{
"metadata": {
"name": "Chapter5(s)"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": "Chapter 5: Crystal Physics"
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 1, Page no:5.72"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\n\n#Copper has FCC structure\nr = 1.273; # Atomic radius in angstrom\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 63.5; # Atomic weight of copper in grams\nn = 4; # No. of atoms per unit cell for FCC\n\n# Calculations\nr1 = r*10**-10; # Radius conversion from angstrom to m\na = (4*r1)/math.sqrt(2); # lattice parameter for FCC\np = (n*A)/(N*a**3); # Density of copper\n\n# Result\nprint 'Lattice Constant a = %3.1e' %a,' m','\\n', 'Density of copper = %3.1f' %p,' kg/m^3';\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Lattice Constant a = 3.6e-10 m \nDensity of copper = 9034.4 kg/m^3\n"
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 2, Page no:5.73"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\n# given intercepts 3,4 and \u221e, the recipocals of intercepts is\n# (1/3):(1/4):(1/\u221e)\n# LCM = 12\n# multiplying by LCM we get miller indices\n# miller indices of a plane are the smallest integers of the reciprocals of its intercerpts\n# therefore miller indices(h k l) is (4 3 0);\n\nh = 4; # miller indice\nk = 3; # miller indice\nl = 0; # miller indice\na = 2; # primitive vector of lattice in angstrom\n\n#Calculations\n\ndhkl = a/math.sqrt((h**2)+(k**2)+(l**2)); #interplanar distance\n\n# Result\nprint 'Miller indices = (4 3 0)';\nprint 'The interplanar distance d = %3.1f' %dhkl,' \u00c5';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Miller indices = (4 3 0)\nThe interplanar distance d = 0.4 \u00c5\n"
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 3, Page no:5.74"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\n# \u03b1-Iron solidifies to BCC structure\n\nr = 1.273; # Atomic radius in angstrom\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 55.85; # Atomic weight of \u03b1-Iron in kilograms\nn = 2; # No. of atoms per unit cell for BCC\np = 7860; # density in kg/m^-3\n\n#Calculations\n\n# p = (n*A)/(N*a^3); density\n\na = ((n*A)/(N*p))**(0.333); # lattice constant\na1 = a*10**10; # m to angstrom conversion\nr = (a1*math.sqrt(3))/4 # atomic radius for BCC\n\n#Output\nprint 'The Radius of the atom = %3.5f' %r,' \u00c5'\nprint 'Note : atomic wt taken as 55.58*10^-3 instead of 55.85 in calculation'",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The Radius of the atom = 1.26955 \u00c5\nNote : atomic wt taken as 55.58*10^-3 instead of 55.85 in calculation\n"
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 4, Page no:5.75"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\nlamda = 1.5418; # wavelength in \u00c5\nh = 1; # miller indice\nk = 1; # miller indice\nl = 1; # miller indice\nn = 1; # given first order\ntheta = 30; # diffraction angle in degrees\n\n# Calculations\ntheta1 = theta*math.pi/180; # degree to radian conversion\n# d = (n*lamda)/(2*sin\u03b8); by Braggs law ------------- 1\n# d = a/sqrt((h^2)+(k^2)+(l^2)); interplanar distance ------------ 2\n# equating 1 and 2\n\na = (n*lamda*math.sqrt((h**2)+(k**2)+(l**2))/(2*math.sin(theta1)))\n\n# Result\nprint 'Interatomic spacing a = %f \u00c5' %a;",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Interatomic spacing a = 2.670476 \u00c5\n"
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 5, page no:5.76"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# variable Declaration\nh1 = 1; # miller indice\nk1 = 1; # miller indice\nl1 = 1; # miller indice\nh0 = 0; # miller indice\nk0 = 0; # miller indice\nl0 = 0; # miller indice\n\n# Calculations\n# dhkl = a/sqrt((h^2)+(k^2)+(l^2)); // interplanar distance\n# assume a = 1(constant) for easier calculation in scilab\n\na = 1;\nd100 = a/math.sqrt((h1**2)+(k0**2)+(l0**2)); # interplanar distance\nd110 = a/math.sqrt((h1**2)+(k1**2)+(l0**2)); # interplanar distance\nd111 = a/math.sqrt((h1**2)+(k1**2)+(l1**2)); # interplanar distance\n\n# Result\nprint 'd100 : d110 : d111 = ','%d ' %d100,':','%3.2f' %d110,':', '%3.2f' %d111;",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "d100 : d110 : d111 = 1 : 0.71 : 0.58\n"
