{
"metadata": {
"name": "",
"signature": "sha256:dc7951185c6d2c2020409b6a4f6420009478010f311e90b1a30732a0039ca5cf"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
" 2: Crystal Structures and X-ray Diffraction"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.1, Page number 2.16"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"r = 1; #assume value of r as 1\n",
" \n",
"#Calculation\n",
"#since 4*r = math.sqrt(3)*a\n",
"a = 4*r/math.sqrt(3); \n",
"R = (a-(2*r))/2; #since 2*r+2*R = a\n",
"R = math.ceil(R*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print \"radius of interstitial sphere is\",R,\"r\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"radius of interstitial sphere is 0.155 r\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.2, Page number 2.17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"r_BCC = 1.258; #atomic radius(Angstrom)\n",
"r_FCC = 1.292; #atomic radius(Angstrom)\n",
"\n",
"#Calculation\n",
"a_BCC = (4*r_BCC)/math.sqrt(3); #In BCC(Angstrom)\n",
"a_BCCm = a_BCC*10**-10; #converting a from Angstrom to m\n",
"V_BCC = a_BCCm**3; #volume of unit cell(m^3)\n",
"n_BCC = ((1/8)*8)+1; #number of atoms per unit cell\n",
"V1_BCC = V_BCC/n_BCC; #volume occupied by 1 atom(m^3)\n",
"a_FCC = 2*math.sqrt(2)*r_FCC; #In FCC(Angstrom)\n",
"a_FCCm = a_FCC*10**-10; #converting a from Angstrom to m\n",
"V_FCC = a_FCCm**3; #volume of unit cell(m^3)\n",
"n_FCC = ((1/2)*6) + ((1/8)*8); #number of atoms per unit cell\n",
"V1_FCC = V_FCC/n_FCC; #volume occupied by 1 atom(m^3)\n",
"delta_V = (V1_BCC - V1_FCC)*100/V1_BCC; #change in volume in %\n",
"delta_V = math.ceil(delta_V*10)/10; #rounding off to 1 decimal\n",
"\n",
"#Result\n",
"print \"decrease of volume during conversion from BCC to FCC is\",delta_V,\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"decrease of volume during conversion from BCC to FCC is 0.5 %\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.3, Page number 2.17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"a = 0.27; #nearest neighbour distance(nm)\n",
"c = 0.494; #height of unit cell(nm)\n",
"N = 6.023*10**26; #avagadro number(k/mol)\n",
"M = 65.37; #atomic weight(kg)\n",
"\n",
"#Calculation\n",
"a_m = a*10**-9; #nearest neighbour distance(m)\n",
"c_m = c*10**-9; #height of unit cell(m)\n",
"V = 3*math.sqrt(3)*(a_m**2)*c_m/2; #volume of unit cell(m^3)\n",
"#if rho is density then mass = V*rho\n",
"#V*rho = 6*M/N\n",
"rho = (6*M)/(N*V); #density(kg/m^3)\n",
"\n",
"#Result\n",
"print \"Volume of the unit cell is\",round(V/1e-29,2),\"*10^-29 m^3\"\n",
"print \"density of Zinc is\",round(rho),\"kg/m^3\"\n",
"print \"answer for density given in the book differs due to rounding off errors\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Volume of the unit cell is 9.36 *10^-29 m^3\n",
"density of Zinc is 6960.0 kg/m^3\n",
"answer for density given in the book differs due to rounding off errors\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.4, Page number 2.18"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"r = 1; #assume value of r as 1\n",
"\n",
"#Calculation\n",
"a = 4*r/math.sqrt(2); #for FCC structure\n",
"R = (a/2)-r; #since 2*r+2*R = a\n",
"R = math.ceil(R*10**4)/10**4; #rounding off to 4 decimals\n",
"\n",
"#Result\n",
"print \"maximum radius of the sphere that can fit into the void is\",R,\"r\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"maximum radius of the sphere that can fit into the void is 0.4143 r\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.5, Page number 2.19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"a = 0.356; #cube edge of diamond(nm)\n",
