{ "metadata": { "name": "", "signature": "sha256:b6d6dfa593701249cd6d305eb45cecde030c3502c19d325045b7e05cf46a035c" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "3: Atomic models" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 3.1, Page number 45" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "N=6.02*10**23; #avagadro number(atoms/mole)\n", "rho=19.3; #density(g/cc)\n", "A=197; #atomic weight(g)\n", "k=8.984*10**9; #value of k(Nm**2/C**2)\n", "Z=79;\n", "Zdash=2;\n", "e=1.6*10**-19; #conversion factor from J to eV\n", "m=2;\n", "v0=8*10**6; \n", "t=2*10**-6; #thickness(m)\n", "\n", "#Calculation\n", "n=N*rho*10**6/A; #number of atoms(per m**3)\n", "b=k*Z*Zdash*e/(m*v0); #impact parameter(m)\n", "f=math.pi*b**2*n*t; #fraction of particles scattered\n", "\n", "#Result\n", "print \"fraction of particles scattered is\",round(f*10**5,1),\"*10**-5\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "fraction of particles scattered is 7.5 *10**-5\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 3.3, Page number 48" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "h=6.626*10**-34; #planck's constant(Js)\n", "c=3*10**8; #velocity of light(m/sec)\n", "e=1.6*10**-19; #conversion factor from J to eV\n", "E=10.5; #energy(eV)\n", "\n", "#Calculation\n", "E=(13.6+E)*e; #energy of photon(J)\n", "lamda=h*c/E; #wavelength(m)\n", "\n", "#Result\n", "print \"wavelength of photon is\",round(lamda*10**9,2),\"nm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "wavelength of photon is 51.55 nm\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 3.4, Page number 49" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "k=8.98*10**9; #value of k(Nm**2/C**2)\n", "e=1.6*10**-19; #conversion factor from J to eV\n", "n=1; #assume\n", "a0=0.53*10**-10; #radius of orbit(m)\n", "\n", "#Calculation\n", "PE=-k*(e**2)/(a0*e*n**2); #potential energy(eV)\n", "E=-13.6/n**2; #energy(eV)\n", "KE=E-PE; #kinetic energy(eV)\n", "\n", "#Result\n", "print \"kinetic energy is\",round(KE,1),\"/n**2 eV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "kinetic energy is 13.5 /n**2 eV\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 3.6, Page number 51" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "Mbyme=1836; \n", "lamda=6562.8; #wavelength for hydrogen(angstrom)\n", "\n", "#Calculation\n", "mew_dashbymew=2*(1+Mbyme)/(1+(2*Mbyme));\n", "lamda_dash=lamda/mew_dashbymew; #wavelength for deuterium(angstrom)\n", "\n", "#Result\n", "print \"wavelength for deuterium is\",int(lamda_dash),\"angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "wavelength for deuterium is 6561 angstrom\n" ] } ], "prompt_number": 14 } ], "metadata": {} } ] }