{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 6: Schroedinger Wave Equation"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example number 1, Page number 211"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "least energy is 37.65 eV\n",
      "answer given in the book is wrong\n"
     ]
    }
   ],
   "source": [
    "#importing modules\n",
    "import math\n",
    "from __future__ import division\n",
    "\n",
    "#Variable declaration    \n",
    "h=6.63*10**-34;        #plancks constant(J sec)\n",
    "m=9.11*10**-31;        #mass(kg)\n",
    "a=10**-10;             #width of box(m)\n",
    "e=1.602*10**-19;       #charge(coulomb)\n",
    "\n",
    "#Calculations\n",
    "E1=(h**2)/(8*m*e*a**2);    #least energy(eV)\n",
    "\n",
    "#Result\n",
    "print \"least energy is\",round(E1,2),\"eV\"\n",
    "print \"answer given in the book is wrong\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example number 2, Page number 211"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "probability of finding the particle is 0.4\n"
     ]
    }
   ],
   "source": [
    "#importing modules\n",
    "import math\n",
    "from __future__ import division\n",
    "\n",
    "#Variable declaration    \n",
    "delta_x=5*10**-10;     #interval(m)\n",
    "a=25*10**-10;          #width(m)\n",
    "\n",
    "#Calculations\n",
    "P=2*delta_x/a;         #probability of finding the particle\n",
    "\n",
    "#Result\n",
    "print \"probability of finding the particle is\",P"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example number 3, Page number 212"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "probability of finding the particle is 19.84 %\n"
     ]
    }
   ],
   "source": [
    "#importing modules\n",
    "import math\n",
    "from __future__ import division\n",
    "from scipy.integrate import quad\n",
    "\n",
    "#Variable declaration    \n",
    "a=1;        #assume\n",
    "\n",
    "#Calculations\n",
    "def zintg(x):\n",
    "    return (2/a)*(1/2)*(1-math.cos(2*math.pi*x/a))\n",
    "\n",
    "P1=quad(zintg,0.45,0.55)[0]\n",
    "\n",
    "#Result\n",
    "print \"probability of finding the particle is\",round(P1*100,2),\"%\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example number 4, Page number 213"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " least energy is 6 eV\n",
      "energy in 2nd excited state is 24 eV\n",
      "energy in 3rd excited state is 54 eV\n",
      "difference of energy between 2nd and 1st excited states is 18 eV\n"
     ]
    }
   ],
   "source": [
    "#importing modules\n",
    "import math\n",
    "from __future__ import division\n",
    "\n",
    "#Variable declaration    \n",
    "h=6.63*10**-34;      #planks constant(Js)\n",
    "m=9.1*10**-31;       #mass(kg)\n",
    "a=2.5*10**-10;       #width(m)\n",
    "e=1.6*10**-19;       #charge(coulomb)\n",
    "n1=1;                      \n",
    "n2=2;\n",
    "n3=3;                #energy states\n",
    "\n",
    "#Calculations\n",
    "E1=n1**2*(h**2)/(8*m*e*a**2);    #least energy(eV)\n",
    "E2=n2**2*E1;                     #energy in 2nd excited state(eV)\n",
    "E3=n3**2*E1;                     #energy in 3rd excited state(eV)\n",
    "delta_E=E2-E1;                   #difference of energy between 2nd and 1st excited states(eV)\n",
    "\n",
    "#Result\n",
    "print \"least energy is\",int(E1),\"eV\"\n",
    "print \"energy in 2nd excited state is\",int(E2),\"eV\"\n",
    "print \"energy in 3rd excited state is\",int(E3),\"eV\"\n",
    "print \"difference of energy between 2nd and 1st excited states is\",int(delta_E),\"eV\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example number 5, Page number 213"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "de-Broglie wavelength of first three energy states are 20 angstrom 10 angstrom 6.67 angstrom\n",
      "energies of first three energy states are 0.38 eV 1.5095 eV 3.396 eV\n",
      "answer in the book varies due to rounding off errors\n"
     ]
    }
   ],
   "source": [
    "#importing modules\n",
    "import math\n",
    "from __future__ import division\n",
    "\n",
    "#Variable declaration    \n",
    "h=6.63*10**-34;      #planks constant(Js)\n",
    "m=9.1*10**-31;       #mass(kg)\n",
    "a=10;                #width(angstrom)\n",
    "e=1.6*10**-19;       #charge(coulomb)\n",
    "n1=1;                      \n",
    "n2=2;\n",
    "n3=3;                #energy states\n",
    "\n",
