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+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 5: The Schrodinger Equation"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.1: Displacement_and_velocity_of_the_object.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear \n",
+"clc\n",
+"disp('Exa-5.1');//The solution involves very complex symbolic equation solving and approximations. Hence only answers are displayed\n",
+"printf('The displacement and velocity of the ball are found out in 2 steps\n step1-before reaching the surface of water and \n step2-Inside water till it rises back to surface\n');\n",
+"printf('The values are as follows: v1(t)=-g*t and y1(t)=H-((g/2)*t^2))\n');\n",
+"printf('In region 2: v2(t)=(-B/m*sqrt(2*H/g))+(B/m-g)*t; y2(t)= H+ HB/mg -B/m*sqrt(2*H/g)+ (B/m-g)');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.2: Solution_for_a_and_b.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear    \n",
+"clc\n",
+"disp('Exa-5.2(a)');\n",
+"h=1.05*10^-34;m=9.11*10^-31;L=10^-10;      // all the values are taken in SI units\n",
+"E1=h^2*%pi^2/(2*m*L^2); E2=4*E1;            //Energies are calculated\n",
+"delE=(E2-E1)/(1.6*10^-19);                  //Difference in energy is converted to eV\n",
+"printf('Energy to be supplied is %.0f eV.\n',delE);\n",
+"disp('Exa-5.2(b)');\n",
+"x1=0.09*10^-10;x2=0.11*10^-10                 //limits of the given region\n",
+"probGnd=(2/L)*integrate('(sin(%pi*x/L)^2)','x',x1,x2);\n",
+"printf('The percentage probablility of finding an electron in the ground state is %.2f.\n',probGnd*100);\n",
+"disp('Exa-5.2(c)');\n",
+"x1=0,x2=0.25*10^-10;\n",
+"probExc=(2/L)*integrate('(sin(2*%pi*x/L)^2)','x',x1,x2);\n",
+"printf('The probablility of finding an electron in the excited state is %.2f.\n',probExc);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.3: Proof_for_average_value_of_x.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear    \n",
+"clc\n",
+"disp('Ex-5.3');\n",
+"x1=0;x2=L;\n",
+"xavg=(2/L)*integrate('sin(%pi*x/L)^2','x',x1,x2);\n",
+"printf('The average value of x is found out to be L/2 which apparently is independent of Qunatum state.');"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}