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-{
- "metadata": {
- "celltoolbar": "Raw Cell Format",
- "name": "",
- "signature": "sha256:3e9b00b8b544a24032a4bb804cb876f45a5efd85913287f396a56723a0eb1a09"
- },
- "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",
- "p=math.cos(45*math.pi/180)\n",
- "r=(1-p)/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": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "critical radius ratio for ligancy 6 = 0.4142\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "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
generated by cgit (git 2.34.1) at 2024-12-19 18:46:24 +0000