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?

Where would a particle in the first excited state (first above ground) of an infinite well mostly likely be found?

An electron in the n = 4 state of a 5 nm wide infinite well makes a transition to the ground state, giving off energy in the form of a photon. What is the photon's wavelength?

Because protons and neutrons are similar in mass, size, and certain other characteristics, a collective term, nucleons, has been coined that encompasses both of these constituents of the atomic nucleus. In many nuclei, nucleons are confined to dimensions of roughly 15 femtometers. Photons emitted by nuclei as the nucleons drop to lower energy levels are known as gamma particles. Their energies arc typically in the MeV range. Why does this make sense?

An electron is trapped in a quantum well (practically infinite). If the lowest-energy transition is to produce a photon of 450 nm wavelength, what should be the well's width?

A classical particle confined to the positive x-axis experiences a force whose potential energy is $U(x)=\frac{1}{x^{2}}-\frac{2}{x}+1 \quad$ (SI units) (a) By finding its minimum value and determining its behaviors at x = 0 and $x=+\infty,$ sketch this potential energy. (b) Suppose the particle has an energy of 0.5 J. Find any turning points. Would the particle be bound?(c) Suppose the particle has an energy of 2.0 J. Find any turning points. Would the particle be bound?

Write out the total wave function $\Psi(x, t)$ for an electron in the n = 3 state of a 10 nm wide infinite well, Other than the symbols x and t, the function should include only numerical values.

A study of classical waves tells us that a standing wave can be expressed as a sum of two traveling waves, Quantum-mechanical traveling waves, are of the form $\Psi(x, t)=A e^{i(k x-\omega t)}$ Show that the infinite well's standing-wave function can be expressed as a sum of two traveling waves.

A classical particle confined to the positive x-axis experiences a force whose potential energy is $U(x)=\frac{1}{x^{2}}-\frac{2}{x}+1 \quad$ (SI units) (a) By finding its minimum value and determining its behaviors at x = 0 and $x=+\infty,$ sketch this potential energy. (b) Suppose the particle has an energy of 0.5 J. Find any turning points. Would the particle be bound?(c) Suppose the particle has an energy of 2.0 J. Find any turning points. Would the particle be bound?

To obtain a rough estimate of the mean lime required for uranium-238 to alpha-decay, let us approximate the combined electrostatic and strong nuclear potential energies by a rectangular potential barrier half as high as the actual 35 MeV maximum potential energy. Alpha particles (mass 4 u) of 4.3 MeV kinetic energy are incident. Let us also assume that the barrier extends from the radius of the nucleus, 7.4 fm, to the point where the electrostatic potential drops to 4.3 MeV (i.e., the classically forbidden region). Because $U \infty 1/r,$ this point is 35/4.3 times the radius of the nucleus, the point at which U(r) is 35 MeV.(a) Use these crude approximations, and the wide-barrier approximation to obtain a value for the time it takes to decay.(b) To gain some appreciation of the difficulties in a theoretical prediction, work the exercise ’‘backward.” Rather than assuming a value for U0, use the known value of the mean time to decay for ura­nium-238 and infer the corresponding value of $U_0.$ Retain all other assumptions.(c) Comment on the sensitivity of the decay time to the height of the potential barrier.