Chapter 7: Quantum Mechanics in Three Dimensions and the Hydrogen Atom

Verify that the solution given in equation (7.6) satisfy differential equations (7.5) as well as the required boundary conditions.

What is a quantum number, and how does it arise?

An electron confinedtoa cubic 3D infinite well 1 nu on aside.

1. What are thethree lowest differentenergies?
2. To how many different states do these three energies correspond?

An electron is trapped in a quantum dot, in which it is continued to a very small region in all three dimensions, If the lowest energy transition is to produce a photon of $\mathbf{450}\mathbf{\text{nm}}$ wavelength, what should be the width of the well (assumed cubic)?

Consider a cubic 3D infinite well.

(a) How many different wave functions have the same energy as the one for which $\left(n_x,n_y,n_z\right)\mathbf{=}\left(5,1,1\right)$?

(b) Into how many different energy levels would this level split if the length of one side were increased by $5\%$ ?

(c) Make a scale diagram, similar to Figure 3, illustrating the energy splitting of the previously degenerate wave functions.

(d) Is there any degeneracy left? If so, how might it be “destroyed”?

Question: An electron is trapped in a cubic 3D well. In the states $\left(n_x,n_y,n_z\right)$= (a) (2,1,1) (b) (1,2,1)(c) (1,1,2), what is the probability of finding the electron in the region $\left(\mathbf{0}⩽\mathbf{x}⩽\mathbf{L},\mathbf{L}\mathbf{/}\mathbf{3}⩽\mathbf{y}⩽\mathbf{2}\mathbf{L}\mathbf{/}\mathbf{3},\mathbf{0}⩽\mathbf{z}⩽\mathbf{L}\right)$. Discus any difference in these results.

Question: Consider a cubic 3D infinite well of side length of L. There are 15 identical particles of mass m in the well, but for whatever reason, no more than two particles can have the same wave function. (a) What is the lowest possible total energy? (b) In this minimum total energy state, at what point(s) would the highest energy particle most likely be found? (Knowing no more than its energy, the highest energy particle might be in any of multiple wave functions open to it and with equal probability.)

Classically, an orbiting charged particle radiates electromagnetic energy, and for an electron in atomic dimensions, it would lead to collapse in considerably less than the wink of an eye.

(a) By equating the centripetal and Coulomb forces, show that for a classical charge -e of mass m held in a circular orbit by its attraction to a fixed charge +e, the following relationship holds

$\mathit{\omega }\mathbf{=}\frac{\mathbf{e}{\mathbf{r}}^{\mathbf{-}\mathbf{3}\mathbf{/}\mathbf{2}}}{\sqrt{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{m}}}$.

(b) Electromagnetism tells us that a charge whose acceleration is a radiates power $\mathit{P}\mathbf{=}{\mathit{e}}^{\mathbf{2}}{\mathit{a}}^{\mathbf{2}}\mathbf{/}\mathbf{6}{\mathit{\epsilon }}_{\mathbf{0}}{\mathit{c}}^{\mathbf{3}}$. Show that this can also be expressed in terms of the orbit radius as $\mathit{P}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{6}}}{\mathbf{96}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{\epsilon }}_{\mathbf{0}}^{\mathbf{3}}{\mathbf{m}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{3}}{\mathbf{r}}^{\mathbf{4}}}$. Then calculate the energy lost per orbit in terms of r by multiplying the power by the period $\mathit{T}\mathbf{=}\mathbf{2}\mathit{\pi }\mathbf{/}\mathit{\omega }$and using the formula from part (a) to eliminate .

(c) In such a classical orbit, the total mechanical energy is half the potential energy, or ${\mathit{E}}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{i}\mathbf{t}}\mathbf{=}\mathbf{-}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{8}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{r}}$. Calculate the change in energy per change in r : $\mathit{d}{\mathit{E}}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{i}\mathbf{t}}\mathbf{/}\mathit{d}\mathit{r}$. From this and the energy lost per obit from part (b), determine the change in per orbit and evaluate it for a typical orbit radius of ${\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}$. Would the electron's radius change much in a single orbit?

(d) Argue that dividing $\mathit{d}{\mathit{E}}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{i}\mathbf{t}}\mathbf{/}\mathit{d}\mathit{r}$ by P and multiplying by dr gives the time required to change r by dr . Then, sum these times for all radii from ${\mathit{r}}_{\mathbf{i}\mathbf{n}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{a}\mathbf{l}}$ to a final radius of 0. Evaluate your result for ${\mathit{r}}_{\mathbf{i}\mathbf{n}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{a}\mathbf{l}}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}$. (One limitation of this estimate is that the electron would eventually be moving relativistically).

In general, we might say that the wavelengths allowed a bound particle are those of a typical standing wave,$\mathit{\lambda }\mathbf{=}\mathbf{2}\mathit{L}\mathbf{/}\mathit{n}$ , where is the length of its home. Given that $\mathit{\lambda }\mathbf{=}\mathit{h}\mathbf{/}\mathit{p}$, we would have $\mathit{p}\mathbf{}\mathbf{=}\mathbf{}\mathit{n}\mathit{h}\mathbf{/}\mathbf{2}\mathit{L}$, and the kinetic energy, ${\mathit{p}}^{\mathbf{2}}\mathbf{/}\mathbf{2}\mathit{m}$, would thus be ${\mathit{n}}^{\mathbf{2}}{\mathit{h}}^{\mathbf{2}}\mathbf{/}\mathbf{8}\mathit{m}{\mathit{L}}^{\mathbf{2}}$. These are actually the correct infinite well energies, for the argumentis perfectly valid when the potential energy is 0 (inside the well) and is strictly constant. But it is a pretty good guide to how the energies should go in other cases. The length allowed the wave should be roughly the region classically allowed to the particle, which depends on the “height” of the total energy E relative to the potential energy (cf. Figure 4). The “wall” is the classical turning point, where there is nokinetic energy left: $\mathit{E}\mathbf{}\mathbf{=}\mathbf{}\mathit{U}$. Treating it as essentially a one-dimensional (radial) problem, apply these arguments to the hydrogen atom potential energy (10). Find the location of the classical turning point in terms of E , use twice this distance for (the electron can be on both on sides of the origin), and from this obtain an expression for the expected average kinetic energies in terms of E . For the average potential, use its value at half the distance from the origin to the turning point, again in terms of . Then write out the expected average total energy and solve for E . What do you obtain

for the quantized energies?

Show that of hydrogen’s spectral series—Lyman, Balmer, Paschen, and so on—only the four Balmer lines of Section 3 are in visible range $\left(400-700nm\right)$