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 6, page no:5.76"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\n# Aluminium is FCC\n\na = 0.405*10**-9; # lattice constant of aluminium\nt = 0.005*10**-2; # thickness of aluminium foil in m\ns = 25*10**-2; # side of square in m\n\n# Calculations\nVUC = a**3; # volume of unit cell\nVal = (s**2)*t # volume of aluminium foil (area*thickness)\nN = Val/VUC # Number if unit cells\n\n# Result\nprint 'Number of unit cells = %3.3e' %N",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Number of unit cells = 4.704e+22\n"
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 7, page no:5.77"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# variable declaration\n# metallic iron changes from BCC to FCC form at 910 degress\nrb = 0.1258*10**-9; # atomic radius of BCC iron atom\nrf = 0.1292*10**-9; # atomic radius of FCC iron atom\n\n# Calculations\n\nab = (4*rb)/(math.sqrt(3)); # lattice constant for BCC\nVbcc = (ab**3)/2; # volume occupied by one BCC atom\naf = (4*rf)/(math.sqrt(2)) # lattice constant for FCC\nVfcc = (af**3)/4; # volume occupied by one FCC atom\ndv = ((Vbcc-Vfcc)/Vbcc)*100 # percentage change in volume\n\n# Result\nprint 'During the structural change the percentage change in volume = %3.4f' %dv;",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "During the structural change the percentage change in volume = 0.4933\n"
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 8, page no:5.78"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math\n#variable declaration\n#Copper Crystallines in FCC structure\n\np = 8960; # Density of copper in kg/m^3\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 63.5; # Atomic weight of copper in kg/mol\nn = 4; # No. of atoms per unit cell for FCC\n\n# Calculations\n\na = ((n*A)/(N*p))**(0.333);\n\n# Result\n\nprint 'Lattice Constant a = %3.4f' %(a*10**10),' \u00c5';\nprint 'atomic wt of copper is taken as 63.5*10^-3 instead of 63.5 in textbook';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Lattice Constant a = 3.6899 \u00c5\natomic wt of copper is taken as 63.5*10^-3 instead of 63.5 in textbook\n"
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 9, page no:5.79"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable declaration\n# (100) planes in rock crystal\nh = 1; # miller indice\nk = 0; # miller indice\nl = 0; # miller indice\na = 2.814 # lattice constant in \u00c5\n\n# Calculations\ndhkl = a/math.sqrt((h**2)+(k**2)+(l**2)); # interplanar distance\n\n#Result\nprint 'd-spacing for (100) plane in rock salt = %3.3f' %dhkl,' \u00c5';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "d-spacing for (100) plane in rock salt = 2.814 \u00c5\n"
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 10, page no:5.79"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\n# FCC structured crystal\n\np = 6250; # Density of crystal in kg/m^3\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 60.2; # molecular weight\nn = 4; # No. of atoms per unit cell for FCC\n\n# Calculations\n\na = ((n*A)/(N*p))**(0.333);\n\n# Result\n\nprint 'Lattice Constant a = %3.3e' %a, 'm';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Lattice Constant a = 4.087e-10 m\n"
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 11, page no:5.80"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\n# (321) plane in simple cubic lattice\nh = 3; # miller indice\nk = 2; # miller indice\nl = 1; # miller indice\na = 4.12 # inter atomic space \u00c5\n\n# Calculations\ndhkl = a/math.sqrt((h**2)+(k**2)+(l**2)); #interplanar distance\n\n# Result\nprint 'd = %3.2f' %dhkl,' \u00c5';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "d = 1.10 \u00c5\n"
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 12, page no:5.81"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\n# BCC structured crystal\n\np = 7860; # Density of iron in kg/m^3\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 55.85; # Atomic weight\nn = 2; # No. of atoms per unit cell for BCC\n\n# Calculations\n\na = ((n*A)/(N*p))**(0.333); #lattice constant\n\n# Result\n\nprint 'Lattice Constant of Fe = %3.3f' %(a*10**10),' \u00c5 ';\nprint 'Note: density of iron is taken as 7.86 instead of 7860 in calculation'",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Lattice Constant of Fe = 2.932 \u00c5 \nNote: density of iron is taken as 7.86 instead of 7860 in calculation\n"
