{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#chapter 3:nuclear models"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.1;pg no:38"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.1, Page:38  \n",
      " \n",
      "\n",
      "The coulomb barrier for the penetration of Th by proton in MeV= 12.96\n",
      "The coulomb barrier for the penetration of Th by alpha particle in MeV= 23.94\n"
     ]
    }
   ],
   "source": [
    "#cal of coulomb barrier for the penetration of Th by alpha particle and proton\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.1, Page:38  \\n \\n\"\n",
    "#Given in cgs units\n",
    "m1=232.;\n",
    "m2=1.;\n",
    "m3=4.;\n",
    "z1=90.;\n",
    "z2=1.;\n",
    "z3=2.;\n",
    "e=4.8*10**-10;# in ergs\n",
    "c=1.4;# nuclear radius constant\n",
    "\n",
    "#Formula: E=(z1*z2*e^2)/(r1+r2)\n",
    "r1=(m1)**(1./3.);\n",
    "r2=(m2)**(1./3.);\n",
    "r3=(m3)**(1./3.);\n",
    "E1=(z1*z2*e*e)/(c*(r1+r2)*10**-13*(1.6*10**-6));\n",
    "print\"The coulomb barrier for the penetration of Th by proton in MeV=\",round(E1,2)\n",
    "E2=(z1*z3*e*e)/(c*(r1+r3)*10**-13*(1.6*10**-6));\n",
    "print\"The coulomb barrier for the penetration of Th by alpha particle in MeV=\",round(E2,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.2;pg no:38"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.2, Page38 \n",
      " \n",
      "\n",
      "\n",
      " The coulomb barrier for the penetration of Th by proton in MeV= 8.84\n",
      "\n",
      " \n",
      " The coulomb barrier for the penetration of Th by alpha particle in MeV= 10.96\n"
     ]
    }
   ],
   "source": [
    "#cal of coulomb barrier for the penetration of Th by alpha particle and proton\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.2, Page38 \\n \\n\"\n",
    "#Given in cgs units\n",
    "m1=112;\n",
    "m2=1;\n",
    "m3=4;\n",
    "m4=66;\n",
    "z1=50;\n",
    "z2=1;\n",
    "z3=2;\n",
    "z4=30;\n",
    "e=4.8*10**-10;# in ergs\n",
    "c=1.4;# nuclear radius constant\n",
    "\n",
    "#Formula: E=(z1*z2*e^2)/(r1+r2)\n",
    "r1=(m1)**(1./3.);\n",
    "r2=(m2)**(1./3.);\n",
    "r3=(m3)**(1./3.);\n",
    "r4=(m4)**(1./3.);\n",
    "E1=(z1*z2*e*e)/(c*(r1+r2)*10**-13*(1.6*10**-6));\n",
    "print\"\\n The coulomb barrier for the penetration of Th by proton in MeV=\",round(E1,2)\n",
    "E2=(z4*z3*e*e)/(c*(r4+r3)*10**-13*(1.6*10**-6));\n",
    "print\"\\n \\n The coulomb barrier for the penetration of Th by alpha particle in MeV=\",round(E2,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.3;pg no:39"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.3, Page:39  \n",
      " \n",
      "\n",
      "The closest distance of approach in fm= 37.9\n"
     ]
    }
   ],
   "source": [
    "#cal of closest distance of approach\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.3, Page:39  \\n \\n\"\n",
    "# Given:\n",
    "E=6;# in MeV\n",
    "z1=79;\n",
    "z2=2;\n",
    "q=4.8*10**-10;\n",
    "# Solution:\n",
    "\n",
    "# At the closest distance of approach, the kineic energy of the alpha particle balances the columb barrier energy.\n",
    "\n",
    "r1=(z1*z2*q*q)/(E*1.6*10**-6);# distance in cm\n",
    "r=r1*10**13;# distance in fm\n",
    "\n",
    "print\"The closest distance of approach in fm=\",round(r,1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.4;pg no:39"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.4, Page:39  \n",
      " \n",
      "\n",
      "The stable nuclied of the isobaric series is Hf atomic no. = 72.6\n"
     ]
    }
   ],
   "source": [
    "#cal of stable nuclied of the isobaric series\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.4, Page:39  \\n \\n\"\n",
    "# Given:\n",
    "A=180.;\n",
    "# Solution:\n",
    "z=(40.*A)/(0.6*(A**(2./3.))+80.);\n",
    "print\"The stable nuclied of the isobaric series is Hf atomic no. =\",round(z,1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.5;pg no:39"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.5, Page:39  \n",
