{ "metadata": { "name": "Chapter1" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 1:Introduction" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.1, Page 12" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#initiation of variable\n", "Mn=1.008665;Mp=1.007276 #Given mass of an electron and a proton in terms of u\n", "\n", "#calculation\n", "Md= Mn-Mp; #mass difference \n", "Md2=Md*931.50; #converting u into Mev/c^2 by multiplying by 931.5 MeV/c^2\n", "\n", "#result\n", "print \"Mass difference in terms of U is\",round(Md,4); \n", "print\"which equals in Mev/c^2. :\",round(Md2,3);" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mass difference in terms of U is 0.0014\n", "which equals in Mev/c^2. : 1.294\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.2, Page 12" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#initiation of variable\n", "Mp=1.007276 ; Me=5.4858*10**-4; #mass of proton and electron in terms of U\n", "\n", "#calculation\n", "Mt=Mp+Me; #Total mass= sum of above masses \n", "\n", "#result\n", "print\"The combined mass of an electron and a proton was found out to be in U.\",round(Mt,3);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The combined mass of an electron and a proton was found out to be in U. 1.008\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.3, Page 13" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#initiation of variable\n", "h=6.621*10**-34 ; c=2.9979*10**8; # h is in J/s and c is in m/s\n", "hc=h*c*((10**9)/(1.6022*10**-19)); #1e=1.602*10^-19 J and 1 m=10^9 nm\n", "\n", "#result\n", "print \"The value of hc in eV.nm is\",round(hc,4); \n", "print 'Hence zero at the end is significant.';\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The value of hc in eV.nm is 1238.8651\n", "Hence zero at the end is significant.\n" ] } ], "prompt_number": 8 } ], "metadata": {} } ] }