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 14, page no:5.82"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# variable Declaration\nr = 0.123*10**-10; # Radius of the atom\n\n# Calculations\na = (4*r)/math.sqrt(3); # Lattice constant in m For a BCC structure\nV = a*a*a; # Volume of BCC\n\n# Result\nprint 'Volume of the unit cell = %3.4e' %V,' m^3';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Volume of the unit cell = 2.2920e-32 m^3\n"
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 15, page no:5.82"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable declaration\na = 0.05; # unit cell edge of an orthorhombic crystal in nm\nb = 0.05; # unit cell edge of an orthorhombic crystal in nm\nc = 0.03; # unit cell edge of an orthorhombic crystal in nm\nIa = 0.025 # intercept on 'a' in nm\nIb = 0.02 # intercept on 'b' in nm\nIc = 0.01 # intercept on 'c' in nm\n\n# Calculations\n\nh = a/Ia; # miller indice h\nk = b/Ib; # miller indice k\nl = c/Ic # miller indice l\n\n# Result\nprint 'Miller indices (h k l) =', '%d' %h,'%d' %k, '%d' %l; ",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Miller indices (h k l) = 2 2 3\n"
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 16, page no:5.83"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Magnesium has HCP structure\n# for HCF(Hexagonal closed packed structure) consider the relation between 'c' and 'a';\n# c/a = sqrt(8/3) = 1.6329 \n\n# Variable Declaration\nr = 0.1605*10**-9; # radius of magnesium atom in m\n\n# Calculations\n\na = 2*r # lattice constant of HCP\nc = a*math.sqrt(float(8)/3); # relation b/w c and a in HCP\nV = (3*(3**0.5))*(a*a*c)/2; #Volume of unit cell in m^3\n\n# Result\nprint 'Volume of the unit cell of magnesium = %g' %V,' m^3';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Volume of the unit cell of magnesium = 1.4033e-28 m^3\n"
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 17, page no:5.84"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math\n\n# Variable Declaration\n# (101),(221) planes in simple cubic lattice\nh1 = 1; # miller indice\nk0 = 0; # miller indice\nl1 = 1; # miller indice\nh2 = 2; # miller indice\nk2 = 2; # miller indice\nl1 = 1; # miller indice\na = 4.2 # inter atomic space \u00c5\n\n# Calculations\n\nd101 = a/math.sqrt((h1**2)+(k0**2)+(l1**2)); # interplanar distance\nd221 = a/math.sqrt((h2**2)+(k2**2)+(l1**2)); # interplanar distance\n\n\n# Result\nprint 'd(101) = %3.4f' %d101,' \u00c5','\\n','d(221) = %3.1f' %d221,' \u00c5 ';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "d(101) = 2.9698 \u00c5 \nd(221) = 1.4 \u00c5 \n"
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 1, page no:5.85"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable declaration\n\n# Copper has FCC structure\na = 3.6; #lattice parameter of copper in \u00c5\n\n# Calculations\n\nr = a*math.sqrt(2)/4; # atomic radius of copper\n\n# Result\nprint 'Atomic Radius of copper = %3.3f' %r,'\u00c5';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Atomic Radius of copper = 1.273 \u00c5\n"
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 2, page no:5.85"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# variable Declaration\n\n# Copper has FCC structure\n\nr = 1.278; # Atomic radius in angstrom\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 63.54; # Atomic weight of copper \nn = 4; # No. of atoms per unit cell for FCC\n\n# Calculations\nr1 = r*10**-10; # Radius conversion from angstrom to m\na = (4*r1)/math.sqrt(2); # lattice parameter for FCC\np = (n*A)/(N*a**3); # Density of copper\n\n# Result\n\nprint ' Density of copper = %3.2f' %p,' kg/m^3';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": " Density of copper = 8934.43 kg/m^3\n"
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 3, page no:5.86"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\nANa = 23; # atomic wt of sodiim\nACl = 35.45 # atomic wt of chlorine\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nn = 4 # No. of atoms per unit cell for FCC\np = 2180; # density in kg/m^-3\n\n# Calculations\n\n# p = (n*A)/(N*a^3); density\nA = ANa+ACl; # atomic wt of NaCl\na = ((n*A)/(N*p))**(0.33333); # lattice constant\nr = a/2 # Distance b/w two adjacent atoms\n\n# Result\nprint 'Distance between two adjacent atoms is r = %3.2e' %r,' m';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Distance between two adjacent atoms is r = 2.81e-10 m\n"
