + "\n", + "#for rock salt\n", + "\n", + "d=a/2 #interplaner spacing\n", + "\n", + "n=1 #first order maximum\n", + "\n", + "l=1.541*10**-10 #wavelength of rock salt crystal\n", + "\n", + "#using Bragg's law\n", + "\n", + "q=math.degrees(math.asin((n*l)/(2*d))) #glancing angl\n", + "\n", + "print\"glancing angle=\",round(q,4),\"degree\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "glancing angle= 33.2038 degree\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.16.1,Page number 1-75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "Ev=1.08 #average energy required to creaet a vacancy\n", + "k=1.38*10**-23 #boltzman constant in J/K\n", + "e=1.6*10**-19 #charge on 1 electron\n", + "\n", + "K=k/e #boltzman constant in eV/K\n", + "\n", + "#for a low concentration of vacancies a relation is\n", + "\n", + "#n=Nexp(-Ev/KT)\n", + "\n", + "#since total no atom is 1 hence N=1\n", + "\n", + "#at 1000k\n", + "\n", + "T1=1000 #temperature\n", + "\n", + "n1=math.exp(-Ev/(K*T1))\n", + "\n", + "#at 500k\n", + "\n", + "T2=500 #temperature\n", + "\n", + "n2=math.exp(-Ev/(K*T2))\n", + "\n", + "v=(n1)/(n2) #ratio of vacancies\n", + "\n", + "print\"ratio of vacancies=\",round(v,4)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio of vacancies= 274234.5745\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.16.2,Page number 1-75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "Ev=1.95 #average energy required to creaet a vacancy\n", + "k=1.38*10**-23 #boltzman constant in J/K\n", + "e=1.6*10**-19 #charge on 1 electron\n", + "K=k/e #boltzman constant in eV/K\n", + "T=500 #temperature\n", + "\n", + "#for a low concentration of vacancies a relation is\n", + "\n", + "#n=Nexp(-Ev/KT)\n", + "\n", + "m=math.exp(-Ev/(K*T)) #ratio of no of vacancies to no of atoms n/N\n", + "\n", + "print\"ratio of no of vacancies to no of atoms=\",\"{0:.3e}\".format(m)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio of no of vacancies to no of atoms= 2.303e-20\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.16.3,Page number 1-76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "Ev=1.8 #average energy required to creaet a vacancy\n", + "k=1.38*10**-23 #boltzman constant in J/K\n", + "e=1.6*10**-19 #charge on 1 electron\n", + "K=k/e #boltzman constant in eV/K\n", + "\n", + "#for a low concentration of vacancies a relation is\n", + "\n", + "#n=Nexp(-Ev/KT)\n", + "\n", + "#ratio of vacancy is n/N assume be r=exp(-Ev/KT)\n", + "\n", + "#since total no atom is 1 hence N=1\n", + "\n", + "#at 1000k\n", + "\n", + "t1=-119 #temperature in degree\n", + "T1=t1+273 #temperature in kelvine\n", + "r1=math.exp(-Ev/(K*T1))\n", + "\n", + "print\"1)ratio of vacancies at -119 degree=\",\"{0:.3e}\".format(r1)\n", + "\n", + "#at 500k\n", + "\n", + "t2=80 #temperature in degree\n", + "\n", + "T2=t2+273 #temperature in kelvine\n", + "\n", + "r2=exp(-Ev/(K*T2))\n", + "\n", + "v=(r1)/(r2) #ratio of vacancies\n", + "\n", + "print\"2)ratio of vacancies at 80 degree=\",\"{0:.3e}\".format(r2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1)ratio of vacancies at -119 degree= 1.399e-59\n", + "2)ratio of vacancies at 80 degree= 2.110e-26\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.16.4,Page number 1-76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#given data\n", + "\n", + "Ev=1.5 #energy of formaton of frankel defect\n", + "k=1.38*10**-23 #boltzman constant in J/K\n", + "e=1.6*10**-19 #charge on 1 electron\n", + "K=k/e #boltzman constant in eV/K\n", + "T=700 #temperature\n", + "N=6.023*10**26 #avogadro's no\n", + "\n", + "#for a low concentration of vacancies a relation is\n", + "\n", + "#n=Nexp(-Ev/KT)\n", + "\n", + "m=math.exp(-Ev/(2*K*T)) #ratio of no of vacancies to no of atoms n/N\n", + "\n", + "qs=5.56 #specific density\n", + "q=5.56*10**3 #real density ke/m**3\n", + "M=0.143 #molecular weight in kg/m**3\n", + "ma=M/N #mass of one molecule\n", + "v=ma/q #vol of one molecule\n", + "\n", + "#v volume containe 1 molecule\n", + "\n", + "#therefore 1 m**3 containe x molecule\n", + "\n", + "x=1./v\n", + "d=m*x #defect per m**3\n", + "dm=d*10**-9 #defect per mm**3\n", + "\n", + "print\"number of frankel defects per mm^3=\",\"{0:.3e}\".format(dm)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "number of frankel defects per mm^3= 9.432e+16\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit