path: root/Engineering_Physics/Chapter_6.ipynb

{
 "metadata": {
  "name": "Chapter 6"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": "Crystallography"
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.1, Page number 185"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the density of diamond\n\n#importing modules\nimport math\n\n#Variable declaration\nr=0.071;      #radius in nm\nN=6.022*10**26;    \n\n#Calculation\nr=r*10**-9;        #converting r from nm to m\n#mass of carbon atom m = 12/N\nm=12/N;\n#mass of diamond M = 8*mass of one carbon atom\nM=8*m;\n#volume of diamond V = (8*r/sqrt(3))^3\nV=(8*r/math.sqrt(3))**3;\nd=M/V;    #density in kg/m^3\nd=math.ceil(d*100)/100;   #rounding off to 2 decimals\n\n#Result\nprint(\"density of diamond in kg/m^3 is\",d);\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('density of diamond in kg/m^3 is', 4520.31)\n"
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.2, Page number 185"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the percentage volume change\n\n#importing modules\nimport math\n\n#Variable declaration\naBCC=0.332;      #lattice constant in nm\naHCP=0.296;      #lattice constant in nm\nc=0.468;      #c in nm\n\n#Calculation\naBCC=aBCC*10**-9;     #converting nm to m\nVbcc=aBCC**3;\naHCP=aHCP*10**-9;     #converting nm to m\nc=c*10**-9;     #converting nm to m\nVhcp=6*(math.sqrt(3)/4)*aHCP**2*c;\nV=Vhcp-Vbcc;\nVch=(V*100)/Vbcc;\nVch=math.ceil(Vch*100)/100;   #rounding off to 2 decimals\n\n#Result\nprint(\"percentage change in volume is\",Vch);\n\n#answer given in the book is wrong",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('percentage change in volume is', 191.12)\n"
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.3, Page number 186"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the density\n\n#importing modules\nimport math\n\n#Variable declaration\nr=1.278;     #atomic radius of Cu in Angstrom\nA=63.54;     #atomic weight of Cu\nn=4;     #for FCC n=4\nNa=6.022*10**26;\n\n#Calculation\nr=r*10**-10;    #converting atomic radius from Angstrom to m\na=2*math.sqrt(2)*r;     \nrho=(n*A)/(Na*a**3);\nrho=math.ceil(rho*100)/100;   #rounding off to 2 decimals\n\n#Result\nprint(\"density of Cu in kg/m^3 is\",rho);\n\n#answer given in the book is wrong",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('density of Cu in kg/m^3 is', 8935.92)\n"
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.4, Page number 186"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the distance between adjacent atoms\n\n#importing modules\nimport math\nimport numpy as np\n\n#Variable declaration\nrho=2180;     #density of NaCl in kg/m^3\nwNa=23;      #atomic weight of Na\nwCl=35.5;      #atomic weight of Cl\nn=4;   #for FCC n=4\nNa=6.022*10**26;\n\n#Calculation\nA=wNa+wCl;     #molecular weight of NaCl\nx=np.reciprocal(3.);\na=((n*A)/(Na*rho))**x;\n\n#Result\nprint(\"interatomic distance in NaCl in m is\",a); \n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('interatomic distance in NaCl in m is', 5.6278114346454509e-10)\n"
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.5, Page number 187"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the interplanar spacing\n\n#importing modules\nimport math\n\n#Variable declaration\na=0.42;    #lattice constant in nm\nh1=1;\nk1=0;\nl1=1;    #indices of the plane (101)\nh2=2;\nk2=2;\nl2=1;    #indices of the plane (221)\n\n#Calculation\na=a*10**-9;    #converting from nm to m\nd1=a/math.sqrt((h1**2)+(k1**2)+(l1**2));     #interplanar spacing for plane (101)\nd1=d1*10**9;    #converting from m to nm\nd1=math.ceil(d1*10**5)/10**5;   #rounding off to 5 decimals\nd2=a/math.sqrt((h2**2)+(k2**2)+(l2**2));     #interplanar spacing for plane (221)\nd2=d2*10**9;    #converting from m to nm\n\n#Result\nprint(\"interplanar spacing for (101) in nm is\",d1);\nprint(\"interplanar spacing for (221) in nm is\",d2);\n\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('interplanar spacing for (101) in nm is', 0.29699)\n('interplanar spacing for (221) in nm is', 0.14)\n"
