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+//chapter 1
+//example 1.4
+//calculate bond energy for NaCl
+//page 15-16
+clear;
+clc;
+//given
+r0=0.236; //in nanometer(interionic distance)
+e=1.6E-19; // in C (charge of electron)
+E_o= 8.85E-12;// absolute premittivity
+N=8; // Born constant
+IE=5.14;// in eV (ionisation energy of sodium)
+EA=3.65;// in eV (electron affinity of Chlorine)
+pi=3.14; // value of pi used in the solution
+//calculate
+r0=r0*1E-9; // since r is in nanometer
+PE=(e^2/(4*pi*E_o*r0))*(1-1/N); // calculate potential energy
+PE=PE/e; //changing unit from J to eV
+printf('\nThe potential energy is\tPE=%.2f eV',PE); 
+NE=IE-EA;// calculation of Net energy
+printf('\nThe net energy is\tNE=%.2f eV',NE);
+BE=PE-NE;// calculation of Bond Energy
+printf('\nThe bond energy is\tBE=%.2f eV',BE);
+// Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer.
+//       (2) There is slight variation in the answer due to round off.