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{
"metadata": {
"name": "",
"signature": "sha256:7b7dfd33b7ab2e69421d0f26bfbfaa20f35bc1084164b445bbbc400215113c08"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 1:Bonding in Solids"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3 , Page no:15"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"r=2; #in angstrom(distance)\n",
"e=1.6E-19; # in C (charge of electron)\n",
"E_o= 8.85E-12;# absolute premittivity\n",
"\n",
"#calculate\n",
"r=2*1*10**-10; # since r is in angstrom\n",
"V=-e**2/(4*3.14*E_o*r); # calculate potential\n",
"V1=V/e; # changing to eV\n",
"\n",
"#result\n",
"print \"\\nThe potential energy is V = \",V,\"J\";\n",
"print \"In electron-Volt V = \",round(V1,3),\"eV\"; \n",
"print \"Note: the answer in the book is wrong due to calculation mistake\";"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The potential energy is V = -1.15153477995e-18 J\n",
"In electron-Volt V = -7.197 eV\n",
"Note: the answer in the book is wrong due to calculation mistake\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4 , Page no:15"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from __future__ import division\n",
"#given\n",
"r0=0.236; #in nanometer(interionic distance)\n",
"e=1.6E-19; # in C (charge of electron)\n",
"E_o= 8.85E-12;# absolute premittivity\n",
"N=8; # Born constant\n",
"IE=5.14;# in eV (ionisation energy of sodium)\n",
"EA=3.65;# in eV (electron affinity of Chlorine)\n",
"pi=3.14; # value of pi used in the solution\n",
"\n",
"#calculate\n",
"r0=r0*1E-9; # since r is in nanometer\n",
"PE=(e**2/(4*pi*E_o*r0))*(1-1/N); # calculate potential energy\n",
"PE=PE/e; #changing unit from J to eV\n",
"NE=IE-EA;# calculation of Net energy\n",
"BE=PE-NE;# calculation of Bond Energy\n",
"\n",
"#result\n",
"print\"The potential energy is PE= \",round(PE,2),\"eV\";\n",
"print\"The net energy is NE= \",round(NE,2),\"eV\";\n",
"print\"The bond energy is BE= \",round(BE,2),\"eV\";\n",
"# Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer.\n",
"# Note: (2) There is slight variation in the answer due to round off."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The potential energy is PE= 5.34 eV\n",
"The net energy is NE= 1.49 eV\n",
"The bond energy is BE= 3.85 eV\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.5 , Page no:16"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"r_0=.41; #in mm(lattice constant)\n",
"e=1.6E-19; #in C (charge of electron)\n",
"E_o= 8.85E-12; #absolute premittivity\n",
"n=0.5; #repulsive exponent value\n",
"alpha=1.76; #Madelung constant\n",
"pi=3.14; # value of pi used in the solution\n",
"\n",
"#calculate\n",
"r=.41*1E-3; #since r is in mm\n",
"Beta=72*pi*E_o*r**4/(alpha*e**2*(n-1)); #calculation compressibility\n",
"\n",
"#result\n",
"print\"The compressibility is\tBeta=\",round(Beta);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The compressibility is\tBeta= -2.50967916144e+15\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.6 , Page no:16"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"r_0=3.56; #in Angstrom\n",
"e=1.6E-19; #in C (charge of electron)\n",
"IE=3.89; #in eV (ionisation energy of Cs)\n",
"EA=-3.61; #in eV (electron affinity of Cl)\n",
"n=10.5; #Born constant\n",
"E_o= 8.85E-12; #absolute premittivity\n",
"alpha=1.763; #Madelung constant\n",
"pi=3.14; #value of pi used in the solution\n",
"\n",
"#calculate\n",
"r_0=r_0*1E-10; #since r is in nanometer\n",
"U=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
"U=U/e; #changing unit from J to eV\n",
"ACE=U+EA+IE; #calculation of atomic cohesive energy\n",
"\n",
"#result\n",
"print\"The ionic cohesive energy is \",round(U),\"eV\";\n",
"print\"The atomic cohesive energy is\",round(ACE),\"eV\";"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The ionic cohesive energy is -6.0 eV\n",
"The atomic cohesive energy is -6.0 eV\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.7 , Page no:17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"r_0=2.81; #in Angstrom\n",
"e=1.6E-19; #in C (charge of electron)\n",
"n=9; #Born constant\n",
"E_o= 8.85E-12; #absolute premittivity\n",
"alpha=1.748; #Madelung constant\n",
"pi=3.14; #value of pi used in the solution\n",
"\n",
"#calculate\n",
"r_0=r_0*1E-10; #since r is in nanometer\n",
"V=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
"V=V/e; #changing unit from J to eV\n",
"V_1=V/2; #Since only half of the energy contribute per ion to the cohecive energy therfore\n",
"\n",
"#result\n",
"print\"The potential energy is V=\",round(V,2),\"eV\";\n",
"print\"The energy contributing per ions to the cohesive energy is \",round(V_1,2),\"eV\";"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The potential energy is V= -7.96 eV\n",
"The energy contributing per ions to the cohesive energy is -3.98 eV\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}
|
