By largely qualitative arguments. Exercise we saw earlier shows that a finite well can hold an $n$th state only if its depth $U_0$ is at least $h^2(n-1)^2 / 8 m L^2$. Show that this result also follows from equation we saw earlier and the accompanying Figure we saw earlier.

Explain to your friend, who is skeptical about energy quantization, the simple evidence provided by distinct colors you sec when you hold a CD (serving as a diffraction grating) near a fluorescent light. It may be helpful to contrast this evidence with the spectrum produced by an incandescent light, which relies on heating a filament to produce a rather nonspecific blackbody spectrum.

A finite well always has at least one bound state. Why does the argument of Exercise we saw earlier fail in the case of a finite well?

Advance an argument based on $p = h/\lambda$ that there is no bound state in a half-infinite well unless $U_0$ is at least $h^2/32 mL^2.$

The deeper the finite well, the more states it holds. In fact, a new state, the nth, is added when the well's depth $U_0$ reaches $h^{2}(n-1)^{2} / 8 m L^{2}.$ (a) Argue that this should be the case based only on k = the shape of the wave inside, and the degree of penetration of the classically forbidden region expected for a state whose energy E is only negligibly below $U_0,$ (b) How many states would be found up to this same "height" in an infinite well.

Whereas an infinite well has an infinite number of bound states, a finite well does not. By relating the well height $U_0$ to the kinetic energy and the kinetic energy, $(through \lambda)$ to n and L, show that the number of bound states is given roughly by ($\sqrt{8 m L^{2} U_{0} / h^{2}}$ Assume that the number is large.)

A 50 eV electron is trapped between electrostatic walls 200 eV high. How far does its wave function extend beyond the walls?

Verify that $A \sin (k x)+B \cos (k x)$ is a solution of equation we saw earlier.

A finite potential energy function U(x) allows $\psi(x),$ the solution of the time-independent Schrödinger equation, to penetrate the classically forbidden region. Without assuming any particular function for U(x), show that $\psi(x)$ must have an inflection point at any value of x where it enters a classically forbidden region.

Verify that solution we saw earlier satisfies the Schrodinger equation in form we saw earlier.

A particle is bound by a potential energy of the form

$$U(x)=\left\{\begin{array}{cl}{0} & {|x|<\frac{1}{2} a} \\ {\infty} & {|x|>\frac{1}{2} a}\end{array}\right.$$

This differs from the infinite well in being symmetric about x = 0, which implies that the probability densities must also be symmetric. Noting that either sine or cosine would fit this requirement and could be made continuous with the zero wave function outside the well, determine the allowed energies and corresponding normalized wave functions for this potential well.

The term interaction is sometimes used interchangeably with force, and other times interchangeably with potential energy. Although force and potential energy certainly aren't the same thing, what justification is there for using the same term to cover both?

A tiny $1 \mu g$ particle is in a 1 cm wide enclosure and takes a year to bounce from one-end to the other and back, (a) How many nodes are there in its enclosure? (b) How would your answer change if the particle were more massive or moving faster?

What is the probability that a particle in the first excited (n=2) state of an infinite well would be found in the middle third of the well? How does this compare with the classical expectation? Why?