{
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"name": "",
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 2:Crystal Structure"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.1 , Page no:40"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"N=6.02*10**26; #in /Kg-molecule (Avogadro's number)\n",
"n=4; #number of molecules per unit cell ofr NaCl\n",
"M=58.5; #in Kg/Kg-molecule (molecular weight of NaCl)\n",
"p=2189; #in Kg/m^3 (density)\n",
"\n",
"#calculate\n",
"a=pow(((n*M)/(N*p)),0.3333333);\n",
"a1=a*10**10; #changing unit to Angstrom\n",
"\n",
"#result\n",
"print\"The lattice constant is a=\",a,\"m\";\n",
"print\"\\t\\t\\ta=\",a1,\"Angstrom\";"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The lattice constant is a= 5.62072226997e-10 m\n",
"\t\t\ta= 5.62072226997 Angstrom\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.2 , Page no:41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from __future__ import division\n",
"\n",
"#given\n",
"N=6.02*10**23; #in /gram-atom (Avogadro's number)\n",
"n=4; #number of atom per unit cell for fcc structure\n",
"M=63.5; #in gram/gram-atom (atomic weight of Cu)\n",
"p=8.96; #in g/cm^3 (density)\n",
"\n",
"#calculate\n",
"a=pow((n*M/(N*p)),0.3333333);\n",
"a=a*1E8; #changing unit from cm to Angstrom\n",
"d=a/math.sqrt(2); #distance infcc lattice\n",
"\n",
"#result\n",
"print\"lattice constant is a=\",a,\"E-08 cm\";\n",
"print\"\\t\\t a=\",round(a,4),\"Angstrom\";\n",
"print\"distance between two nearest Cu atoms is d=\",round(d,2),\"Angstrom\";"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"lattice constant is a= 3.61113576149 E-08 cm\n",
"\t\t a= 3.6111 Angstrom\n",
"distance between two nearest Cu atoms is d= 2.55 Angstrom\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.3 , Page no:41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from __future__ import division\n",
"\n",
"#given\n",
"N=6.02E26; #in /Kg-atom (Avogadro's number)\n",
"n=2; #number of molecules per unit cell for bcc lattice\n",
"M=55.85; #in Kg/Kg-atom (atomic weight of Iron)\n",
"p=7860; #in Kg/m^3 (density)\n",
"\n",
"#calculate\n",
"a=pow((n*M/(N*p)),0.33333);\n",
"a1=a*1E10; #changing unit to Angstrom\n",
"\n",
"#result\n",
"print\"The lattice constant is a=\",a,\"m\";\n",
"print\"\\t\\t\\ta=\",round(a1,3),\"Angstrom\";"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The lattice constant is a= 2.86928355016e-10 m\n",
"\t\t\ta= 2.869 Angstrom\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.4 , Page no:42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from __future__ import division\n",
"\n",
"#given\n",
"N=6.02*10**26; #in /Kg-atom (Avogadro's number)\n",
"n=2; #number of molecules per unit cell for bcc lattice\n",
"M=6.94; #in Kg/Kg-atom (atomic weight of Iron)\n",
"p=530; #in Kg/m^3 (density)\n",
"\n",
"#calculate\n",
"a=pow((n*M/(N*p)),0.33333);\n",
"a1=a*1E10; #changing unit to Angstrom\n",
"\n",
"#result\n",
"print\"The lattice constant is a=\",a,\"m\";\n",
"print\"\\t\\t\\ta=\",round(a1,3),\"Angstrom\";"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The lattice constant is a= 3.51776567326e-10 m\n",
"\t\t\ta= 3.518 Angstrom\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.5 , Page no:42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from __future__ import division\n",
"\n",
"#given\n",
"N=6.02*10**23; #in /gram-molecule (Avogadro's number)\n",
"M=58.5; #in gram/gram-molecule (atomic weight of NaCl)\n",
"p=2.17; #in g/cm^3 (density)\n",
"\n",
"#calculate\n",
"#since V=M/p\n",
"#(1/d)^-3=2N/V=2Np/M\n",
"#therefore d= (M/2Np)^-3\n",
"d=pow((M/(2*N*p)),0.33333333);\n",
"d1=d*1*10**8; #changing unit from cm to Angstrom\n",
"\n",
"#result\n",
"print\"The distance between two adjacent atoms of NaCl is d=\",d,\"m\";\n",
"print\"\\t\\t\\t\\t\\t\\t d=\",round(d1,3),\"Angstrom\";"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The distance between two adjacent atoms of NaCl is d= 2.81853408124e-08 m\n",
"\t\t\t\t\t\t d= 2.819 Angstrom\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.6 , Page no:43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from __future__ import division\n",
"\n",
"#given\n",
"r_Na=0.98; #in Angstrom (radius of sodium ion)\n",
"r_Cl=1.81; #in Angstrom (radius of chloride ion)\n",
"M_Na=22.99; #in amu (atomic mass of sodium)\n",
"M_Cl=35.45; #in amu (atomic mass of chlorine)\n",
"\n",
"#calculate\n",
"a=2*(r_Na+r_Cl); #lattice parameter\n",
"#PF=volume of ions present in the unit cell/volume of unit cell\n",
"PF=((4*(4/3)*3.14)*r_Na**3+(4*(4/3)*3.14)*r_Cl**3)/a**3;\n",
"#Density=mass of unit cell/volume of unit cell\n",
"p=4*(M_Na+M_Cl)*1.66E-27/(a*1E-10)**3;\n",
"p1=p*1E-3; #changing unit to gm/cm^-3\n",
"\n",
"#result\n",
"print\"Lattice constant is a=\",round(a,3),\"Angstrom\";\n",
"print\"Packing fraction is =\",round(PF,3); \n",
"print\"Density is p=\",round(p),\"Kg/m^3\";\n",
"print\"Density is p=\",round(p1,2),\"g/cm^3\";"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Lattice constant is a= 5.58 Angstrom\n",
"Packing fraction is = 0.662\n",
"Density is p= 2233.0 Kg/m^3\n",
"Density is p= 2.23 g/cm^3\n"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}