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.6, Page number 187"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To identify the axial intercepts\n\n#Variable declaration\nh1=1;\nk1=0;\nl1=2;    #indices for plane (102)\nh2=2;\nk2=3;\nl2=1;    #indices for plane (231)\nh3=3;\nk3=-1;\nl3=2;    #indices for plane (31'2)\n\n#Calculation\n#intercepts made by the plane is a/h, b/k, c/l\n#for plane (102) intercepts are a/1=a, b/0=infinite, c/2\n#for plane (231) intercepts are a/2, b/3, c/1=c\n#for plane (31'2) intercepts are a/3=a, b/-1=-b, c/2\n\n#Result\nprint(\"for plane (102) intercepts are a/1=a, b/0=infinite, c/2\");\nprint(\"for plane (231) intercepts are a/2, b/3, c/1=c\");\nprint(\"for plane (312) intercepts are a/3=a, b/-1=-b, c/2\");\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "for plane (102) intercepts are a/1=a, b/0=infinite, c/2\nfor plane (231) intercepts are a/2, b/3, c/1=c\nfor plane (312) intercepts are a/3=a, b/-1=-b, c/2\n"
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.7, Page number 188"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the angle between 2 planes\n\n#importing modules\nimport math\n\n#Variable declaration\nu1=1;\nv1=1;\nw1=1;    #indices for plane (111)\nu2=2;\nv2=1;\nw2=2;    #indices for plane (212)\n\n#Calculation\nA=u1*u2+v1*v2+w1*w2;    \nB1=math.sqrt((u1**2)+(v1**2)+(w1**2));\nB2=math.sqrt((u2**2)+(v2**2)+(w2**2));\nB=A/(B1*B2);\nB=math.ceil(B*10**4)/10**4;   #rounding off to 4 decimals\ntheta=math.acos(B);    #angle in radian\ntheta=theta*57.2957795;     #converting radian to degrees\ntheeta=math.ceil(theta*10**3)/10**3;   #rounding off to 3 decimals\ndeg=int(theta);     #converting to degrees\nt=60*(theta-deg);\nmi=int(t);        #converting to minutes\nsec=60*(t-mi);      #converting to seconds\nsec=math.ceil(sec*10**2)/10**2;   #rounding off to 2 decimals\n\n#Result\nprint(\"angle between the planes in degrees is\",theeta);\nprint(\"angle between the planes is\",deg,\"degrees\",mi,\"minutes\",sec,\"seconds\");\n\n#answer given in the book is wrong",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('angle between the planes in degrees is', 15.783)\n('angle between the planes is', 15, 'degrees', 46, 'minutes', 57.85, 'seconds')\n"
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.8, Page number 188"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#sketching the crystallographic planes",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.9, Page number 189"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the lattice constant\n\n#importing modules\nimport math\n\n#Variable declaration\nd=0.2338;     #interplanar distance in nm\nh=-1;\nk=1;\nl=1;   #indices of the plane (1'11)\n\n#Calculation\nd=d*10**-9;       #converting from nm to m\na=d*math.sqrt((h**2)+(k**2)+(l**2));\na=a*10**9;      #converting lattice constant from m to nm\na=math.ceil(a*10**5)/10**5;   #rounding off to 5 decimals\n\n#Result\nprint(\"lattice constant in nm is\",a);\n\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('lattice constant in nm is', 0.40496)\n"
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.10, Page number 189"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To show that for a simple cubic system d100:d110:d111=sqrt(6):sqrt(3):sqrt(2)\n\n#importing modules\nimport math\n\n#variable declaration\nh1=1;\nk1=0;\nl1=0;    #indices for plane (100)\nh2=1;\nk2=1;\nl2=0;    #indices for plane (110)\nh3=1;\nk3=1;\nl3=1;    #indices for plane (111)\n\n#Calculation\n#d=a/math.sqrt((h**2)+(k**2)+(l**2))\n#d100=a/math.sqrt((h1**2)+(k1**2)+(l1**2))\nx1=math.sqrt((h1**2)+(k1**2)+(l1**2));\n#d100=a/x1 = a/1 = a\n#d110=a/math.sqrt((h2**2)+(k2**2)+(l2**2))\nx2=math.sqrt((h2**2)+(k2**2)+(l2**2));\nx2=math.ceil(x2*10**4)/10**4;   #rounding off to 4 decimals\n#d110=a/x2 = a/sqrt(2)\n#d111=a/math.sqrt((h3**2)+(k3**2)+(l3**2))\nx3=math.sqrt((h3**2)+(k3**2)+(l3**2));\nx3=math.ceil(x3*10**4)/10**4;   #rounding off to 4 decimals\n#d111=a/x3 = a/sqrt(3)\n#hence d100:d110:d111=a:a/sqrt(2):a/sqrt(3)\n#multiplying RHS by sqrt(6) we get d100:d110:d111=sqrt(6):sqrt(3):sqrt(2)\n\n#Result\nprint(\"value of x1 is\",x1);\nprint(\"value of x2 is\",x2);\nprint(\"value of x3 is\",x3);\nprint(\"d100:d110:d111=sqrt(6):sqrt(3):sqrt(2)\");",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('value of x1 is', 1.0)\n('value of x2 is', 1.4143)\n('value of x3 is', 1.7321)\nd100:d110:d111=sqrt(6):sqrt(3):sqrt(2)\n"