      " \n",
      "\n",
      "The stable nuclied of the isobaric series is Sr atomic no. = 38.0\n",
      "\n",
      " Hence the nuclides of z<38 fall on the left of the limb of B vs Z parabola while the nuclides of z>38 fall on the right limb of the parabola.\n"
     ]
    }
   ],
   "source": [
    "#cal of stable nuclied of the isobaric series is Sr atomic no\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.5, Page:39  \\n \\n\"\n",
    "# Given:\n",
    "A=87.;\n",
    "\n",
    "# Solution:\n",
    "z=(40.*A)/(0.6*(A**(2./3.))+80.);\n",
    "print\"The stable nuclied of the isobaric series is Sr atomic no. =\",round(z)\n",
    "# nereast integer is 38\n",
    "print\"\\n Hence the nuclides of z<38 fall on the left of the limb of B vs Z parabola while the nuclides of z>38 fall on the right limb of the parabola.\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.7;pg no:40"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.7, Page:40  \n",
      " \n",
      "\n",
      "\n",
      " The binding energy for the last proton in 12C in (MeV)= 15.92\n",
      "\n",
      " The binding energy for the last neutron in 12C in (MeV)= 9.43\n",
      "\n",
      " The binding energy for the last proton in 28Si in (MeV)= 11.55\n",
      "\n",
      " The binding energy for the last neutron in 28Si in (MeV)= 17.2\n"
     ]
    }
   ],
   "source": [
    "#cal of binding energy for the last proton,neutron\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.7, Page:40  \\n \\n\"\n",
    "# Given:\n",
    "B=11.009305;\n",
    "C1=12;\n",
    "C2=11.001433;\n",
    "p=1.0078;\n",
    "n=1.0087;\n",
    "Al=26.981535;\n",
    "Si1=27.976927;\n",
    "Si2=26.986705;\n",
    "# Solution:\n",
    "m1=(B+p-C1);#(a)\n",
    "E1=m1*931;# of last proton in C in MeV\n",
    "print\"\\n The binding energy for the last proton in 12C in (MeV)=\",round(E1,2)\n",
    "\n",
    "m2=(C2+n-C1);#(b)\n",
    "E2=m2*931;# of last neutron in C in MeV\n",
    "print\"\\n The binding energy for the last neutron in 12C in (MeV)=\",round(E2,2)\n",
    "\n",
    "m3=(Al+p-Si1);#(c)\n",
    "E3=m3*931;# of last proton in Si in MeV\n",
    "print\"\\n The binding energy for the last proton in 28Si in (MeV)=\",round(E3,2)\n",
    "\n",
    "m4=(Si2+n-Si1);#(d)\n",
    "E4=m4*931;# of last neutron in Si in MeV\n",
    "print\"\\n The binding energy for the last neutron in 28Si in (MeV)=\",round(E4,2)\n",
    "\n",
    "# Note: There is a calculation error in the textbook for the (b) part."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.8;pg no:41"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.8, Page: 41 \n",
      " \n",
      "\n",
      "\n",
      " The binding energy for N(14) in (MeV)= 10.27\n",
      "\n",
      " The binding energy for O(16) in (MeV)= 7.16\n"
     ]
    }
   ],
   "source": [
    "#cal of binding energy for N,O\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.8, Page: 41 \\n \\n\"\n",
    "# Given:\n",
    "d=2.014102;\n",
    "C=12;\n",
    "a=4.002603;\n",
    "N=14.003074;\n",
    "O=15.994915;\n",
    "\n",
    "# Solution:\n",
    "m1=(C+d-N);\n",
    "E1=m1*931;# The binding energy for N(14)\n",
    "print\"\\n The binding energy for N(14) in (MeV)=\",round(E1,2)\n",
    "\n",
    "m2=(C+a-O);\n",
    "E2=m2*931;#The binding energy for O(16) \n",
    "print\"\\n The binding energy for O(16) in (MeV)=\",round(E2,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.9;pg no:41"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.9, Page:41  \n",
      " \n",
      "\n",
      "\n",
      " The binding energy in (MeV)= 18.95\n"
     ]
    }
   ],
   "source": [
    "#cal of binding energy\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.9, Page:41  \\n \\n\"\n",
    "# Given:\n",
    "D=-1.997042;\n",
    "n=1.0087;\n",
    "# Solution:\n",
    "m=(D+2.*n);\n",
