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diff --git a/backup/Modern_Physics_version_backup/Chapter14.ipynb b/backup/Modern_Physics_version_backup/Chapter14.ipynb deleted file mode 100755 index a7beb792..00000000 --- a/backup/Modern_Physics_version_backup/Chapter14.ipynb +++ /dev/null @@ -1,241 +0,0 @@ -{ - "metadata": { - "name": "Chapter14" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 14:Elementary Particles" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.2, Page 451" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "mvo=1116.0;mp=938.0;mpi=140.0; #mass of various particles\n", - "\n", - "#calculation\n", - "Q=(mvo-mp-mpi); #Q value of energy\n", - "Pp=100.0;Ppi=100; #momentum of various particles\n", - "Kp=5.0;Kpi=38-Kp; #kinetic energy of particles\n", - "\n", - "#result\n", - "print \"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\n", - "Q=105.2 # The Q value for the given decay\n", - "Muc2=105.80344 #mass energy\n", - "\n", - "#calculation\n", - "Ke= Q**2/(2*Muc2); #Ke=Ee-mec2;\n", - "\n", - "#result\n", - "print \"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\n", - "from __future__ import division\n", - "from sympy.solvers import solve\n", - "from sympy import Symbol\n", - "from sympy import *\n", - "from math import sqrt\n", - "mkc2=494.0; mpic2=135.0;mec2=0.5;# mass of various particles\n", - "\n", - "#calculation\n", - "Q1=mkc2-mpic2-mec2; #Q of reaction\n", - "# the neutrino has negligible energy\n", - "x = symbols('x')\n", - "k=solve((x**2+135.0**2)**(0.5)+x-494,x);# assigning the Q to sum of energies and simplifying\n", - "\n", - "print \"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\n", - "mpi_=140;mp=938;mKo=498;mLo=1116; #mass of various particles\n", - "\n", - "#calculation\n", - "Q1= mpi_+mp-mKo-mLo; #Q value of reaction 1\n", - "mK_=494.0;mpio=135.0; \n", - "Q2=mK_+mp-mLo-mpio; #Q value of reaction 2\n", - "\n", - "#result\n", - "print\"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\n", - "mpic2=135.0; #mass energy of pi particle\n", - "\n", - "#calculation\n", - "Q=-mpic2;\n", - "mp=938.0;mpi=135.0;\n", - "Kth=(-Q)*((4*mp)+mpi)/(2*(mp)); #threshold energy\n", - "\n", - "#result\n", - "print\"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\n", - "mpc2=938.0; #rest energy of proton\n", - "\n", - "#result\n", - "Q=mpc2+mpc2-(4*mpc2); #Q value of reaction \n", - "Kth=(-Q)*(6*mpc2/(2*mpc2)); # thershold kinetic energy\n", - "\n", - "#result\n", - "print \"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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