{ "metadata": { "name": "", "signature": "sha256:4a31773c970db99aecfd55acb0ef8a97d14b54e9f5682909d8824d63b10be118" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter2 : Atomic model & bonding in solids" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "example-2.1, page no-28" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#given\n", "#atomic no. of gold\n", "Z=79\n", "#kinetic energy of alpha particle\n", "E=7.68*1.6*(10)**(-13) #J because [1MeV=1.6*(10)**(-13)]\n", "e=1.6*10**(-19) #C\n", "E0=8.854*10**(-12) #F/m\n", "#the distance of closest approach is given by:\n", "d0=2*e*Z*e/(4*(math.pi)*E0*E) #m\n", "print \"The closest approach of alpha particle is %.2ef m\" %d0" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The closest approach of alpha particle is 2.96e-14f m\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "example-2.2, page no-29" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "from numpy import *\n", "#given\n", "#IN THE RUTHERFORD SCATTERING EXPERIMENT\n", "#the no of particles scattered at\n", "theta1=(pi)/2 #radians\n", "#is\n", "N90=44 #per minute\n", "#the number of particles scattered particales N is given by\n", "#N=C*(1/(sin(theta/2))**4) where C is propotionality constant\n", "#solving above equation for C\n", "C=N90*(sin(theta1/2))**4 \n", "# now to find the no of particles scatering at 75 and 135 degrees\n", "theta2=75*(pi)/180 #radians\n", "N75=C*(1/(sin(theta2/2))**4) #per minute\n", "theta3=135*(pi)/180 #radians\n", "N135=C*(1/(sin(theta3/2))**4) #per minute\n", "print \"The no of particles scattered at 75 and 135 degrees are %d per minute and %d per minutes\" %(N75,N135)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The no of particles scattered at 75 and 135 degrees are 80 per minute and 15 per minutes\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "example-2.3, page no-32" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#given\n", "#mass of electron\n", "m=9.11*10**(-31) #kg\n", "#charge on an electron\n", "e=1.6*10**(-19) #C\n", "#plank's constant\n", "h=6.62*10**(-34)\n", "E0=8.85*10**(-12) \n", "#NO OF ELECTRONS SHELLS IN HYDROZEN ATOm\n", "n=1\n", "#atomic number of hydrogen\n", "Z=1\n", "#radius of first orbit of hydrogen is given by\n", "r1=n**2*E0*h**2/((pi)*m*Z*e**2) #m\n", "print \"The radius of the first orbit of the electron in the hydrogen atom %.2e\"%(r1)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The radius of the first orbit of the electron in the hydrogen atom 5.29e-11\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "example-2.4, page no-32" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#given\n", "#mass of electron\n", "m=9.11*10**(-31) #kg\n", "#charge on an electron\n", "e=1.6*10**(-19) #C\n", "#plank's constant\n", "h=6.62*10**(-34)\n", "E0=8.85*10**(-12) \n", "#NO OF ELECTRONS SHELLS IN HYDROZEN ATOm\n", "n=1\n", "#atomic number of hydrogen\n", "Z=1\n", "#ionization potential energy of hydrogen atom is given by\n", "E=m*Z**2*e**4/(8*(E0)**2*h**2*n**2) #J\n", "#energy in eV\n", "EV=E/e #eV\n", "print \"The ionization potential for hydrogen atom is %0.2f V\" %(EV)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The ionization potential for hydrogen atom is 13.59 V\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "example-2.6, page no-36" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#given\n", "#uncertainity in the momentum\n", "deltap=10**-27 #kg ms**-1\n", "#according to uncertainity principle\n", "#deltap* deltax >=h/(2*(pi))\n", "#we know that \n", "h=6.626*10**-34 #Js\n", "#here instead of inequality we are using only equality just for notation otherwise it is greater than equal to as mentioned above\n", "#now deltax is given by\n", "deltax=h/(2*(pi)*deltap) #m\n", "print \"The minimum uncertainity is %.2e m\"%(deltax)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The minimum uncertainity is 1.05e-07 m\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "example-2.10, page no- 57" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#given\n", "#ionization potential of hydrogen\n", "E1=13.6 #eV\n", "#when \n", "n=3\n", "E3=-E1/n**2 #eV\n", "#when \n", "n=5\n", "E5=-E1/n**2 #eV\n", "print \"Energy of 3rd and 5th orbits are %0.2f eV and %0.2f eV\"%(E3,E5)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Energy of 3rd and 5th orbits are -1.51 eV and -0.54 eV\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "example-2.11, page no-59" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#given\n", "#dipole moment og HF is\n", "DM=6.375*10**(-30) #Cm\n", "#intermolecular distance\n", "r=0.9178*10**(-10) #m\n", "#charge on an electron\n", "e=1.67*10**(-19) #C\n", "#since the HF posses ionic characters\n", "#so\n", "#Hf in fully ionic state has dipole moment as\n", "DM2=r*e #Cm\n", "#percentage ionic characters\n", "percentage=DM/DM2*100 #%\n", "print \"The percentage ionic character is %0.2f approx.\"%(percentage)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The percentage ionic character is 41.59 approx.\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "example-2.12, page no-60" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#given\n", "#elctronegativity of In\n", "EnIn=1.5\n", "#elctronegativity of As\n", "EnAs=2.2\n", "#elctronegativity of Ga\n", "EnGa=1.8\n", "#for InAs\n", "ionic_charater1=(1-exp((-0.25)*(EnAs-EnIn)**2))*100 #in %\n", "#for GaAs\n", "ionic_charater2=(1-exp((-0.25)*(EnAs-EnGa)**2))*100 # in %\n", "print \"Ionic character in InAs and GaAs are %0.1f %% and %0.1f %%\"%(ionic_charater1,ionic_charater2)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Ionic character in InAs and GaAs are 11.5 % and 3.9 %\n" ] } ], "prompt_number": 30 } ], "metadata": {} } ] }