What is the product of the uncernainties determined in Exercises we saw earlier? Discuss the result.

The potential energy shared by two atoms in a diatomic molecule, depicted in Figure we saw earlier. is often approximated by the fairly simple function $U(x)=\left(a / x^{12}\right)$ $\left(b / x^6\right)$, where constants $a$ and $b$ depend on the atoms involved. In Section we saw earlier, it is said that near its minimum value, it can be approximated by an even simpler function-it should ""look like"" a parabola. (a) In terms of $a$ and $b$, find the minimum potential energy $U\left(x_0\right)$ and the separation $x_0$ at which it occurs. (b) The parabolic approximation $U_p(x)=U\left(x_0\right)+$ $\frac{1}{2} \kappa\left(x-x_0\right)^2$ has the same minimum value at $x_0$ and the same first derivative there (i.e., 0 ). Its second derivative is $\kappa$. the spring constant of this Hooke's law potential energy. In terms of $a$ and $b$, what is the spring constant of $U(x)$ ?

It was shown that the infinite well energies follow simply from $\lambda=h / p;$ the formula for kinetic energy, $p^2/2m;$ and a famous standing-wave condition, $\lambda=2 L / n.$ The arguments are perfectly valid when the potential energy is 0 (inside the well) and L is strictly constant, but they can also be useful in other cases. The length L allowed the wave should be roughly the distance between the classical turning points, where there is no kinetic energy left. Apply these arguments to the oscillator potential energy, $U(x)=\frac{1}{2} \kappa x^{2}.$ Find the location x of the classical turning point in terms of E; use twice this distance for L; then insert this into the infinite well energy formula, so that E appears on both sides. Thus far, the procedure really only deals with kinetic energy. Assume, as is true for a classical oscillator, that there is as much potential energy, on average, as kinetic energy. What do you obtain for the quantized energies?

To a good approximation, the hydrogen chloride molecule, HCl, behaves vibrationally as a quantum harmonic oscillator of spring constant 480 N/m and with effective oscillating mass just that of the lighter atom, hydrogen. If it were in its ground vibrational state, what wavelength photon would be just right to bump this molecule up to its next-higher vibrational energy state?

Air is mostly $N_2,$ diatomic nitrogen, with an effective spring constant of $2.3 \times 10^3 N/m,$ and an effective oscilating mass of half the atomic mass. For roughly what temperatures should vibration contribute to its heat capacity?

A 2 kg block oscillates with an amplitude of 10 cm on a spring of force constant 120 N/m. (a) In which quantum state is the block? (b) The block has a slight electric charge and drops to a lower energy level by generating a photon. What is the minimum energy decrease possible, and what would be the corresponding fractional change in energy?

Just what is stationary in a stationary state? The particle? Something else?

For the harmonic oscillator potential energy $U = \frac{1}{2}{kx^2},$ the ground-state wave function is $\psi(x)=A e^{-(\sqrt{m \kappa} / 2 \hbar) x^{2}}$ and its energy is $\frac{1}{2} \hbar \sqrt{\kappa / m}$

(a) Find the classical turning points for a particle with this energy.

(b) The Schrodinger equation says that $\psi(x)$ and its second derivative should be of the opposite sign when E > U and of the same sign when E< U. These two regions are divided by the classical turning points. Verify the relationship between $\psi(x)$ and its second derivative for the ground-state oscillator wave function.

Using equation we saw earlier, find the energy of a particle confined to a finite well whose walls are half the height of the ground-state infinite well energy, $E_1$. (A calculator or computer able to solve equations numerically may be used. but this happens to be a case where an exact answer can be deduced without too much trouble.)

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.

Using the result of Exercise we saw earlier and a computer or calculator able to solve (transcendental) equations, find the two bound-state energies for a well in which $U_0=4\left(\pi^2 \hbar^2 / 2 m L^2\right)$.

Write solutions to the Schrodinger equation appropriate in the various regions, impose required continuity conditions, and obtain the energy quantization condition: $\sqrt{E} \cot \left(\frac{\sqrt{2 m E}}{\hbar} L\right)=-\sqrt{U_{0}-E}$

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?