"aw = 12.01; #atomic weight of carbon(kg)\n",
"N = 6.023*10**26; #avagadro's number\n",
"\n",
"#Calculation\n",
"a_m = a*10**-9; #cube edge of diamond(m)\n",
"n = 8/(a_m**3); #number of atoms(per m^3)\n",
"n = n/10**29;\n",
"n = math.ceil(n*10**3)/10**3; #rounding off to 3 decimals\n",
"n = n*10**29;\n",
"M = aw/N; #mass of 1 carbon atom(kg)\n",
"rho = M*n;\n",
"\n",
"#Result\n",
"print \"number of atoms per m^3 is\",n\n",
"print \"density of diamond is\",int(rho),\"kg/m^3\"\n",
"print \"answer for density given in the book differs due to rounding off errors\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"number of atoms per m^3 is 1.774e+29\n",
"density of diamond is 3537 kg/m^3\n",
"answer for density given in the book differs due to rounding off errors\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.6, Page number 2.19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"MW = 23+35.5; #molecular weight of NaCl(gm)\n",
"N = 6.023*10**23; #avagadro number(mol-1)\n",
"rho = 2.18; #density of NaCl(gm/cm^3)\n",
"V = 1; #volume of unit cube(cm^3)\n",
"\n",
"#Calculation\n",
"M = MW/N; #mass of NaCl molecule(gm)\n",
"n = rho/M; #number of molecules per unit volume\n",
"n = 2*n; #since NaCl is diatomic(atoms/cm^3)\n",
"#length of edge of unit cube is n*a\n",
"#volume V = n^3*a^3 = 1 cm^3 \n",
"a = (V/n)**(1/3); #distance between two adjacent atoms(cm)\n",
"a = a*10**8; #distance between two adjacent atoms(A)\n",
"a = math.ceil(a*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print \"distance between two adjacent atoms is\",a,\"Angstrom\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"distance between two adjacent atoms is 2.814 Angstrom\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.7, Page number 2.20"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"AW = 63.5; #atomic weight of Cu(gm/mol)\n",
"N = 6.023*10**23; #avagadro's number(mol-1)\n",
"r = 1.278; #atomic radius(A)\n",
"n = 4; #number of atoms in unit cell\n",
"\n",
"#Calculation\n",
"r = r*10**-8; #atomic radius(cm)\n",
"M = AW/N; #mass of each Copper atom(gm)\n",
"a = 4*r/math.sqrt(2); #lattice constant(cm)\n",
"m = n*M; #mass of unit cell(gm)\n",
"rho = m/a**3; #density of copper crystal(gm/cm^3)\n",
"rho = math.ceil(rho*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print \"density of Copper crystal is\",rho,\"gm/cm^3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"density of Copper crystal is 8.929 gm/cm^3\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.8, Page number 2.21"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"r = 0.1249; #atomic radius(nm)\n",
"PF = 0.68; #packing factor for BCC\n",
"\n",
"#Calculation\n",
"a = 4*r/math.sqrt(3); #lattice constant(nm)\n",
"a_m = a*10**-9; #lattice constant(m)\n",
"V = a_m**3; #volume of unit cell(m^3)\n",
"FV = 1-PF; #free volume\n",
"FV1 = FV*V; #free volume per unit cell(m^3)\n",
"\n",
"#Result\n",
"print \"free volume per unit cell is\",round(FV1/1e-30,4),\"*10^-30 m^3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"free volume per unit cell is 7.6795 *10^-30 m^3\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.11, Page number 2.27"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"#in (100) plane the total number of atoms are n\n",
"n = (1/4)*4;\n",
"\n",
"#Calculation\n",
"#in (111) plane, area = (1/2)*base*height = (1/2)*a*sqrt(2)*a*sqrt(2)*cosd(30)\n",
"x = math.cos(30);\n",
"#area = (1/2)*a*sqrt(2)*a*sqrt(2)*x = 0.866*a^2 = 0.58/a^2\n",
"n1 = (1/360)*60*3;\n",
"\n",
"#Result\n",
"print \"number of atoms in (100) plane are\",n;\n",
"#A = a^2. number of atoms per mm^2 is n/a^2\n",
