Chemistry 3730 Fall 1998 Test 1 Solutions

1. See the lab manual for explanations of these calculations.

   

2. must be at least as large as m. Therefore so that the minimum value of is .
3. Since the box is one-dimensional, this is a problem in two coordinates, and :

   

   The solutions must satisfy

   

4. We must first define the wavefunction:

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

   

   For n=1:

   > int(psi(1,x)^2,x=3*L/8..5*L/8);

   

   > evalf(%);

   

   For n=2:

   > int(psi(2,x)^2,x=3*L/8..5*L/8);

   

   > evalf(%);

   

   The probability is much greater for n=1 than for n=2 because in the latter case the range of integration includes a wavefunction node.

5. I have already defined the wavefunction and I did all the calculations in the same worksheet so I didn't have to do that again.

   To calculate :

   > int(psi(1,x)*(x-L/2)*psi(1,x),x=0..L);

   

   :

   > int(psi(1,x)*(x-L/2)^2*psi(1,x),x=0..L);

   

   :

   > int(psi(1,x)*(x-L/2)^3*psi(1,x),x=0..L);

   

   Note that the odd moments ( with q odd) are all zero. This is a general property of distributions which are symmetric about their mean.

6. The kinetic energy is , the particle-in-a-box energy:

   

   The transition energy from level n to n+1 is

   

   This is enormously smaller than the kinetic energy so quantum effects (like the discreteness of the energy levels) should not be detectable.

7. Define the (approximate) wavefunction:

   > phi:=x->A*x*(L-x);

   

   The wavefunction is normalized if the integral of the probability density is 1:

   > N := int(phi(x)^2,x=0..L);

   

   > solve(N=1,A);

   

   Either one of these is an acceptable normalization factor. The sign is not physically significant.