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{
"metadata": {
"name": "",
"signature": "sha256:f3a8bba44ff010c0f1004529fad8de811f19168b5752929eb1e0272cfc5fc53c"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"4: Crystallography"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 4.2, Page number 70"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"d=2180; #density of NaCl(kg/m^3)\n",
"M=23+35.5; #Molecular weight of NaCl(gm)\n",
"Na=6.02*10**26; #Avgraodo no(per kg mole)\n",
"n=4; #for f.c.c\n",
"\n",
"#calculation\n",
"a=(n*M/(Na*d))**(1/3); #lattice constant(m)\n",
"d=a/2; #distance(m)\n",
"d=d*10**10; #distance(angstrom)\n",
"d=math.ceil(d*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print \"distance between two adajcent atoms is\",d,\"angstrom\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"distance between two adajcent atoms is 2.815 angstrom\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 4.3, Page number 70"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"d=2.163; #density(gm/cm^3)\n",
"M=58.45; #molecular weight(gm)\n",
"Na=6.02*10**23; #Avgraodo no.(molecules/gm mole)\n",
"\n",
"#calcualtion\n",
"n=Na/M; #no. of molecules(per gram)\n",
"n=n*d; #no. of molecules(per cm^3) \n",
"n=2*n; #no. of atom(per cm^3)\n",
"n=n**(1/3); #no. of atoms in a row 1cm long\n",
"d=1/n; #spacing between atoms(cm)\n",
"\n",
"#Result\n",
"print \"spacing between atoms is\",round(d/1e-8,2),\"angstrom\"\n",
"print \"answer in the book varies due to rounding off errors\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"spacing between atoms is 2.82 angstrom\n",
"answer in the book varies due to rounding off errors\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 4.4, Page number 74"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"r=1.278; #radius(A.U)\n",
"n=4; #structure is f.c.c\n",
"M=63.54; #atomic weight(gm)\n",
"Na=6.02*10**23; #Avgraodo no.(per gm mole)\n",
"\n",
"#calculation \n",
"a=4*r/(math.sqrt(2)); #lattice constant(A.U)\n",
"V=a**3; #volume(cm^3)\n",
"rho=n*M/(Na*V); #density(gm/cm^3)\n",
"rho=rho*(10**8)**3 #density(gm/m^3)\n",
"\n",
"#Result\n",
"print \"Density is\",round(rho,2),\"g/m^3\"\n",
"print \"answer in the textbook varies due to rounding off errors\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Density is 8.94 g/m^3\n",
"answer in the textbook varies due to rounding off errors\n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 4.10, Page number 85"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Vaiable declaration\n",
"r=1.746; #atomic radius(AU)\n",
"\n",
"#calulation\n",
"a=4*r/math.sqrt(2); #lattice constant(AU)\n",
"#for (200)\n",
"h=2;k=0;l=0;\n",
"d=a/math.sqrt(h**2+k**2+l**2); #interplanar spacing(AU)\n",
"#for (220)\n",
"h=2;k=2;l=0; \n",
"d1=a/math.sqrt(h**2+k**2+l**2); #interplanar spacing(AU)\n",
"#for (111)\n",
"h=1;k=1;l=1;\n",
"d2=a/math.sqrt(h**2+k**2+l**2); #interplanar spacing(AU)\n",
"d=math.ceil(d*10**4)/10**4; #rounding off to 4 decimals\n",
"d1=math.ceil(d1*10**3)/10**3; #rounding off to 3 decimals\n",
"d2=math.ceil(d2*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print \"spacing for (200) is\",d,\"A.U\"\n",
"print \"answer in the book varies in 3rd decimal due to rounding off errors\"\n",
"print \"spacing for (220) is\",d1,\"A.U\"\n",
"print \"spacing for (111) is\",d2,\"A.U\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"spacing for (200) is 2.4693 A.U\n",
"answer in the book varies in 3rd decimal due to rounding off errors\n",
"spacing for (220) is 1.746 A.U\n",
"spacing for (111) is 2.852 A.U\n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 4.11, Page number 86"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#calculation\n",
"#for (i)\n",
"l=1;m=0;n=0;\n",
"p=0;q=1;r=0;\n",
"d=math.acos(((l*p)+(m*q)+(n*r))/(math.sqrt(l**2+m**2+n**2)*math.sqrt(p**2+q**2+r**2))); #angle between 2 normals(radian)\n",
"theta=math.degrees(d); #angle between 2 normals(degrees) \n",
"\n",
"#for (ii)\n",
"l=1;m=2;n=1;\n",
"p=1;q=1;r=1;\n",
"d1=math.acos(((l*p)+(m*q)+(n*r))/(math.sqrt(l**2+m**2+n**2)*math.sqrt(p**2+q**2+r**2))); #angle between 2 normals(radian)\n",
"theta1=math.degrees(d1); #angle between 2 normals(degrees) \n",
"deg=int(theta1); #angle(degrees)\n",
"t=60*(theta1-deg);\n",
"m=int(t); #angle(minutes)\n",
" \n",
"#Result\n",
"print \"angle between the normal to pair of miller incdices (100) and (010) is\",theta,\"degrees\"\n",
"print \"angle between the normal to pair of miller incdices (121) and (111) is\",deg,\"degrees\",m,\"minutes\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"angle between the normal to pair of miller incdices (100) and (010) is 90.0 degrees\n",
"angle between the normal to pair of miller incdices (121) and (111) is 19 degrees 28 minutes\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 4.13, Page number 87"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#variable declaration\n",
"a=3.61*10**-7; #lattice constant(mm)\n",
"\n",
"#Calcualtion\n",
"#for plane (100)\n",
"SA=a*a; #surface area(mm^2)\n",
"tamc=2; #total atoms included\n",
"ans=tamc/SA; #number of atoms per mm^2\n",
"#for (ii) plane (110)\n",
"A=a*(math.sqrt(2)*a); #area of the plane(mm^2)\n",
"tamc=2; #total atoms included according to sketch\n",
"ans1=tamc/A; #number of atoms per mm^2\n",
"#for (iii) plane (111)\n",
"A=0.866*a*a; #area of the plane(mm^2)\n",
"tamc=2; #total atoms included according to sketch\n",
"ans2=tamc/A; #number of atoms per mm^2\n",
"\n",
"#Result\n",
"print \"atoms per mm^2 for (100) is\",round(ans/1e+13,3),\"*10**13 atoms/mm^2\"\n",
"print \"atoms per mm^2 for (110) is\",round(ans1/1e+13,3),\"*10**13 atoms/mm^2\"\n",
"print \"atoms per mm^2 for (111) is\",round(ans2/1e+13,3),\"*10**13 atoms/mm^2\"\n",
"print \"answer for plane (111) given in the book is wrong\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"atoms per mm^2 for (100) is 1.535 *10**13 atoms/mm^2\n",
"atoms per mm^2 for (110) is 1.085 *10**13 atoms/mm^2\n",
"atoms per mm^2 for (111) is 1.772 *10**13 atoms/mm^2\n",
"answer for plane (111) given in the book is wrong\n"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}
|