"print \"number of atoms per mm^2 is 1/a^2\";\n",
"print \"number of atoms in (110) plane is 1\";\n",
"#in (110) plane, area is sqrt(2)*a*a = sqrt(2)*a^2\n",
"print \"unit area contains 1/(sqrt(2)*a^2) = 0.707/a^2 atoms/mm^2\";\n",
"print \"total number of atoms in (111) plane is\",n1\n",
"#number of atoms per unit area is 0.5/(0.866*a^2)\n",
"print \"number of atoms per unit area is 0.58/a^2 atoms/mm^2\";"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"number of atoms in (100) plane are 1.0\n",
"number of atoms per mm^2 is 1/a^2\n",
"number of atoms in (110) plane is 1\n",
"unit area contains 1/(sqrt(2)*a^2) = 0.707/a^2 atoms/mm^2\n",
"total number of atoms in (111) plane is 0.5\n",
"number of atoms per unit area is 0.58/a^2 atoms/mm^2\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.12, Page number 2.28"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"h1 = 1;\n",
"k1 = 1;\n",
"l1 = 0; #for (110) plane\n",
"h2 = 2;\n",
"k2 = 1;\n",
"l2 = 2; #for (212) plane\n",
"r = 0.1278; #atomic radius(nm)\n",
"\n",
"#Calculation\n",
"r = r*10**-9; #atomic radius(m) \n",
"x1 = math.sqrt(h1**2+k1**2+l1**2);\n",
"a = 4*r/x1; #nearest neighbouring distance(m)\n",
"a = a*10**9; #nearest neighbouring distance(nm)\n",
"d_110 = a/x1; #interplanar spacing for (110),(nm)\n",
"x2 = math.sqrt(h2**2+k2**2+l2**2);\n",
"d_212 = a/x2; #interplanar spacing for (212),(nm)\n",
"d_212 = math.ceil(d_212*10**4)/10**4; #rounding off to 4 decimals\n",
"\n",
"#Result\n",
"print \"interplanar spacing for (110) is\",d_110,\"nm\"\n",
"print \"interplanar spacing for (212) is\",d_212,\"nm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"interplanar spacing for (110) is 0.2556 nm\n",
"interplanar spacing for (212) is 0.1205 nm\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.13, Page number 2.29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"h1 = 1;\n",
"k1 = 0;\n",
"l1 = 0; #for (100) plane\n",
"h2 = 1;\n",
"k2 = 1;\n",
"l2 = 0; #for (110) plane \n",
"h3 = 1;\n",
"k3 = 1;\n",
"l3 = 1; #for (111) plane\n",
"\n",
"#Calculation\n",
"x1 = math.sqrt(1/(h1**2+k1**2+l1**2)); #for (100) plane\n",
"x2 = math.sqrt(1/(h2**2+k2**2+l2**2)); #for (110) plane \n",
"x3 = math.sqrt(1/(h3**2+k3**2+l3**2)); #for (111) plane\n",
"x2 = math.ceil(x2*10**2)/10**2; #rounding off to 2 decimals\n",
"x3 = math.ceil(x3*10**2)/10**2; #rounding off to 2 decimals\n",
"\n",
"#Result\n",
"print \"ratio of the seperation between successive lattice planes is %d:%.2f:%.2f\"%(x1,x2,x3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"ratio of the seperation between successive lattice planes is 1:0.71:0.58\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.14, Page number 2.30"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Calculation\n",
"#plane intercepts at a,b/2,3*c\n",
"#therefore intercepts are (1 1/2 3)\n",
"x = 1;\n",
"y = 1/2;\n",
"z = 3;\n",
"def lcm(x, y):\n",
" if y > z:\n",
" greater = y\n",
" else:\n",
" greater = z\n",
"\n",
" while(True):\n",
" if((greater % y == 0) and (greater % z == 0)):\n",
" lcm = greater\n",
" break\n",
" greater += 1\n",
"\n",
" return lcm\n",
"\n",
"lcm = lcm(y,z); \n",
"x_dash = (1/x)*lcm; #miller index of x plane\n",
"y_dash = (1/y)*lcm; #miller index of y plane \n",
"z_dash = (1/z)*lcm; #miller index of z plane\n",
"\n",
"#Result\n",
"print \"miller indices of the plane are (\",x_dash,y_dash,z_dash,\")\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"miller indices of the plane are ( 3.0 6.0 1.0 )\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.15, Page number 2.37"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"d = 0.282; #lattice spacing(nm)\n",
"n = 1; #first order\n",
