Chemistry 3730 assignment 8 solutions

1. We need to first enter the wavefunctions:
   

   > psi2s := (r,theta,phi) ->
     1/8*sqrt(2/(Pi*a0^3))*(2-r/a0)*exp(-r/(2*a0));

   

   > psi3s := (r,theta,phi) ->
     1/243*sqrt(3/(Pi*a0^3))*(27-18*r/a0+2*(r/a0)^2)
     *exp(-r/(3*a0));

   

   
   

   Maple will need to know the sign of a0 to solve the problems in this assignment.
   

   > assume(a0>0);

   
   

   In the 2s state,
   

   > int(int(int(
     evalc(conjugate(psi2s(r,theta,phi))*r*psi2s(r,theta,phi))
     *r^2*sin(theta),theta=0..Pi),phi=0..2*Pi),r=0..infinity);

   

   
   

   Note that the use of evalc() and conjugate() are not strictly required in this and the following case since the wavefunctions are real.

   In the 3s state,
   

   > int(int(int(
     evalc(conjugate(psi3s(r,theta,phi))*r*psi3s(r,theta,phi))
     *r^2*sin(theta),theta=0..Pi),phi=0..2*Pi),r=0..infinity);

   

   
   

   As discussed in introductory chemistry courses, the shells get bigger (the electrons spend more time away from the nucleus) as the principal quantum number increases. The increase in average distance from the nucleus ( ) reflects this.

2. 1. Let's start with the 3s radial probability density (r.p.d.):
      

      > rpd3s := r -> N3s*r^2*(27-18*r/a0+2*(r/a0)^2)^2
        *exp(-2*r/(3*a0));

      

      
      

      Here, N3s is a normalization factor which we must determine:
      

      > solve(int(rpd3s(r),r=0..infinity)=1,N3s);

      

      > N3s:=";

      

      
      

      Now, for the state ( has an identical r.p.d.):
      

      > rpd3d2 := r -> N3d2*r^2*(r/a0)^4*exp(-2*r/(3*a0));

      

      > solve(int(rpd3d2(r),r=0..infinity)=1,N3d2);

      

      > N3d2:=";

      

      
      

   2. For the 3s state:
      

      > plot(subs(a0=1,rpd3s(r)),r=0..40);

      

      
      

      Note that I chose the upper limit of the plot command large enough that I could see the whole function.

      For the state:
      

      > plot(subs(a0=1,rpd3d2(r)),r=0..40);

      

      
      

   3. We want to find the places where the 3s r.p.d. is zero:
      

      > solve(rpd3s(r)=0,r);

      

      

      
      

      The radial nodes are
      

      > rn1:="[3];

      

      > rn2 := ""[4];

      

      > rn0:=0;

      

      
      

   4. rn1 is the last node.
      

      > int(rpd3s(r),r=rn1..infinity);

      

      > evalf(");