    "#Calculations\n",
    "lamda1=2*a/n1;       #de-Broglie wavelength of first energy state(angstrom)\n",
    "lamda2=2*a/n2;       #de-Broglie wavelength of second energy state(angstrom)\n",
    "lamda3=2*a/n3;       #de-Broglie wavelength of third energy state(angstrom)\n",
    "E1=n1**2*(h**2)/(8*m*e*(a*10**-10)**2);    #energy in 1st excited state(eV)\n",
    "E2=n2**2*E1;                     #energy in 2nd excited state(eV)\n",
    "E3=n3**2*E1;                     #energy in 3rd excited state(eV)\n",
    "\n",
    "#Result\n",
    "print \"de-Broglie wavelength of first three energy states are\",int(lamda1),\"angstrom\",int(lamda2),\"angstrom\",round(lamda3,2),\"angstrom\"\n",
    "print \"energies of first three energy states are\",round(E1,2),\"eV\",round(E2,4),\"eV\",round(E3,3),\"eV\"\n",
    "print \"answer in the book varies due to rounding off errors\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example number 6, Page number 214"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "energy in 1st excited state is 14.7 *10**-19 J\n",
      "mass is 9.3 *10**-31 kg\n",
      "quantum state is 10.4\n",
      "as n is not an integer, En is not permitted value of energy\n"
     ]
    }
   ],
   "source": [
    "#importing modules\n",
    "import math\n",
    "from __future__ import division\n",
    "\n",
    "#Variable declaration    \n",
    "h=6.63*10**-34;      #planks constant(Js)\n",
    "a=0.2*10**-9;        #width(m)\n",
    "e=1.602*10**-19;     #charge(coulomb)\n",
    "n5=5;                #energy state\n",
    "E5=230*e;            #energy 0f 5th state(J)\n",
    "En=10**3*e;          #energy(eV)\n",
    "\n",
    "#Calculations\n",
    "E1=E5/n5**2;         #energy in 1st excited state(eV)\n",
    "m=h**2/(8*E1*a**2);    #mass(kg)\n",
    "n=math.sqrt(En/E1);    #quantum state\n",
    "\n",
    "#Result\n",
    "print \"energy in 1st excited state is\",round(E1*10**19,1),\"*10**-19 J\"\n",
    "print \"mass is\",round(m*10**31,1),\"*10**-31 kg\"\n",
    "print \"quantum state is\",round(n,1)\n",
    "print \"as n is not an integer, En is not permitted value of energy\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example number 7, Page number 225"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 48,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "reflection coefficient in 1st case is 0.05\n",
      "answer given in the book is wrong\n",
      "transmission coefficient in 1st case is 0.95\n",
      "for E=0.025, E<V. so transmission coefficient is 0 and reflection coefficient is 1\n",
      "reflection coefficient in 3rd case is 1\n",
      "transmission coefficient in 3rd case is 0\n"
     ]
    }
   ],
   "source": [
    "#importing modules\n",
    "import math\n",
    "from __future__ import division\n",
    "\n",
    "#Variable declaration \n",
    "e=1.6*10**-19;     #charge(coulomb)\n",
    "E1=0.04*e;         #energy(J)\n",
    "V=0.03*e;          #energy barrier(J) \n",
    "E2=0.025*e;        #energy(J)\n",
    "E3=0.03*e;         #energy(J)\n",
    "m=1;               #assume \n",
    "k1=1;              #assume\n",
    "\n",
    "#Calculations\n",
    "x=math.sqrt(E1-V);\n",
    "y=math.sqrt(E1+V);\n",
    "R1=((math.sqrt(E1)-x)/(math.sqrt(E1)+y))**2;     #reflection coefficient\n",
    "T1=1-R;                                          #transmission coefficient\n",
    "k2=math.sqrt(2*m*(E3-V));                       \n",
    "R2=((k1-k2)/(k1+k2))**2;                         #reflection coefficient\n",
    "T2=4*k1*k2/(k1+k2)**2;                           #transmission coefficient \n",
    "\n",
    "#Result\n",
    "print \"reflection coefficient in 1st case is\",round(R1,2)\n",
    "print \"answer given in the book is wrong\"\n",
    "print \"transmission coefficient in 1st case is\",round(T1,2)\n",
    "print \"for E=0.025, E<V. so transmission coefficient is 0 and reflection coefficient is 1\"\n",
    "print \"reflection coefficient in 3rd case is\",int(R2)\n",
    "print \"transmission coefficient in 3rd case is\",int(T2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example number 8, Page number 226"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 54,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "value of E/V is 1.03\n"
     ]
    }
   ],
   "source": [
    "#importing modules\n",
    "import math\n",
    "from __future__ import division\n",
    "\n",
    "#Variable declaration    \n",
    "T=0.5;            #transmission coefficient \n",