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 4, page no:5.87"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\nr = 1.273; # Atomic radius in angstrom\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 55.85 ; # Atomic weight of Fe \nn = 2; # No. of atoms per unit cell for BCC\np = 7860; # density in kg/m^-3\n\n# Calculations\n\n# p = (n*A)/(N*a^3); density\n\na = ((n*A)/(N*p))**(0.33333); # lattice constant\na1 = a*10**10; # m to angstrom conversion\nr = (a1*math.sqrt(3))/4 # atomic radius for BCC\n\n# Result\nprint 'The Radius of the Fe = %3.3f' %r,' \u00c5';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The Radius of the Fe = 1.242 \u00c5\n"
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 5, page no:5.88"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\nN = 6.023*10**26; # Avagadros number in atoms/kilomole\nA = 119; # Atomic weight of pottasium bromide\nn = 4; # No. of atoms per unit cell for FCC\np = 2700; # density in kg/m^-3\n\n# Calculations\n\n# p = (n*A)/(N*a^3); density\n\na = ((n*A)/(N*p))**(0.33333); # lattice constant\na1 = a*10**10; # m to angstrom conversion\n\n# Output\nprint 'Lattice constant = %3.1f' %a1,' \u00c5';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Lattice constant = 6.6 \u00c5\n"
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 6, page no:5.88"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\na = 4.3*10**-10; # Lattice constant in \u00c5\np = 960; # Density of crystal in kg/m^3\nA = 23; # Atomic wt\nN = 6.023*10**26; # avogadros no in atoms/kilomole\n\n# Calculations\n\nn = (p*N*(a**3))/A; # No. of atoms per unit cell\n\n# result\nprint 'No. of atoms per unit cell = %3.0f' %n,' (BCC)';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "No. of atoms per unit cell = 2 (BCC)\n"
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 7, page no:5.89"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration \n\n# given crystal has BCC structure\nr = 1.2*10**-10; # atomic radius in m\n\n# Calculations\n\na = (4*r)/math.sqrt(3); # lattice constant\nV = a**3; # volume of cell\n\n# Result\nprint 'Volume of the cell = %3.3e' %V,' m^3';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Volume of the cell = 2.128e-29 m^3\n"
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 8, page no:5.89"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\na = 4*10**-10; # lattice constant of the crystal\nh = 1 # miller indice\nk = 0 # miller indice\nl = 0 # miller indice\n\n# Calculations\n\n# in fig consider (100) plane. the no of atoms in plane ABCD\nN = 4*(float(1)/4); # Number of atoms\np = N/(a*a); # planar atomic density in atoms/m^2\np1 = p*10**-6 # planar atomic density in atoms/mm^2\n\n# Result\nprint 'planar atomic density = %3.2e' %p1,' atoms/mm^2';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "planar atomic density = 6.25e+12 atoms/mm^2\n"
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 9, page no:5.90"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\n# in fig 5(b) the given plane is parallel to X and Z axes.Thus,its numerical intercepts on these axes is infinity\n#The numerical intercept on y axis is 1/2. Thus the numerical intercepts of plane is (\u221e 1/2 \u221e)\nprint ' Miller indices of plane shown in fig 5.(b) = (0 2 0)';\n# in fig 5(c) the given plane is parallel to Z axis.Thus its numerical intercept on z axis is infinity\n# The numerical intercept on x axis is 1 and y axis is 1/2. this numerical intercepts on plane is (1 1/2 \u221e )\nprint ' Miller indices of plane shown in fig 5.(c) = (1 2 0)'\n# in fig 5(d) the given plane is parallel to Z axis.Thus its numerical intercept on z axis is infinity\n# The numerical intercept on x axis is 1/2 and y axis is 1/2. this numerical intercepts on plane is (1/2 1/2 \u221e )\nprint ' Miller indices of plane shown in fig 5.(d) = (2 2 0)'",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": " Miller indices of plane shown in fig 5.(b) = (0 2 0)\n Miller indices of plane shown in fig 5.(c) = (1 2 0)\n Miller indices of plane shown in fig 5.(d) = (2 2 0)\n"
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 11, page no:5.91"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\n\n# (311) plane in simple cubic lattice\nh = 3; # miller indice\nk = 1; # miller indice\nl = 1; # miller indice\na = 2.109*10**-10 # lattice constant in m\n\n# Calculations\ndhkl = a/math.sqrt((h**2)+(k**2)+(l**2)); # interplanar distance\n\n# Result\nprint 'd = %3.3e' %dhkl,' m';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "d = 6.359e-11 m\n"
