Question: Section 7.5 argues that knowing all three components of would violate the uncertainty principle. Knowing its magnitude and one component does not. What about knowing its magnitude and two components? Would be left any freedom at all and if so, do you think it would be enough to satisfy the uncertainly principle?

Question: Verify the correctness of the normalization constant of the radial wave function given in Table 7.4 as

$\frac{\mathbf{1}}{{\left(2a_0\right)}^{\mathbf{3}\mathbf{/}\mathbf{2}}\sqrt{\mathbf{3}}{\mathbf{a}}_{\mathbf{0}}}$

A mathematical solution of the azimuthal equation (7-22) is $\mathbf{\Phi }\left(\phi \right)\mathbf{=}{\mathbf{Ae}}^{\sqrt{\mathbf{-}\mathbf{D\phi }}}\mathbf{+}{\mathbf{Be}}^{{}^{\sqrt{\mathbf{-}\mathbf{D\phi }}}}$ , which applies when $\mathbf{D}$ is negative, (a) Show that this simply cannot meet itself smoothly when it finishes a round trip about the z-axis. The simplest approach is to consider $\mathbf{\phi }\mathbf{=}\mathbf{0}$ and $\mathbf{\phi }\mathbf{=}\mathbf{2}\mathbf{\pi }$. (b) If $\mathbf{D}$ were $\mathbf{0}$, equation (7-22) would say simply that the second derivative $\mathbf{\Phi }\left(\phi \right)$of is $\mathbf{0}$ . Argue than this too leads to physically unacceptable solution, except in the special case of $\mathbf{\Phi }\mathbf{\left(}\mathbf{\phi }\mathbf{\right)}$ being constant, which is covered by the ${\mathbf{m}}_{\mathbf{l}}\mathbf{=}\mathbf{0}$ , case of solutions (7-24).

For the cubic 3D infinite well wave function

$\mathit{\psi }\left(x,y,z\right)\mathbf{=}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\frac{{\mathbf{n}}_{\mathbf{x}}\mathbf{\pi }\mathbf{x}}{\mathbf{L}}\mathit{s}\mathit{i}\mathit{n}\frac{{\mathbf{n}}_{\mathbf{y}}\mathbf{\pi }\mathbf{y}}{\mathbf{L}}\mathit{s}\mathit{i}\mathit{n}\frac{{\mathbf{n}}_{\mathbf{z}}\mathbf{\pi }\mathbf{z}}{\mathbf{L}}$ show that the correct normalization constant is $\mathit{A}\mathbf{=}{\left(2/L\right)}^{\mathbf{3}\mathbf{/}\mathbf{2}}$.

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).

Using the functions given in Table 7.4, verify that for the more circular electron orbit in hydrogen $\mathbf{(}\mathit{i}\mathbf{.}\mathit{e}\mathbf{.}\mathbf{,}\mathit{l}\mathbf{=}\mathit{n}\mathbf{-}\mathbf{1}\mathbf{)}$, the radial probability is of the form

$\mathit{P}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{\propto }{\mathit{r}}^{\mathbf{2}}\mathit{n}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{r}\mathbf{/}\mathbf{n}{\mathbf{a}}_{\mathbf{o}}}$

Show that the most probable radius is given by

${\mathit{r}}_{\mathbf{m}\mathbf{o}\mathbf{s}\mathbf{t}\mathbf{p}\mathbf{r}\mathbf{o}\mathbf{b}\mathbf{a}\mathbf{b}\mathbf{l}\mathbf{e}}\mathbf{=}{\mathit{n}}^{\mathbf{2}}{\mathit{a}}_{\mathbf{o}}$