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authorkinitrupti2017-05-12 18:40:35 +0530
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+{
+ "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": {}
+ }
+ ]
+} \ No newline at end of file