Added(A)/Deleted(D) following books

 A  Principles_of_Physics_by_F.J.Bueche/Chapter10_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter10_2.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter11_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter13_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter14_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter15_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter16_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter17_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter18_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter19_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter1_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter1_2.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter20_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter21_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter22_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter23_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter24_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter25_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter26_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter27_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter2_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter2_2.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter3_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter3_2.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter4_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter4_2.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter5_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter5_2.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter6_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter6_2.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter7_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter7_2.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter8_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter9_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/Chapter9_2.ipynb
A  Principles_of_Physics_by_F.J.Bueche/chapter12_1.ipynb
A  Principles_of_Physics_by_F.J.Bueche/screenshots/11.2_1.png
A  Principles_of_Physics_by_F.J.Bueche/screenshots/24.4_1.png
A  Principles_of_Physics_by_F.J.Bueche/screenshots/8.4_1.png

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diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter26_1.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter26_1.ipynb
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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 26: Energy levels and Spectra"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex26.1:pg-1208"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The ionization energy of the hydrogen atom is E= 13.6  eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 26_1\n",
+    "import math \n",
+    "  \n",
+    "#To find the ionization energy of the hydrogen atom\n",
+    "e=13.6        #units in eV\n",
+    "print \"The ionization energy of the hydrogen atom is E=\",round(e,1),\" eV\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex26.2:pg-1209"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The wavelength of fourth line in Paschen series is= 1002.0  nm\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 26_2\n",
+    "import math  \n",
+    "  #To find the wavelength of fourth line in Paschen series\n",
+    "n1=3     #Units in constant\n",
+    "n2=7     #Units in constant\n",
+    "r=1.0974*10**7         #units in meter**-1\n",
+    "lamda=round((1/r)*((n1**2*n2**2)/(n2**2-n1**2))*10**9)       #Units in nm\n",
+    "print \"The wavelength of fourth line in Paschen series is=\",round(lamda),\" nm\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex26.3:pg-1209"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The energy associated with line 1 is E1= -54.4  eV\n",
+      "The energy associated with line 2 is E2= -13.6  eV\n",
+      "The energy associated with line 3 is E3= -6.04  eV\n",
+      "\n",
+      "The first line of balmer series is lamda= 163.0  nm and belongs to the ultraviolet region\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 26_3\n",
+    "import math \n",
+    "  \n",
+    "  #To draw the energy level diagram and the find the first line of balmer type series\n",
+    "n=1\n",
+    "e1=-54.4/n**2      #units in ev\n",
+    "n=2\n",
+    "e2=-54.4/n**2      #units in ev\n",
+    "n=3\n",
+    "e3=-54.4/n**2      #units in ev\n",
+    "print \"The energy associated with line 1 is E1=\",round(e1,1),\" eV\\nThe energy associated with line 2 is E2=\",round(e2,1),\" eV\\nThe energy associated with line 3 is E3=\",round(e3,2),\" eV\\n\"\n",
+    "e1=1       #units in eV\n",
+    "e2=7.6       #Units in eV\n",
+    "lamda1=1240         #units in nm\n",
+    "lamda=(e1/e2)*lamda1        #Units in nm\n",
+    "print \"The first line of balmer series is lamda=\",round(lamda),\" nm and belongs to the ultraviolet region\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex26.4:pg-1209"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The longest wavelength of light capable of ionizing hydrogen atom is lamda= 91.2  nm\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 26_4\n",
+    "import math \n",
+    "  \n",
+    "  #To find the longest wavelength of light capable of ionizing hydrogen atom\n",
+    "  #First method\n",
+    "R=1.097*10**7          #Units in meter**-1\n",
+    "lamda=(1/R)*10**9                 #Units in meters\n",
+    "  #Second method\n",
+    "E=13.6        #units in eV\n",
+    "e1=1        #units in eV\n",
+    "lamda3=1240.0         #Units in eV\n",
+    "lamda2=(e1/E)*(lamda3)        #Units in nm\n",
+    "print \"The longest wavelength of light capable of ionizing hydrogen atom is lamda=\",round(lamda2,1),\" nm\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex26.5:pg-1210"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The energy difference between n=1 and n=2 level is E= 17713.0  eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 26_5\n",
+    "import math \n",
+    "  \n",
+    "  #To find  the energy difference between the n is 1 and n is 2 level\n",
+    "e1=1        #Units in eV\n",
+    "lamda2=1240        #Units in eV\n",
+    "lamda3=0.07       #Units in eV\n",
+    "e2=lamda2/lamda3        #Units in eV\n",
+    "e=e2-e1            #Units in eV\n",
+    "print \"The energy difference between n=1 and n=2 level is E=\",round(e),\" eV\"\n",
+    "  #In textbook answer is prinred wrong as E=18000 eV the correct answer is E=17713 eV \n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.11"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}