{ "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": {} } ] }