    "E=m*931.;\n",
    "print\"\\n The binding energy in (MeV)=\",round(E,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.10;pg no:41"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.10, Page:41  \n",
      " \n",
      "\n",
      "\n",
      " The neutron seperation energy in (MeV)= 7.3596\n",
      "\n",
      " The proton seperation energy in (MeV)= 7.9647\n"
     ]
    }
   ],
   "source": [
    "#cal of seperation energy\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.10, Page:41  \\n \\n\"\n",
    "# Given:\n",
    "mH=1.007825;\n",
    "mn=1.008665;\n",
    "M1=207.97666;# mass of Pb 208\n",
    "M2=206.97590;# mass of Pb 207\n",
    "M3=206.97739;# mass of Tl 207\n",
    "\n",
    "# Solution:\n",
    "\n",
    "B1=((82*1.007825+126*1.008665)-207.97666)*931;# binding energy for Pb 208\n",
    "B2=((82*1.007825+125*1.008665)-206.97590)*931;# binding energy for Pb 207\n",
    "B3=((81*1.007825+126*1.008665)-206.97739)*931;# binding energy for Tl 207\n",
    "Sn=B1-B2;# neutron seperation energy\n",
    "Sp=B1-B3;# proton seperation energy\n",
    "\n",
    "print\"\\n The neutron seperation energy in (MeV)=\",round(Sn,4)\n",
    "print\"\\n The proton seperation energy in (MeV)=\",round(Sp,4)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.11;pg no:42"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.11, Page:42  \n",
      " \n",
      "\n",
      " The neutron seperation energy in (MeV)= 12.41\n",
      " The proton seperation energy in (MeV)= 8.79\n"
     ]
    }
   ],
   "source": [
    "#cal of seperation energy\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.11, Page:42  \\n \\n\"\n",
    "# Given:\n",
    "mH=1.007825;\n",
    "mn=1.008665;\n",
    "M1=22.98977;# mass of Na 23\n",
    "M2=21.994435;# mass of Na 22\n",
    "M3=21.991385;# mass of Ne 22\n",
    "# Solution:\n",
    "\n",
    "m1=((11.*1.007825+12.*1.008665)-M1);\n",
    "m2=((11.*1.007825+11.*1.008665)-M2);\n",
    "m3=((10.*1.007825+12.*1.008665)-M3);\n",
    "Sn=(m1-m2)*931.;# neutron seperation energy\n",
    "Sp=(m1-m3)*931.;# proton seperation energy\n",
    "\n",
    "print\" The neutron seperation energy in (MeV)=\",round(Sn,2)\n",
    "print\" The proton seperation energy in (MeV)=\",round(Sp,2)\n",
    "\n",
    "# Note: The answers are given in the form of atomic mass units where as in the question its asked for energies."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.12;pg no:43"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.12, Page:43  \n",
      " \n",
      "\n",
      "\n",
      " The excited level density in (MeV)= 894.29\n",
      "\n",
      " The level spacing in (keV)= 1.12\n",
      "\n",
      " The nuclear temperature in (MeV)= 2.0\n"
     ]
    }
   ],
   "source": [
    "#cal of excited level density,level spacing and nuclear temperature\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.12, Page:43  \\n \\n\"\n",
    "# Given:\n",
    "C=0.3;# in MeV^-1\n",
    "a=2.0;# in MeV\n",
    "E=8; # in MeV\n",
    "import math\n",
    "# Solution:\n",
    "d=C*(math.exp(2*((2*8)**(0.5))));# excited level density\n",
    "s=(1/d)*1000;# level spacing\n",
    "nT=(E/a)**(0.5);# nuclear temperature\n",
    "print\"\\n The excited level density in (MeV)=\",round(d,2)\n",
    "print\"\\n The level spacing in (keV)=\",round(s,2)\n",
    "print\"\\n The nuclear temperature in (MeV)=\",round(nT,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.13;pg no:43"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.13, Page:43  \n",
      " \n",
      "\n",
      "The expression for 1st, 2nd, and 3rd excited states are K times respectively. 5.0 12.0 21.0\n"
     ]
    }
   ],
   "source": [
    "#cal of expression for 1st, 2nd, and 3rd excited states\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.13, Page:43  \\n \\n\"\n",
    "# Given:\n",
    "I0=3./2.;# ground state of spin\n",
    "\n",
    "# Solution:\n",
    "I1=I0+1.;\n",
    "I2=I0+2.;\n",
