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Diffstat (limited to 'Modern_Physics/Chapter14.ipynb')
| -rwxr-xr-x | Modern_Physics/Chapter14.ipynb | 250 |
1 files changed, 71 insertions, 179 deletions
diff --git a/Modern_Physics/Chapter14.ipynb b/Modern_Physics/Chapter14.ipynb index 4fff498c..a7beb792 100755 --- a/Modern_Physics/Chapter14.ipynb +++ b/Modern_Physics/Chapter14.ipynb @@ -1,7 +1,6 @@ { "metadata": { - "name": "", - "signature": "sha256:bb0d2851c786dd28ae3ad8cbd1e6a7fc5db7cc1384cc2031f2bb6380ed7515e2" + "name": "Chapter14" }, "nbformat": 3, "nbformat_minor": 0, @@ -13,7 +12,7 @@ "level": 1, "metadata": {}, "source": [ - "Chapter 14: Nuclear Physics Apllications" + "Chapter 14:Elementary Particles" ] }, { @@ -21,43 +20,23 @@ "level": 2, "metadata": {}, "source": [ - "Example 14.1, page no. 505" + "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", - "#Variable declaration\n", - "\n", - "MLi = 7.016003 #atomic mass of Lithium\n", - "MH = 1.007825 #atomic mass of Hydrogen\n", - "MHe = 4.002603 #atomic mass of Helium\n", - "c2 = 931.50 #Square of speed of light (MeV/u)\n", - "\n", - "#Calculation\n", - "\n", - "Q = (MLi + MH - 2*MHe) * c2\n", - "\n", - "#Results\n", - "\n", - "print \"(a) The Q value is\",round(Q,1),\"MeV.\"\n", - "\n", - "\n", - "\n", - "#Variable declaration\n", - "\n", - "Kincident = 0.6 #kinetic energy of incident particle (MeV)\n", - "\n", - "#Calculation\n", - "\n", - "Kproducts = Q + Kincident\n", - "\n", - "#Results\n", - "\n", - "print \"(b) The kinetic energy of the products is\",round(Kproducts,1),\"MeV.\"" + "#result\n", + "print \"The kinetic energy of the particles Kp and Kpi are\", Kp,\" MeV and\",Kpi,\" MeV respectively\"" ], "language": "python", "metadata": {}, @@ -66,46 +45,33 @@ "output_type": "stream", "stream": "stdout", "text": [ - "(a) The Q value is 17.3 MeV.\n", - "(b) The kinetic energy of the products is 17.9 MeV.\n" + " The kinetic energy of the particles Kp and Kpi are 5.0 MeV and 33.0 MeV respectively\n" ] } ], - "prompt_number": 6 + "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 14.2, page no. 509" + "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", - "#Variable declaration\n", - "\n", - "A = 27 #Atomic number of Aluminum\n", - "d = 2.7 #density of aluminum\n", - "Av = 6.02 * 10 ** 23 #Avogadro number nuclei/mol\n", - "sigma = 2.0 * 10 **-31 #capture cross section (m^2)\n", - "x = 0.3 * 10 ** -3 #thickness of the foil(m)\n", - "R0 = 5.0 * 10 ** 12 #rate of incident particles(neutrons/cm^2.s)\n", - "\n", - "#Calculation\n", - "\n", - "n = Av * d / A * 10**6\n", - "R = R0 * sigma * x * n\n", - "\n", - "#Result\n", - "\n", - "print \"The number of neutrons captured by 1.0cm^2/s is \",round(R/10**7,1),\"X 10^7 neutrons/cm^2.s\"\n", - "\n", - "\n" + "#result\n", + "print \"The maximum kinetic energy in MeV is\",round(Ke,3);\n" ], "language": "python", "metadata": {}, @@ -114,40 +80,39 @@ "output_type": "stream", "stream": "stdout", "text": [ - "The number of neutrons captured by 1.0cm^2/s is 1.8 X 10^7 neutrons/cm^2.s\n" + "The maximum kinetic energy in MeV is 52.3\n" ] } ], - "prompt_number": 9 + "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 14.3, page no. 512" + "Example 14.4, Page 455" ] }, { "cell_type": "code", "collapsed": false, "input": [ - "\n", - "#Variable declaration\n", - "\n", - "A = 235 #atomic number of uranium\n", - "m = 10**3 #mass of 235U\n", - "Av = 6.02*10**23#avogadro number (nuclei/mol)\n", - "Q = 208 #disintegration energy per event (MeV)\n", - "\n", - "#Calculation\n", - "\n", - "N = Av * m / A\n", - "E = N * Q\n", - "\n", - "#Result\n", - "\n", - "print \"The disintegration energy is\",round(E/10**26,2),\"X 10^26 MeV.\"" + "#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": {}, @@ -156,83 +121,34 @@ "output_type": "stream", "stream": "stdout", "text": [ - "The disintegration energy is 5.33 X 10^26 MeV.