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 12, page no:5.92"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\nh = 1; # miller indice\nk = 1; # miller indice\nl = 0; # miller indice\nd = 2.86*10**-10 # interplanar distance in m\n\n# Calculations\na = d*math.sqrt((h**2)+(k**2)+(l**2)); # interplanar distance\n\n# Result\nprint 'Lattice constant a = %3.3e' %a,' m';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Lattice constant a = 4.045e-10 m\n"
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 13, page no:5.93"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\nh1 = 1;\nh0 = 0;\nk0 = 0;\nl0 = 0;\nl1 = 1;\n\n# Calculations\n\n# we know that dhkl = a/sqrt( h^2 + k^2 + l^2)\n# let sqrt( h^2 + k^2 + l^2) = p\np101 = math.sqrt( h1**2 + k0**2 + l1**2);\np100 = math.sqrt( h1**2 + k0**2 + l0**2);\np001 = math.sqrt( h0**2 + k0**2 + l1**2);\n\n# Result\nprint 'd101 : d100 : d001 :: a/%3.4f' %p101,' : ','a/%d' %p100,':',' a/%d ' %p001;",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": " d101 : d100 : d001 :: a/1.4142 : a/1 : a/1 \n"
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 14, page no:5.93"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# if a plane cut intercepts of lengths l1,l2,l3 the on three crystal axes ,then\n# l1 : l2 : l3 = pa : pq :rc\n# where a,b and c are primitive vectors of the unit cell and p,q and r are numbers related to miller indices (hkl) of plane by relation\n# 1/p : 1/q : 1/r = h : k : l\n# since, the crystal is simple cubic a = b = c and given that h = 1, k = 1 and l = 1\n# p : q : r = 1/h : 1/k : 1/l = 1/1 : 1/1 : 1/1 \n# p : q : r = 1 : 1 : 1\n# similarly l1 : l2 : l3 = 1a : 1a : 1a\nprint 'ratio of intercepts on the three axes by (111) plane is l1 : l2 : l3 = 1 : 1 : 1';\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "ratio of intercepts on the three axes by (111) plane is l1 : l2 : l3 = 1 : 1 : 1\n"
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 15, page no:5.94"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\n\nr = 1.246*10**-10; # atomic radius in m\nh1 = 1 # miller indice\nh2 = 2 # miller indice\nk0 = 0 # miller indice\nk1 = 1 # miller indice\nk2 = 2 # miller indice\nl0 = 0 # miller indice\nl1 = 1 # miller indice\n\n# Calculations\na = (4*r)/math.sqrt(2); # lattice constant\nd111 = a/math.sqrt((h1**2)+(k1**2)+(l1**2)); # interplanar distance\nd200 = a/math.sqrt((h2**2)+(k0**2)+(l0**2)); # interplanar distance\nd220 = a/math.sqrt((h2**2)+(k2**2)+(l0**2)); # interplanar distance\n\n# Result\nprint 'd111 = %3.3e' %d111,' m','\\n' 'd200 = %3.4e' %d200,' m','\\n''d220 = %3.3e' %d220,' m';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "d111 = 2.035e-10 m \nd200 = 1.7621e-10 m \nd220 = 1.246e-10 m\n"
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 16, page no:5.95"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# variable Declaration\n# the intercept along X-axis be c1 = a\n# the intercept along Y-axis be c2 = b/2 and\n# the intercept along Z-axis be c3 = 3c\n# Therefore, p = c1/a = a/a = 1\n# q = c2/b = (b/2)/b = 1/2\n# r = c3/c = (3c)/c = 3\n# therefore h = 1/p = 1\n# k = 1/q = 2\n# l = 1/r = 1/3\n# lcm of 1 1 and 3 = 3\nh = 1\nk = 2\nl = float(1)/3\ns = 3 ; # lcm\nh1= s*h\nk1= s*k\nl1= s*l;\n\n# Result\nprint '(h k l) =', '%d' %h1,' %d' %k1,'%3.0f' %l1;",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "(h k l) = 3 6 1\n"
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Addl_Example 17, page no:5.96"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math;\n\n# Variable Declaration\nd = 1.3*10**-10 # interplanar distance\nn = 1; # given first order\ntheta = 23; # Bragg reflection angle in degrees\n\n# Calculations\ntheta1 = theta*math.pi/180; # degree to radian conversion\n# d = (n*lamda)/(2*sin\u03b8); by Braggs law ------------- 1\nlamda = (2*d*math.sin(theta1)/n)\n\n# Result\nprint 'Wavelength of X-ray = %3.4f' %(lamda*10**10),' \u00c5';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Wavelength of X-ray = 1.0159 \u00c5\n"
}
],
"prompt_number": 40
},
{
"cell_type": "code",
"collapsed": false,
"input": "",
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}