    "I3=I0+3.;\n",
    "K=1.;# Assumed as some constant\n",
    "# Formula: E=(h^2/(2*I))*((I*(I+1))-I0*(I0+1))\n",
    "# Consider   K=(h^2/(2*I))=1\n",
    "\n",
    "E1=K*((I1*(I1+1.))-(I0*(I0+1.)));# For 1 excited state\n",
    "\n",
    "E2=K*((I2*(I2+1.))-(I0*(I0+1.)));# For 2 excited state\n",
    "\n",
    "E3=K*((I3*(I3+1.))-(I0*(I0+1.)));# For 3 excited state\n",
    "\n",
    "print\"The expression for 1st, 2nd, and 3rd excited states are K times respectively.\",E1,E2,E3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.14;pg no:44"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.14, Page:44  \n",
      " \n",
      "\n",
      "\n",
      " The energy of state 4 (+) in (keV)= 146.67\n",
      "\n",
      " The energy of state 6 (+) in (keV)= 308.0\n",
      "\n",
      " The energy of state 8 (+) in (keV)= 528.0\n",
      "\n",
      " The energy of state 10 (+) in (keV)= 806.67\n"
     ]
    }
   ],
   "source": [
    "#cal of energy of state 4,6,8,10\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.14, Page:44  \\n \\n\"\n",
    "# Given:\n",
    "E2=44;# in keV\n",
    "\n",
    "# Solution:\n",
    "E4=E2*((4.*5.)/(2.*3.));# for part (a)\n",
    "E6=E2*((6.*7.)/(2.*3.));# for part (b)\n",
    "E8=E2*((8.*9.)/(2.*3.));# for part (c)\n",
    "E10=E2*((10.*11.)/(2.*3.));# for part (d)\n",
    "\n",
    "print\"\\n The energy of state 4 (+) in (keV)=\",round(E4,2)\n",
    "print\"\\n The energy of state 6 (+) in (keV)=\",round(E6,2)\n",
    "print\"\\n The energy of state 8 (+) in (keV)=\",round(E8,2)\n",
    "print\"\\n The energy of state 10 (+) in (keV)=\",round(E10,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.15;pg no:44"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.15, Page:44  \n",
      " \n",
      "\n",
      "n= 11\n",
      "\n",
      " For the required level of energy 525 keV nearest even integer is = & spin is (+) 12.0\n"
     ]
    }
   ],
   "source": [
    "#cal of energy\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.15, Page:44  \\n \\n\"\n",
    "# Given:\n",
    "E2=44;# in keV\n",
    "En=525;# in keV\n",
    "\n",
    "# Solution:\n",
    "n=(En)/E2;\n",
    "# \n",
    "print\"n=\",n\n",
    "print\"\\n For the required level of energy 525 keV nearest even integer is = & spin is (+)\",round(n+1,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##example 3.16;pg no:45"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 3.16, Page:45  \n",
      " \n",
      "\n",
      "n1= 3.31818181818\n",
      "\n",
      " For the required level of energy 146 keV nearest even integer is =& spin is (+) 4.32\n",
      "\n",
      " \n",
      " n2= 6.90909090909\n",
      "\n",
      " For the required level of energy 304 keV nearest even integer is =& spin is (+) 6.91\n",
      "\n",
      " \n",
      " n3= 11.6818181818\n",
      "\n",
      " For the required level of energy 514 keV nearest even integer is =& spin is (+) 12.68\n"
     ]
    }
   ],
   "source": [
    "#cal of  energy\n",
    "#intiation of all variables\n",
    "# Chapter 3\n",
    "print\"Example 3.16, Page:45  \\n \\n\"\n",
    "# Given:\n",
    "E2=44.;# in keV\n",
    "En1=146.;# in keV\n",
    "En2=304.;# in keV\n",
    "En3=514.;# in keV\n",
    "# Solution:\n",
    "n1=(En1)/E2;\n",
    "n2=(En2)/E2;\n",
    "n3=(En3)/E2;\n",
    "print\"n1=\",n1\n",
    "print\"\\n For the required level of energy 146 keV nearest even integer is =& spin is (+)\",round(n1+1.,2)\n",
    "print\"\\n \\n n2=\",n2\n",
    "print\"\\n For the required level of energy 304 keV nearest even integer is =& spin is (+)\",round(n2,2)\n",
    "print\"\\n \\n n3=\",n3\n",
    "print\"\\n For the required level of energy 514 keV nearest even integer is =& spin is (+)\",round(n3+1.,2)\n",
    "\n",
    "#Note: In the last part (c) the answer given in the textbook is 8(+). But the correct answer is 12(+)"
   ]
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