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{
"metadata": {
"name": "",
"signature": "sha256:358f4a01b294025154ed5b68aa4867e70b7293d6e35ba5318010f0ecabaea268"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"1: Bonding in Solids"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.1, Page number 1.4"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"IE_1 = 502 #first ionisation energy(kJ/mol)\n",
"EA_B = -335 #electron affinity for B atom(kJ/mol)\n",
"e = 1.602*10**-19\n",
"r = 0.3 #inter ionic seperation(nm)\n",
"epsilon0 = 8.85*10**-12 #permittivity of free space(C/N-m)\n",
"\n",
"#Calculation\n",
"r = r*10**-9 #inter ionic seperation(m)\n",
"N = 6.022*10**23*10**-3\n",
"E = (-e**2*N)/(4*math.pi*epsilon0*r) #electrostatic attraction energy(kJ/mol)\n",
"dE = IE_1+EA_B+E; #net change in energy per mole\n",
"\n",
"#Result\n",
"print \"electrostatic attraction energy is\",int(E),\"kJ/mol\"\n",
"print \"net change in energy is\",int(dE),\"kJ/mol\"\n",
"print \"A+B- molecule will be stable\"\n",
"print \"answer for net change,dE given in the book is wrong\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"electrostatic attraction energy is -463 kJ/mol\n",
"net change in energy is -296 kJ/mol\n",
"A+B- molecule will be stable\n",
"answer for net change,dE given in the book is wrong\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.2, Page number 1.5"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"IP_K = 4.1 #IP of K(eV)\n",
"EA_Cl = 3.6 #EA of Cl(eV)\n",
"e = 1.602*10**-19\n",
"epsilon0 = 8.85*10**-12 \n",
"\n",
"#Calculation\n",
"delta_E = IP_K - EA_Cl #energy required(eV)\n",
"#if their total energy is 0, delta_E = Ec\n",
"Ec = delta_E\n",
"R = e/(4*math.pi*epsilon0*Ec) #seperation between ion pair(m)\n",
"R = R*10**9 #seperation between ion pair(nm)\n",
"R = math.ceil(R*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print \"energy required to form K+Cl- ion pair is\",delta_E,\"eV\"\n",
"print \"seperation between ion pair is\",R,\"nm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"energy required to form K+Cl- ion pair is 0.5 eV\n",
"seperation between ion pair is 2.881 nm\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.3, Page number 1.6"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"e = 1.602*10**-19\n",
"EA = 3.65 #electron affinity of Cl(eV)\n",
"IP = 5.14 #ionisation energy of Na(eV)\n",
"epsilon0 = 8.85*10**-12\n",
"r0 = 236 #equilibrium distance(pm)\n",
"\n",
"#Calculation\n",
"r0 = r0*10**-12 #equilibrium distance(m)\n",
"V = (-e**2)/(4*math.pi*epsilon0*r0) ##potential energy(J)\n",
"V = V/e #potential energy(eV)\n",
"Ue = V\n",
"BE = -Ue - IP + EA #bond energy(eV)\n",
"BE = math.ceil(BE*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print \"bond energy for NaCl molecule is\",BE,\"eV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"bond energy for NaCl molecule is 4.614 eV\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 1.4, Page number 1.19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"e = 1.602*(10**-19)\n",
"epsilon0 = 8.85*(10**-12)\n",
"r0 = 0.281 #equilibrium seperation(nm)\n",
"A = 1.748 #Madelung constant\n",
"n = 9 #born repulsive exponent\n",
"\n",
"#Calculation\n",
"r0 = r0*10**-9 #equilibrium seperation(m)\n",
"CE = (A*e**2)*(1-(1/n))/(4*math.pi*epsilon0*r0) #cohesive energy(J)\n",
"CE = CE/e #cohesive energy(eV)\n",
"CE = math.ceil(CE*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print \"cohesive energy of NaCl is\",CE,\"eV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"cohesive energy of NaCl is 7.966 eV\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}
|
