{ "metadata": { "name": "", "signature": "sha256:e984fee9b841dd6e9b7eedf1533b0a0d297cd9f484c047f051ce48a09b156826" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter2-Nuclear Engineering" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1-pg54" ] }, { "cell_type": "code", "collapsed": false, "input": [ "## Example 2.1\n", "import math\n", "#determine atoms in deuterium\n", "## Given data\n", "atom_h = 6.6*10**24; ## Number of atoms in Hydrogen\n", "## Using the data given in Table II.2, Appendix II for isotropic abundance of deuterium\n", "isoab_H2 = 0.015; ## Isotropic abundance of deuterium\n", "## Calculation\n", "totatom_d=(isoab_H2*atom_h)/100.;\n", "## Result\n", "print\"%s %.2e %s \"%('\\n Number of deuterium atoms = ',totatom_d,'');\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Number of deuterium atoms = 9.90e+20 \n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2-pg54" ] }, { "cell_type": "code", "collapsed": false, "input": [ "## Example 2.2\n", "import math\n", "#determine atomic weight of oxygen\n", "## Given data \n", "## Using the data given in the example 2.2\n", "atwt_O16 = 15.99492; ## Atomic weight of O-16 isotope\n", "isoab_O16 = 99.759; ## Abundance of O-16 isotope\n", "atwt_O17 = 16.99913; ## Atomic weight of O-17 isotope\n", "isoab_O17 = 0.037; ## Abundance of O-17 isotope\n", "atwt_O18 = 17.99916; ## Atomic weight of O-18 isotope\n", "isoab_O18 = 0.204; ## Abundance of O-18 isotope\n", "## Calculation\n", "atwt_O=(isoab_O16*atwt_O16 + isoab_O17*atwt_O17 + isoab_O18*atwt_O18)/100.;\n", "## Result\n", "print\"%s %.2f %s \"%('\\n Atomic Weight of Oxygen = ',atwt_O,'');\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Atomic Weight of Oxygen = 16.00 \n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex3-pg55" ] }, { "cell_type": "code", "collapsed": false, "input": [ "## Example 2.3\n", "import math\n", "#determine rest mass energy of electron\n", "## Given data\n", "me = 9.1095*10**(-28); ## Mass of electron in grams\n", "c = 2.9979*10**10; ## Speed of light in vacuum in cm/sec\n", "## Calculation\n", "rest_mass = me*c**2;\n", "## Result\n", "print\"%s %.2e %s \"%('\\n Rest mass energy of electron = ',rest_mass,' ergs\\n');\n", "print('Expressing the result in joules')\n", "## 1 Joule = 10^(-7)ergs\n", "rest_mass_j = rest_mass*10**(-7);\n", "print\"%s %.2e %s \"%('\\n Rest mass energy of electron = ',rest_mass_j,' joules\\n');\n", "print('Expressing the result in MeV')\n", "## 1 MeV = 1.6022*10^(-13)joules\n", "rest_mass_mev = rest_mass_j/(1.6022*10**(-13));\n", "print\"%s %.2f %s \"%('\\n Rest mass energy of electron = ',rest_mass_mev,' MeV\\n');\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Rest mass energy of electron = 8.19e-07 ergs\n", " \n", "Expressing the result in joules\n", "\n", " Rest mass energy of electron = 8.19e-14 joules\n", " \n", "Expressing the result in MeV\n", "\n", " Rest mass energy of electron = 0.51 MeV\n", " \n" ] } ], "prompt_number": 4 } ], "metadata": {} } ] }