"theta = (8+(35/60))*(math.pi/180); #glancing angle(radian)\n",
"theeta = 90; #maximum value possible(degrees)\n",
"\n",
"#Calculation\n",
"theeta = theeta*math.pi/180; #maximum value possible(radian) \n",
"d = d*10**-9; #lattice spacing(m)\n",
"lamda = 2*d*math.sin(theta)/n; #wavelength of X-rays(m)\n",
"lamda_nm = lamda*10**9; #wavelength of X-rays(nm)\n",
"lamda_nm = math.ceil(lamda_nm*10**4)/10**4; #rounding off to 4 decimals\n",
"n1 = 2*d*math.sin(theeta)/lamda; #maximum order of diffraction possible\n",
"\n",
"#Result\n",
"print \"wavelength of X-rays is\",lamda_nm,\"nm\"\n",
"print \"maximum order of diffraction possible is\",int(n1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"wavelength of X-rays is 0.0842 nm\n",
"maximum order of diffraction possible is 6\n"
]
}
],
"prompt_number": 35
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.16, Page number 2.37"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"lamda = 1.5; #wavelength(A.U)\n",
"d = 1.6; #interplanar spacing(A.U)\n",
"theta = 90; #maximum glancing angle possible(degrees)\n",
"\n",
"#Calculation\n",
"theta = theta*math.pi/180; #maximum glancing angle possible(radian)\n",
"n = 2*d*math.sin(theta)/lamda; #maximum possible diffraction order\n",
"\n",
"#Result\n",
"print \"maximum possible diffraction order is\",int(n)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"maximum possible diffraction order is 2\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.17, Page number 2.38"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"h = 1;\n",
"k = 1;\n",
"l = 1; #for (111) plane\n",
"theta = 30; #glancing angle(degrees)\n",
"n = 1; #first order\n",
"lamda = 1.5418; #wavelength of X-rays(A)\n",
"\n",
"#Calculation\n",
"theta = theta*math.pi/180; #glancing angle(radian)\n",
"x = math.sqrt(h**2+k**2+l**2);\n",
"lamda = lamda*10**-10; #wavelength of X-rays(m)\n",
"d = lamda/(2*math.sin(theta));\n",
"a = d*x; #interatomic spacing(m)\n",
"a = a*10**10; #interatomic spacing(A)\n",
"a = math.ceil(a*10**4)/10**4; #rounding off to 4 decimals\n",
"\n",
"#Result\n",
"print \"the interatomic spacing is\",a,\"Angstrom\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the interatomic spacing is 2.6705 Angstrom\n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.18, Page number 2.38"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"d100 = 0.28; #lattice constant(nm)\n",
"h = 1;\n",
"k = 1;\n",
"l = 0; #for (110) plane\n",
"n = 2; #second order\n",
"lamda = 0.071; #wavelength of X-rays(nm)\n",
"\n",
"#Calculation\n",
"lamda_m = lamda*10**-9; #wavelength of X-rays(m)\n",
"d110 = d100/math.sqrt(h**2+k**2+l**2); #interatomic spacing(nm)\n",
"d110 = d110*10**-9; #interatomic spacing(m)\n",
"theta = math.asin(n*lamda_m/(2*d110)); #glancing angle(radian)\n",
"theta = theta*180/math.pi; #glancing angle(degrees)\n",
"\n",
"#Result\n",
"print \"the glancing angle is\",int(theta),\"degrees\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the glancing angle is 21 degrees\n"
]
}
],
"prompt_number": 46
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.19, Page number 2.39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"a = 0.38; #lattice constant(nm)\n",
"h = 1;\n",
"k = 1;\n",
"l = 0; #for (110) plane\n",
"\n",
"#Calculation\n",
"d = a/math.sqrt(h**2+k**2+l**2);\n",
"d = math.ceil(d*10**2)/10**2; #rounding off to 2 decimals\n",
"\n",
"#Result\n",
"print \"the distance between (110) planes is\",d,\"nm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the distance between (110) planes is 0.27 nm\n"
]
}
],
"prompt_number": 48