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.11, Page number 190"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To find the ratio of intercepts in a simple cubic crystal\n\n#variable declaration\nh=2;\nk=3;\nl=1;    #indices for plane (231)\n\n#Calculation\n#intercepts made by the plane is a/h, b/k, c/l\n#for a cubic unit cell, a=b=c\n#for plane (231) intercepts are a/2, a/3, a/1 = a\n#ratio of the intercepts is 1/2:1/3:1\n#LCM is 6. multiplying by LCM, we get ratio l1:l2:l3 = 3:2:6\n\n#Result\nprint(\"l1:l2:l3 = 3:2:6\");",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "l1:l2:l3 = 3:2:6\n"
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.12, Page number 190"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To find the lengths of intercepts along Y and Z axes\n\n#variable declaration\nh=1;\nk=2;\nl=3;    #indices for plane (123)\nl1=0.8;       #l1 in armstrong\na=0.8;       #a in armstrong\nb=1.2;       #b in armstrong\nc=1.5;       #c in armstrong\n\n#Calculation\n#intercepts made by the plane is a/h, b/k, c/l\n#for plane (123) intercepts are a/1 = a, b/2, c/3\n#ratio of the intercepts l1:l2:l3 = a:b/2:c/3\n#thus 0.8:l2:l3 = 0.8:1.2/2:1.5/3\nl2=1.2/2;     #l2 in armstrong\nl3=1.5/3;     #l3 in armstrong\n\n#Result\nprint(\"value of l2 in armstrong is\",l2);\nprint(\"value of l3 in armstrong is\",l3);",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('value of l2 in armstrong is', 0.6)\n('value of l3 in armstrong is', 0.5)\n"
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.13, Page number 191"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the nearest neighbour distance\n\n#Calculation\n#in simple cubic unit cell, corner atom is the nearest neighbour to another corner atom. \n#Hence nearest neighbour distance is a.\n#in BCC the body centered atom is the nearest neighbour to a corner atom.\n#the distance between body centered atom and corner atom is 2r\n#but r=sqrt(3)*a/4\n#distance = 2*sqrt(3)*a/4 = sqrt(3)*a/2\n#in FCC the face centered atom is the nearest neighbour to a corner atom.\n#the distance between face centered atom and corner atom is 2r\n#but r = a/sqrt(8)\n#distance = 2*a/sqrt(8) = a/sqrt(2)\n\n#Result\nprint(\"in simple cubic unit cell nearest neighbour distance is a\");\nprint(\"in body centered cubic unit cell nearest neighbour distance is sqrt(3)*a/2\");\nprint(\"in face centered cubic unit cell nearest neighbour distance is a/sqrt(2)\");",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "in simple cubic unit cell nearest neighbour distance is a\nin body centered cubic unit cell nearest neighbour distance is sqrt(3)*a/2\nin face centered cubic unit cell nearest neighbour distance is a/sqrt(2)\n"
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.14, Page number 191"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the spacing of the lattice plane\n\n#importing modules\nimport math\n\n#variable declaration\na=2.04;     #lattice parameter in armstrong\nh=2;\nk=1;\nl=2;    #indices for plane (212)\n\n#Calculation\na=a*10**-10;    #converting from armstrong to m\nd=a/math.sqrt((h**2)+(k**2)+(l**2));\nd=d*10**10;    #converting from m to armstrong\nd=math.ceil(d*10**3)/10**3;   #rounding off to 3 decimals\n\n#Result\nprint(\"interplanar distance in armstrong is\",d);\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('interplanar distance in armstrong is', 0.681)\n"
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.15, Page number 191"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the number of atoms per unit cell\n\n#importing modules\nimport math\n\n#variable declaration\nr=1.278;      #radius of Cu in armstrong\nM=63.54;     #atomic weight of Cu\nrho=8980;    #density in kg/m^3\nNa=6.022*10**26;\n\n#Calculation\nr=r*10**-10;    #radius in m\na=math.sqrt(8)*r;\nn=(rho*Na*a**3)/M;\n\n#Result\nprint(\"interatomic distance in m is\",a);\nprint(\"number of atoms per Cu unit cell is\",int(n));",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('interatomic distance in m is', 3.6147298654256317e-10)\n('number of atoms per Cu unit cell is', 4)\n"
