SIUnits.jl Sample usage (Solving problem from Griffiths QM chapter 5)

gistfile1.txt

{
 "metadata": {
  "language": "Julia",
  "name": "SIUnits Demo"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "using TexExtensions\nusing SIUnits",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 11
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "# Constants\n\u0127 = 1.05457 * 10.0^(-34) * (Joule*Second)\n# Electron mass\nm = 9.18938 * 10.0^(-31) * (KiloGram)\n# Avogadro's number\nN_a = 6.02214129 * 10.0^23 * Mole^(-1);",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 45
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Griffiths 5.16"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "# Given Quantities\n    # Mass-density\n    \u03c1_copper = 9.86*(Gram/CentiMeter^3)\n    # Atomic Weight \n    M_copper = 63.5*(Gram/Mole);",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": "a)"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "# Assumptions\nq = 1",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "\\[1\\]",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 13,
       "text": "1"
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "# Calculating the Fermi Energy of Copper\nN = N_a\nV = M_copper/\u03c1_copper\n# Equation 5.42\n\u03c1 = (N*q)/V\n# From Equation 5.43\nE_f = (\u0127^2/(2m)) * (3*\u03c1*\u03c0^2)^(2//3)\nas(E_f, ElectronVolt)",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "\\[7.446879565794478\\;\\text{eV}\\]",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 14,
       "text": "7.446879565794478 eV"
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": "b)"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "# Calculate the velocity\nv = sqrt(2*(E_f/m))",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "\\[1.6114403910119676\\times 10^{6}\\;\\frac{\\text{m} }{\\text{s} }\\]",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 25,
       "text": "1.6114403910119676e6 m s\u207b\u00b9"
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "# For comparison, the speed of light\nc = 299_792_458*Meter/Second\nv/c",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "\\[0.005375186559936633\\]",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 29,
       "text": "0.005375186559936633"
      }
     ],
     "prompt_number": 29
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": "c)"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "# Constants\nkB = 1.38065 * 10.0^(-23) * Joule/Kelvin;",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 33
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "# Calculate the Fermi Temperature\nT = E_f/kB",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "\\[86417.38257559626\\;\\text{K} \\]",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 34,
       "text": "86417.38257559626 K "
      }
     ],
     "prompt_number": 34
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": "d)"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "# Equation 5.46\nP=((3\u03c0^2)^(2/3)*\u0127^2)/(5m) * \u03c1^(5//3)",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "\\[4.462711654221244\\times 10^{10}\\;\\frac{\\text{kg} }{\\text{m} \\;\\text{s}^{2}}\\]",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 44,
       "text": "4.462711654221244e10 kg m\u207b\u00b9s\u207b\u00b2"
      }
     ],
     "prompt_number": 44
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Problem 5.17"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "B=(5//3)P",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "\\[7.437852757035406\\times 10^{10}\\;\\frac{\\text{kg} }{\\text{m} \\;\\text{s}^{2}}\\]",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 45,
       "text": "7.437852757035406e10 kg m\u207b\u00b9s\u207b\u00b2"
      }
     ],
     "prompt_number": 45
    }
   ],
   "metadata": {}
  }
 ]
}