Schroedinger Equation Problems (Serway et al. p.224)

2. (a) (psi)(psi)* = (A^2) [cos(2 pi x / L)]^2 = (probability density)

total integral[(psi)(psi)*]dx = (A^2) integral[cos^2(2 pi x / L)] dx

=1 if the integral is over all x values, corresponding to 1 particle present.

In this case, only x-values between –L/4 and +L/4 contribute to the integral,

and they correspond to (2 pi x / L) in the range from -pi/2 to +pi/2 .

This is a complete cycle of the cos^2 function, whose average value is 0.5

and whose integral is the area of a rectangle of height = mean value,

in other words: (mean value)(range of x) = (1/2)[L/4 – (-L/4)] = L/4

Therefore, (A^2) (L/4) = 1 , giving A = (1/2)L^(1/2)

3. psi is of form A sin(kx) where k = 5E10 = (2 pi)/lambda

1. lambda = (2 pi)/(5E10) = 1.26E-10 m
2. momentum p = h/lambda = 5.27E-24 kg m/s
3. Since electron is free, energy = kinetic energy = p^2 / 2m

= (5.27E-24)^2 / [(2)(9.11E-31)] = 1.53E-17 J = 95 eV

12. hf = hc/(lambda) = (6.63E-34)(3.0E8)/(694.3E-9) = 2.86E-19 J

= E2 – E1 = {[(pi hbar / L)^2] / 2m}{2^2 – 1^2} for an infinite-potential square well

Therefore: [(pi hbar / L)^2]/2m = 2.86E-19 / {4 – 1} = 9.54E-20

(pi hbar / L) = [3.09E-10][(2)(9.11E-31)]^(1/2) = 4.17E-25

L = (pi hbar)/ 4.17E-25 = (h/2)/ 4.17E-25 = 7.95E-10 m