{
"metadata": {
"name": "",
"signature": "sha256:64e1fbee0e1d9b8157cae12a98a7773b847a3c5e842a9d3f124a5485fe931875"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 2: Crystal Structure"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.1, Page number 2.16"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from math import sqrt \n",
"\n",
"#Variable declaration\n",
"#Assuming r=1 for simpliciy in calculations\n",
"r = 1\n",
"\n",
"#Calculations\n",
"a = (4*r)/sqrt(3)\n",
"#Let R be the radius of interstitial sphere that can fit into the void,therefore,\n",
"R = (a-2*r)/2 \n",
"\n",
"#Result\n",
"print \"The maximum radius of interstitial sphere is\",round(R,3),\"r\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum radius of interstitial sphere is 0.155 r\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.2, Page number 2.17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from math import sqrt\n",
"\n",
"#Variable declaration\n",
"r1 = 1.258*10**-10 #atomic radius(m)\n",
"r2 = 1.292*10**-10 #atomic radius(m)\n",
"\n",
"#Calculations\n",
"#In BCC\n",
"a_bcc = (4*r1)/sqrt(3)\n",
"v_bcc = a_bcc**3 #volume of unit cell(m^3)\n",
"n1 = ((1./8.)*8.)+1\n",
"V1 = v_bcc/n1 #volume occupied by 1 atom(m^3)\n",
"\n",
"#In FCC\n",
"a_fcc = 2*sqrt(2)*r2\n",
"v_fcc = a_fcc**3 #volume of unit cell(m^3)\n",
"n2 = ((1./2.)*6.)+((1./8.)*8.)\n",
"V2 = v_fcc/n2 #volume occupied by 1 atom(m^3)\n",
"\n",
"del_v = ((V1-V2)/V1)*100 #change in volume\n",
"\n",
"#Result\n",
"print \"During the conversion of iron from BCC to FCC, the decrease in volume is\",round(del_v,1),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"During the conversion of iron from BCC to FCC, the decrease in volume is 0.5 %\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.3, Page number 2.17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from math import sqrt\n",
"\n",
"#Variable declaration\n",
"a = 0.27*10**-9 #nearest neighbour distance(m)\n",
"c = 0.494*10**-9 #height of unit cell(m)\n",
"M = 65.37 #atomic weight of zinc\n",
"N = 6.023*10**26 #Avogadro's number(k/mol)\n",
"\n",
"#Calculations\n",
"V = (3*sqrt(3)*a**2*c)/2 #volume of unit cell\n",
"rho = (6*M)/(N*V) #density of crystal\n",
"\n",
"#Results\n",
"print \"Volume of unit cell =\",round((V/1E-29),2),\"*10^29 m^3\"\n",
"print \"Density of zinc =\",round(rho),\"kg/m^3\"\n",
"print \"\\nThe solution differs because of rounding-off of the digits in the textbook\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Volume of unit cell = 9.36 *10^29 m^3\n",
"Density of zinc = 6960.0 kg/m^3\n",
"\n",
"The solution differs because of rounding-off of the digits in the textbook\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.4, Page number 2.18"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from math import sqrt\n",
"\n",
"#Varaible declaration\n",
"#Let r be the radius of atom and R be the radius of sphere\n",
"#For simplicity in calculations, let us assume r =1\n",
"r = 1\n",
"\n",
"#Calculations\n",
"#For FCC structure\n",
"a = (4*r)/sqrt(2)\n",
"R = (a/2)-r\n",
"\n",
"print \"Maximum radius of sphere =\",round(R,3),\"r\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum radius of sphere = 0.414 r\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.5, Page number 2.19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration\n",
"a = 0.356*10**-9 #cube edge(m)\n",
"M = 12.01 #atomic weight of carbon\n",
"N = 6.023*10**26 #Avogadro's number(k/mol)\n",
"na = 1.77*10**29 #no. of atoms per meter cube\n",
"\n",
"#Calculations\n",
"#Diamond has 2 interpenetrating FCC lattices. Since each FCC unit cell has 4 atoms, the total no. of atoms per unit cell is 8\n",
"n = 8/a**3\n",
"m = M/N\n",
"rho = m*na\n",
"\n",
"#Result\n",
"print \"Number of atoms =\",round((n/1E+29),3),\"*10^29\"\n",
"print \"The density of diamond is\",round(rho,1),\"kg/m^3(Calculation mistake in the textbook)\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Number of atoms = 1.773 *10^29\n",
"The density of diamond is 3529.4 kg/m^3(Calculation mistake in the textbook)\n"
]
}
],
"prompt_number": 58
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.6, Page number 2.19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration\n",
"rho = 2.18 #density of NaCl(gm/cm^3)\n",
"N = 6.023*10**23 #Avogadro's number(/mol)\n",
"\n",
"#Caculations\n",
"w = 23+35.5 #molecular weight of NaCl\n",
"m = w/N #mass of NaCl molecule\n",
"nm = rho/m #no. of molecules per unit volume(molecule/cm^3)\n",
"#Since NaCl is diatomic\n",
"n = 2*nm\n",
"#Let a be the distance between adjacent atoms in NaCl and\n",
"# n be the no. of atoms along the edge of the cube\n",
"#length of an edge = na\n",
"#volume of unit cube = n^3*a^3\n",
"a = (1/n)**(1./3.)\n",
"\n",
"#Result\n",
"print \"The distance between two adjacent atoms is\",round((a/1E-8),2),\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The distance between two adjacent atoms is 2.81 A\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.7, Page number 2.20"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from math import sqrt\n",
"\n",
"#Variable declaration\n",
"w = 63.5 #atomic weight of copper\n",
"r = 1.278*10**-8 #atomic rdius(m)\n",
"N = 6.023*10**23 #Avogadro's number(/mol)\n",
"\n",
"#Calculations\n",
"m = w/N #mass of each copper atom(gm)\n",
"#Since copper has FCC structure lattice constant\n",
"a = (4*r)/sqrt(2) \n",
"n = 4 #no. of atoms in unit cell of FCC structure\n",
"M = n*m #mass of unit cell\n",
"rho = M/a**3 #density\n",
"\n",
"#Result\n",
"print \"Density of copper =\",round(rho,2),\"gm/cm^3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Density of copper = 8.93 gm/cm^3\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}