\n" + " The value of maximum kinetic enrgy for pi-meson and positron are 266 MeV & [228.553643724696] MeV\n" ] } ], - "prompt_number": 12 + "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 14.4, page no. 513" + "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", - "\n", - "#Variable declaration\n", - "\n", - "A = 235 #atomic mass of Uranium\n", - "Av = 6.02 * 10**23 # Avagadro's number\n", - "Q = 208 # disintegration energy per event (MeV)\n", - "\n", - "#Calculation\n", - "\n", - "N = Av / A *10**3 #No of nuclei in 1kg of 235U\n", - "E = N*Q #disintegration energy (MeV)\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", - "\n", - "print \"The disintegration energy is\",round(E/10**26,2),\" X 10^26 MeV.\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The disintegration energy is 5.33 X 10^26 MeV.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.5, page no. 513" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration\n", - "\n", - "rBa = 6.2 * 10 **-15 #nuclear radius of Ba(m)\n", - "rKr = 5.4 * 10**-15 #nuclear radius of Kr(m)\n", - "Z1 = 56 #atomic number of Ba\n", - "Z2 = 36 #atomic number of Kr\n", - "k = 1.440 * 10 **-9 #Coulomb constant (eV.nm)\n", - "\n", - "#Calculation\n", - "\n", - "r = rBa + rKr\n", - "U = k * Z1 * Z2 /( round(r/10**-15) * 10**-15)\n", - "\n", - "#Result\n", - "\n", - "print \"The potential energy of the two nuclei is\",round(U/10**6),\"MeV.\" " + "print\"The Q values of reactions 1 and 2 are\", Q1,\" MeV and\",Q2,\"MeV\";" ], "language": "python", "metadata": {}, @@ -241,52 +157,34 @@ "output_type": "stream", "stream": "stdout", "text": [ - "The potential energy of the two nuclei is 242.0 MeV.\n" + "The Q values of reactions 1 and 2 are -536 MeV and 181.0 MeV\n" ] } ], - "prompt_number": 17 + "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 14.6, page no. 519" + "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", - "\n", - "#variable declaration\n", - "\n", - "e = 1.6 * 10 ** -19 #electron charge (C)\n", - "r = 1.0 * 10**-14 #separation (m)\n", - "k = 8.99 * 10**9 #Coulomb constant(N.m^2/C^2)\n", - "\n", - "#Calculation\n", - "\n", - "U = k * e**2/ r\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", - "\n", - "print \"(a)The height of the potential barrier is\",round(U/e/10**6,2),\"MeV.\"\n", - "\n", - "\n", - "#Variable Declaration\n", - "\n", - "kB = 1.38 * 10**-23 #Boltzmann's constant (J/K)\n", - "\n", - "#Calculation\n", - "\n", - "T = (U/2.0)*(2.0/3.0)/(kB+.07*10**-23)\n", - "\n", - "#Result\n", - "\n", - "print\"(b)The effective temperature is\",round(T/10**8,1),\"X 10^8 K.\"" + "print\"The threshold kinetic energy in MeV is\",round(Kth,3);" ], "language": "python", "metadata": {}, @@ -295,39 +193,33 @@ "output_type": "stream", "stream": "stdout", "text": [ - "(a)The height of the potential barrier is 0.14 MeV.\n", - "(b)The effective temperature is 5.3 X 10^8 K.\n" + "The threshold kinetic energy in MeV is 279.715\n" ] } ], - "prompt_number": 22 + "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Example 14.7, page no. 530" + "Example 14.7, Page 460" ] }, { "cell_type": "code", "collapsed": false, "input": [ + "#initiation of variable\n", + "mpc2=938.0; #rest energy of proton\n", "\n", - "import math\n", - "\n", - "#variable declaration\n", - "\n", - "mu = 55 #mu for x-rays \n", - "\n", - "#Calculation\n", - "\n", - "x = math.log(2)/mu\n", - "\n", - "#Result\n", + "#result\n", + "Q=mpc2+mpc2-(4*mpc2); #Q value of reaction \n", + "Kth=(-Q)*(6*mpc2/(2*mpc2)); # thershold kinetic energy\n", "\n", - "print \"The half value thickness for lead is\",round(x/10**-2,2),\"X 10^-2 cm.\"" + "#result\n", + "print \"The threshold kinetic energy in MeV is\",round(Kth,3);" ], "language": "python", "metadata": {}, @@ -336,11 +228,11 @@ "output_type": "stream", "stream": "stdout", "text": [ - "The half value thickness for lead is 1.26 X 10^-2 cm.\n" + "The threshold kinetic energy in MeV is 5628.0\n" ] } ], - "prompt_number": 24 + "prompt_number": 21 } ], "metadata": {} |