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diff --git a/backup/Modern_Physics_version_backup/Chapter14_1.ipynb b/backup/Modern_Physics_version_backup/Chapter14_1.ipynb deleted file mode 100755 index 84af5d28..00000000 --- a/backup/Modern_Physics_version_backup/Chapter14_1.ipynb +++ /dev/null @@ -1,146 +0,0 @@ -{ - "metadata": { - "name": "MP-14" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": "Elementary Particles" - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": "Example 14.2 Page 451" - }, - { - "cell_type": "code", - "collapsed": false, - "input": "#initiation of variable\nmvo=1116.0;mp=938.0;mpi=140.0; #mass of various particles\n\n#calculation\nQ=(mvo-mp-mpi); #Q value of energy\nPp=100.0;Ppi=100; #momentum of various particles\nKp=5.0;Kpi=38-Kp; #kinetic energy of particles\n\n#result\nprint \"The kinetic energy of the particles Kp and Kpi are\", Kp,\" MeV and\",Kpi,\" MeV respectively\"", - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": " The kinetic energy of the particles Kp and Kpi are 5.0 MeV and 33.0 MeV respectively\n" - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": "Example 14.3 Page 453" - }, - { - "cell_type": "code", - "collapsed": false, - "input": "#initiation of variable\nQ=105.2 # The Q value for the given decay\nMuc2=105.80344 #mass energy\n\n#calculation\nKe= Q**2/(2*Muc2); #Ke=Ee-mec2;\n\n#result\nprint \"The maximum kinetic energy in MeV is\",round(Ke,3);\n", - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": "The maximum kinetic energy in MeV is 52.3\n" - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": "Example 14.4 Page 455" - }, - { - "cell_type": "code", - "collapsed": false, - "input": "#initiation of variable\nfrom __future__ import division\nfrom sympy.solvers import solve\nfrom sympy import Symbol\nfrom sympy import *\nfrom math import sqrt\nmkc2=494.0; mpic2=135.0;mec2=0.5;# mass of various particles\n\n#calculation\nQ1=mkc2-mpic2-mec2; #Q of reaction\n# the neutrino has negligible energy\nx = symbols('x')\nk=solve((x**2+135.0**2)**(0.5)+x-494,x);# assigning the Q to sum of energies and simplifying\n\nprint \"The value of maximum kinetic enrgy for pi-meson and positron are\",266,\"MeV &\",k,\" MeV\";", - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": " The value of maximum kinetic enrgy for pi-meson and positron are 266 MeV & [228.553643724696] MeV\n" - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": "Example 14.5 Page 457" - }, - { - "cell_type": "code", - "collapsed": false, - "input": "#initiation of variable\nmpi_=140;mp=938;mKo=498;mLo=1116; #mass of various particles\n\n#calculation\nQ1= mpi_+mp-mKo-mLo; #Q value of reaction 1\nmK_=494.0;mpio=135.0; \nQ2=mK_+mp-mLo-mpio; #Q value of reaction 2\n\n#result\nprint\"The Q values of reactions 1 and 2 are\", Q1,\" MeV and\",Q2,\"MeV\";", - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": "The Q values of reactions 1 and 2 are -536 MeV and 181.0 MeV\n" - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": "Example 14.6 Page 459" - }, - { - "cell_type": "code", - "collapsed": false, - "input": "#initiation of variable\nmpic2=135.0; #mass ennergy of pi particle\n\n#calculation\nQ=-mpic2;\nmp=938.0;mpi=135.0;\nKth=(-Q)*((4*mp)+mpi)/(2*(mp)); #threshold energy\n\n#result\nprint\"The threshold kinetic energy in MeV is\",round(Kth,3);", - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": "The threshold kinetic energy in MeV is 279.715\n" - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": "Example 14.7 Page 460" - }, - { - "cell_type": "code", - "collapsed": false, - "input": "#initiation of variable\nmpc2=938.0; #rest energy of proton\n\n#result\nQ=mpc2+mpc2-(4*mpc2); #Q value of reaction \nKth=(-Q)*(6*mpc2/(2*mpc2)); # thershold kinetic energy\n\n#result\nprint \"The threshold kinetic energy in MeV is\",round(Kth,3);", - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": "The threshold kinetic energy in MeV is 5628.0\n" - } - ], - "prompt_number": 21 - } - ], - "metadata": {} - } - ] -}
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