added books

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": {}