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.16, Page number 192"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the Miller indices \n\n#variable declaration\na=0.429;\nb=1;\nc=0.379;    #intercepts of an orthorhombic crystal\n\n#Calculation\n#ratio of intercepts are 0.214:1:0.188 = (a/0.429)*0.214:1:(c/0.379)*0.188 = a/2:b:c/2\n#thus the coefficients are 1/2:1:1/2. inverses are 2,1,2.\n#thus miller indices for the first plane are (212)\n#ratio of intercepts are 0.858:1:0.754 = (a/0.429)*0.0.858:1:(c/0.379)*0.754 = 2a:b:2c\n#thus the coefficients are 2:1:2. inverses are 1/2,1,1/2. LCM is 2. multiplying with LCM we get 1,2,1\n#thus miller indices for the second plane are (121)\n#ratio of intercepts are 0.429:infinite:0.126 = (a/0.429)*0.429:infinite:(c/0.379)*0.126 = a:infiniteb:c/3\n#thus the coefficients are 1:infinte:1/3. inverses are 1,0,3.\n#thus miller indices for the third plane are (103)\n\n#Result\nprint(\"miller indices for the first plane are (212)\");\nprint(\"miller indices for the second plane are (121)\");\nprint(\"miller indices for the third plane are (103)\");\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "miller indices for the first plane are (212)\nmiller indices for the second plane are (121)\nmiller indices for the third plane are (103)\n"
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.17, Page number 193"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the number of atoms per unit area\n\n#importing modules\nimport math\nimport numpy as np\n\n#variable declaration\nh1=1;\nk1=0;\nl1=0;    #indices of the first plane (100)\nh2=1;\nk2=1;\nl2=0;    #indices of the second plane (110)\nh3=1;\nk3=1;\nl3=1;    #indices of the third plane (111)\n\n#Calculation\nn_1=np.reciprocal(4.);\nn_2=np.reciprocal(2.);\nn_3=np.reciprocal(6.);\nn1=(n_1*4)+1;    #number of atoms per unit cell in (100)\n#number of atoms per m^2 is 2/a**2. but a=sqrt(8)*r.\n#hence number of atoms per m^2 is 1/(4*r**2)\nn2=(n_1*4)+(2*n_2);   #number of atoms per unit cell in (110)\n#number of atoms per m^2 is 1/a*sqrt(2)*a. but a=sqrt(8)*r.\n#hence number of atoms per m^2 is 1/(8*sqrt(2)*r**2)\nn3=(n_3*3)+(3*n_2);   #number of atoms per unit cell in (111)\n#number of atoms per m^2 is 2/(sqrt(3)/4)*a**2. but a=4*r.\n#hence number of atoms per m^2 is 1/(2*sqrt(3)*r**2)\n\n#Result\nprint(\"number of atoms per unit cell in (100)\",n1);\nprint(\"number of atoms per m^2 is 1/(4*r**2)\");\nprint(\"number of atoms per unit cell in (110)\",n2);\nprint(\"number of atoms per m^2 is 1/(8*sqrt(2)*r**2)\");\nprint(\"number of atoms per unit cell in (111)\",n3);\nprint(\"number of atoms per m^2 is 1/(2*sqrt(3)*r**2)\");\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "('number of atoms per unit cell in (100)', 2.0)\nnumber of atoms per m^2 is 1/(4*r**2)\n('number of atoms per unit cell in (110)', 2.0)\nnumber of atoms per m^2 is 1/(8*sqrt(2)*r**2)\n('number of atoms per unit cell in (111)', 2.0)\nnumber of atoms per m^2 is 1/(2*sqrt(3)*r**2)\n"
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example number 6.18, Page number 194"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#To calculate the atomic packing fraction and ionic packing fraction of NaCl\n\n#importing modules\nimport math\n\n#variable declaration\nr=0.97;    #radius of Na+ ion in armstrong\nR=1.81;    #radius of Cl- ion in armstrong\n\n#Calculation\n#atomic packing factor=packing density PD\n#PD=Volume of atoms/Volume of unit cell\n#volume of unit cell=a**3\n#volume of atoms=number of atoms*volume of 1 atom = 4*(4/3)*math.pi*r**3\n#but r=a/sqrt(8). hence PD = 4*(4/3)*math.pi*(a/(2*sqrt(2)))**3*(1/a**3) = 0.74\n#atomic packing factor = 0.74\nr=r*10**-10;   #radius of Na+ ion in m\nR=R*10**-10;   #radius of Cl- ion in m\nVna = (4*4*math.pi*r**3)/3;    #volume of Na atoms\nVcl = (4*4*math.pi*R**3)/3;    #volume of Cl atoms \nV=(2*(r+R))**3;      #volume of unit cell\nIPF=(Vna+Vcl)/V;     #ionic packing factor\nIPF=math.ceil(IPF*10**4)/10**4;   #rounding off to 4 decimals\n\n#Result\nprint(\"atomic packing factor = 0.74\");\nprint(\"ionic packing factor of NaCl crystal is\",IPF);",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "atomic packing factor = 0.74\n('ionic packing factor of NaCl crystal is', 0.6671)\n"
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "",
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}