Chemistry 3730 assignment 9 solutions

1. The particle-in-a-box wavefunctions are
   

   > psi := (n,L,x) -> sqrt(2/L)*sin(n*Pi*x/L);

   

   
   

   The perturbing potential is of the form
   

   > V := x -> A*sin(2*Pi*x/Lambda);

   

   
   

   We will need to tell Maple that n is an integer:
   

   > assume(n,integer);

   
   

   We now compute the energy shift caused by the perturbation:
   

   > deltaE := int(psi(n,L,x)*V(x)*psi(n,L,x),x=0..L);

   

   > simplify(deltaE);

   

   
   

2. The 1s wavefunction is
   

   > psi1s := (r,theta,phi) -> sqrt(1/(Pi*a0^3))*exp(-r/a0);

   

   
   

   We will treat the term as a perturbation:
   

   V := r -> -e^2/(4*Pi*epsilon0)*k/r^2;

   

   
   

   The energy correction for the 1s state is calculated by
   

   > assume(a0>0);
   > deltaE:=int(int(int(
     psi1s(r,theta,phi)*V(r)*psi1s(r,theta,phi)*r^2*sin(theta),
     r=0..infinity),theta=0..Pi),phi=0..2*Pi);

   

   
   

   The ionization energy is just the energy of the ground state (give or take the sign). The energy of the 1s state is . This can be rewritten in terms of :
   

   > E1s := -hbar^2/(2*a0^2*mu);

   

   
   

   The ratio of the perturbation to the ground state energy is
   

   > ratio := deltaE/E1s;

   

   
   

   We want this ratio to be less than . This requires k to be less than . Since is approximately equal to the mass of the electron, we use this value. The rest of the constants are obtained from tables. We get . One way to think of this number is that the deviation from the Coulomb law, if there is one, wouldn't become apparent until two particles were within of each other. This distance is very small on an atomic scale so the deviation wouldn't have much effect on chemical properties.