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.20, Page number 2.39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"#area of (110) plane is a*sqrt(2)*a = sqrt(2)*a^2\n",
"theta = 30; #glancing angle(degrees)\n",
"\n",
"#Calculation\n",
"theta = theta*math.pi/180; #glancing angle(radian)\n",
"n = (1/4)*4; #number of atoms\n",
"x = math.cos(theta);\n",
"#area of (111) plane is (a/sqrt(2))*x*a*sqrt(2)\n",
"#hence area is (sqrt(3)/2)*a^2\n",
"n1 = 3*(1/6); #number of atoms\n",
"\n",
"#Result\n",
"print \"area of (110) plane contains\",n,\"atom\"\n",
"print \"density of lattice points is 1/(sqrt(2)*a^2)\"\n",
"print \"area of (111) plane contains\",n1,\"atom\"\n",
"#density of lattice points is (1/2)/(sqrt(3)*a^2/2)\n",
"print \"density of lattice points is 1/(sqrt(3)*a^2)\"\n",
"#density of lattice points (111) plane : (110) plane is 1/(sqrt(3)*a^2) : 1/(sqrt(2)*a^2) = sqrt(2):sqrt(3)\n",
"print \"density of lattice points (111) plane : (110) plane is sqrt(2):sqrt(3)\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"area of (110) plane contains 1.0 atom\n",
"density of lattice points is 1/(sqrt(2)*a^2)\n",
"area of (111) plane contains 0.5 atom\n",
"density of lattice points is 1/(sqrt(3)*a^2)\n",
"density of lattice points (111) plane : (110) plane is sqrt(2):sqrt(3)\n"
]
}
],
"prompt_number": 49
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.21, Page number 2.40"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"n = 2; #second order\n",
"h = 1;\n",
"k = 1;\n",
"l = 0; #plane (110)\n",
"lamda = 0.065; #wavelength of X-rays(nm)\n",
"a = 0.26; #axial length(nm)\n",
"\n",
"#Calculation\n",
"lamda_m = lamda*10**-9; #wavelength of X-rays(m)\n",
"a_m = a*10**-9; #axial length(m)\n",
"x = math.sqrt(h**2+k**2+l**2);\n",
"theta = math.asin(n*lamda_m*x/(2*a_m)); #glancing angle(radian)\n",
"theta = theta*180/math.pi; #glancing angle(degrees)\n",
"deg = int(theta); #glancing angle(degrees)\n",
"t = 60*(theta-deg);\n",
"mint = int(t); #glancing angle(minutes)\n",
"\n",
"#Result\n",
"print \"the glancing angle is\",deg,\"degrees\",mint,\"minutes\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the glancing angle is 20 degrees 42 minutes\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.22, Page number 2.41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"n = 1; #first order\n",
"h = 1;\n",
"k = 1;\n",
"l = 1; #for (111) plane\n",
"lamda = 1.54; #wavelength(A.U)\n",
"theta = 19.2; #Bragg's angle(degrees)\n",
"\n",
"#Calculation\n",
"lamda_m = lamda*10**-10; #wavelength(m)\n",
"theta = theta*math.pi/180; #Bragg's angle(radian)\n",
"d = n*lamda_m/(2*math.sin(theta)); #interplanar spacing(m)\n",
"a = d*math.sqrt(h**2+k**2+l**2); #cube edge(m)\n",
"a_AU = a*10**10; #cube edge(A.U)\n",
"a_AU = math.ceil(a_AU*10**4)/10**4; #rounding off to 4 decimals\n",
"\n",
"#Result\n",
"print \"the cube edge is\",a_AU,\"A.U\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the cube edge is 4.0554 A.U\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.23, Page number 2.41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"n = 1; #first order\n",
"h = 1;\n",
"k = 1;\n",
"l = 1; #for (111) plane\n",
"lamda = 1.54; #wavelength(A.U)\n",
"theta = 19.2; #Bragg's angle(degrees)\n",
"\n",
"#Calculation\n",
"lamda_m = lamda*10**-10; #wavelength(m)\n",
"theta = theta*math.pi/180; #Bragg's angle(radian)\n",
"d = n*lamda_m/(2*math.sin(theta)); #interplanar spacing(m)\n",
"a = d*math.sqrt(h**2+k**2+l**2); #cube edge(m)\n",
"\n",
"#Result\n",
"print \"the cube edge is\",round(a/1e-10,3),\"*10^-10 m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the cube edge is 4.055 *10^-10 m\n"
]
}
],
"prompt_number": 9
}
],
"metadata": {}
}
]
}