Question

In nacl molecule, the interaction potential between na ion and cl ion fits the following equation:

V(r)=-e²/r+ α.e^-r/q

^{in this equation, the first term hammer is the coulomb potential,the second term results from the distribution of other electrons in the system.}

^{The equilibrium position between the two ions is where the potential is minimum:}

f(r)=-dV/dr=e²/r²- α/q . e^-r/q  =0   the root of this equation gives the r₀ bond length.experimental for NaCl  α=1.09*10³ eV,q=0.33 A and e² =14.4 A eV

Write the c ++ program that calculates the bond length in A using the constant (using NEWTON-RAPHSON-METHOD) please could you give detail about solution?

Answer 1

`Hey,

Note: Brother if you have any queries related the answer please do comment. I would be very happy to resolve all your queries.

The function is 14.4/(x*x)-1090.0*exp(-x/0.33)/0.33;

So, df/dx=14.4*(-3)/x^3+1090/0.33*exp(-x/0.33)/0.33=-43.2/x^3+10900000*exp(-x/0.33)

which can be further simplified to

(10900000.0*exp(-(100.0*x)/33.0))/1089.0 - 144.0/(5.0*x*x*x);

#include<iostream>
#include<cmath>
#define EPSILON 0.001
using namespace std;
  
// An example function whose solution is determined using
double func(double x)
{
return 14.4/(x*x)-1090.0*exp(-x/0.33)/0.33;
}
  
double derivFunc(double x)
{
return (10900000.0*exp(-(100.0*x)/33.0))/1089.0 - 144.0/(5.0*x*x*x);
}
  
// Function to find the root
void newtonRaphson(double x)
{
double h = func(x) / derivFunc(x);
while (abs(h) >= EPSILON)
{
h = func(x)/derivFunc(x);

// x(i+1) = x(i) - f(x) / f'(x)   
x = x - h;
}
  
cout << "The bond length r0 : " << x<<"A"<<endl;
}
  
// Driver program to test above
int main()
{
double x0 = 1; // Initial values assumed
newtonRaphson(x0);
return 0;
}

Kindly revert for any queries

Thanks.