Assignment 10 solutions

1. Define the integrals:

   > HAA := R ->
   > e^2/(4*Pi*epsilon0*a0)*(exp(-2*R/a0)*(1+a0/R)-a0/R-1/2);

   

   > HAB := R ->
   > -e^2/(4*Pi*epsilon0*a0)*(SAB(R)/2+exp(-R/a0)*(1+R/a0));

   

   > SAB := R -> exp(-R/a0)*(1+R/a0+(R/a0)^2/3);

   

   > Ep := R -> (HAA(R)+HAB(R))/(1+SAB(R));

   

   > Em := R -> (HAA(R)-HAB(R))/(1-SAB(R));

   

   > assume(a0>0);
   > limit(Ep(R),R=infinity);

   

   > limit(Em(R),R=infinity);

   

   This is exactly the ground-state energy of a hydrogen atom. This makes sense because as , a hydrogen molecular ion should separate into a hydrogen atom and a proton.

2. > Veffp := R -> Ep(R) + e^2/(4*Pi*epsilon0*R);

   

   > Veffm := R -> Em(R) + e^2/(4*Pi*epsilon0*R);

   

   > e := 1.6022e-19;

   

   > epsilon0 := 8.8542e-12;

   

   > a0 := 5.292e-11;

   

   > plot({Veffp,Veffm},0..7*a0,-3e-18..1e-18);

   

   To figure out which curve is which:

   > evalf(Veffp(1e-10));

   

   > evalf(Veffm(1e-10));

   

   The bonding state is represented by the + effective potential energy curve (Veffp).

3. To find a minimum in , set the derivative with respect to R equal to zero and solve:

   > Req:=fsolve(diff(Veffp(R),R)=0,R,5e-11..1.5e-10);

   

4. > k := evalf(subs(R=Req,1/2*diff(Veffp(R),R$2)));

   

   > mH := 1.0078e-3/6.0221e23;

   

   > mu := mH/2;

   

   > omega0 := sqrt(k/mu);

   

   > hbar := 6.6261e-34/(2*Pi);

   

   > E0 := evalf(1/2*hbar*omega0);

   

5. > evalf(limit(Veffp(R),R=infinity)-Veffp(Req)+E0);