    "a=1;\n",
    "b=6;\n",
    "c=1;\n",
    "\n",
    "#Calculations\n",
    "k1byk2=(b+math.sqrt((b**2)-(4*a*c)))/(2*a);  \n",
    "x=k1byk2**2;\n",
    "EbyV=x/(x-1);        #value of E/V\n",
    "\n",
    "#Result\n",
    "print \"value of E/V is\",round(EbyV,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example number 9, Page number 226"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 40,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "transmission probabilities in 1st case are 4.5 *10**-5 and 1.4 *10**-4\n",
      "transmission probabilities in 2nd case are 2.1 *10**-9 and 1.25 *10**-9\n",
      "answer for transmission probability in 2nd case given in the book is wrong\n"
     ]
    }
   ],
   "source": [
    "#importing modules\n",
    "import math\n",
    "from __future__ import division\n",
    "\n",
    "#Variable declaration    \n",
    "m=9.1*10**-31;       #mass(kg)\n",
    "a1=5*10**-10;         #width(m)\n",
    "a2=10*10**-10;         #width(m)\n",
    "e=1.6*10**-19;       #charge(coulomb)\n",
    "V0=5;                #energy barrier(eV)\n",
    "E1=1;                #energy of electron(eV)\n",
    "E2=2;\n",
    "chi=1.054*10**-34;   #plancks constant(Js)\n",
    "\n",
    "#Calculations\n",
    "beta1=math.sqrt(2*m*(V0-E1)*e/chi**2);      #value of beta(m-1)\n",
    "x1=int(-2*a1*beta1);\n",
    "beta2=math.sqrt(2*m*(V0-E2)*e/chi**2);      #value of beta(m-1)\n",
    "x2=round(-2*a1*beta2,1);\n",
    "T1=math.exp(x1);               #transmission probability in 1st case\n",
    "T2=math.exp(x2);               #transmission probability in 1st case\n",
    "x3=int(-2*a2*beta1);\n",
    "x4=round(-2*a2*beta1,1);\n",
    "T1dash=math.exp(x3);               #transmission probability in 2nd case\n",
    "T2dash=math.exp(x4);               #transmission probability in 1st case\n",
    "\n",
    "#Result\n",
    "print \"transmission probabilities in 1st case are\",round(T1*10**5,1),\"*10**-5 and\",round(T2*10**4,1),\"*10**-4\"\n",
    "print \"transmission probabilities in 2nd case are\",round(T1dash*10**9,1),\"*10**-9 and\",round(T2dash*10**9,2),\"*10**-9\"\n",
    "print \"answer for transmission probability in 2nd case given in the book is wrong\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example number 10, Page number 227"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 59,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "transmission coefficient is 4 math.exp ( -1.625 )\n"
     ]
    }
   ],
   "source": [
    "#importing modules\n",
    "import math\n",
    "from __future__ import division\n",
    "\n",
    "#Variable declaration    \n",
    "m=9.1*10**-31;       #mass(kg)\n",
    "a=10**-10;           #width(m)\n",
    "e=1.6*10**-19;       #charge(coulomb)\n",
    "V0=5;                #energy barrier(eV)\n",
    "E=2.5;               #energy of electron(eV)\n",
    "chi=1.05*10**-34;    #plancks constant(Js)\n",
    "\n",
    "#Calculations\n",
    "x=16*E*(V0-E)/V0**2;\n",
    "y=-2*a*math.sqrt(2*m*(V0-E)*e/chi**2);\n",
    "#T=x*math.exp(y);     #transmission coefficient\n",
    "\n",
    "#Result\n",
    "print \"transmission coefficient is\",int(x),\"math.exp (\",round(y,3),\")\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example number 11, Page number 227"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 76,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "average lifetime of nucleus is 3.7 *10**17 years\n",
      "answer given in the book is wrong\n"
     ]
    }
   ],
   "source": [
    "#importing modules\n",
    "import math\n",
    "from __future__ import division\n",
    "\n",
    "#Variable declaration    \n",
    "P=10**21;        #probability(T per sec)\n",
    "m=4*1.6*10**-27;       #mass(kg)\n",
    "a=2*10**-14;           #width(m)\n",
    "e=1.67*10**-19;        #charge(coulomb)\n",
    "V0=30;                 #energy barrier(eV)\n",
    "E=4.2;                 #energy of electron(eV)\n",
    "chi=1.05*10**-34;      #plancks constant(Js)\n",
    "\n",
    "#Calculations\n",
    "x=P*16*E*(V0-E)/V0**2;\n",
    "y=-2*a*math.sqrt(2*m*(V0-E)*10**6*e/chi**2);\n",
    "T=x*math.exp(y);     #transmission coefficient\n",
    "tow=1/T;             #average lifetime of nucleus(years) \n",
    "\n",
    "#Result\n",
    "print \"average lifetime of nucleus is\",round(tow/10**17,1),\"*10**17 years\"\n",
    "print \"answer given in the book is wrong\""
   ]
  }
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