Dear John, note that 23.85 Å is equal to 2.385 nm, while the observed 4.3 nm is approximately two times larger.
There is a simple error in your calculation that exactly fixes the factor of two. Note that the actual calculation you should have done has a radius proportional to $1/m$ and the correct $m$ that you should substitute is the reduced mass of the two-body problem governing the relative position of the two particles.
http://en.wikipedia.org/wiki/Reduced_mass
The reduced mass is $m_1 m_2 / (m_1+m_2)$.
Now, the important point is that an exciton is not a bound state of the effective electron and a superheavy nucleus: instead, it is a bound state of an effective electron and an effective hole - a larger counterpart of the positronium (an electron-positron bound state).
http://en.wikipedia.org/wiki/Exciton
Assuming that both the electron and hole masses are equal, 0.26 $m_0$, the reduced (and still also effective) mass is 0.26/2 $m_0$ = 0.13 $m_0$, and the resulting $a$ is twice as big as your result, 4.77 nm - assuming that your arithmetics is right.
The deviation from 4.3 nm is not too large but I can only handwave if I were trying to pinpoint the most important source of the discrepancy. It could be a different effective mass of the hole; finite-size effects caused by the fact that the silicon atoms were not quite uniformly distributed inside the exciton, and so on.
Update
Oh, in fact, I noticed that your properties table does include a special figure of the effective hole's mass and it differs from the electron mass: 0.38 $m_0$. So the reduced mass is
$$\frac{0.38\times 0.26}{0.38+0.26} m_0 = \frac{0.0988}{0.64} m_0 = 0.154 m_0 $$
and the calculated radius is
$$ \frac{11.7}{0.154} \times 0.53\ Å = 40.3\ Å. $$
Well, this is 7 percent too small, much like the previous one was 7 percent too big. ;-)
Hydrogen atoms with composite heavy fermions
Concerning your second question, as you clearly realize, the calculated radius of the "atom" with such "heavy electrons" would be much smaller than the ordinary atom. This also proves that the assumptions of such a calculation fail: the heavy fermions (in condensed matter physics) are the result of the collective action of many atoms on the electron and its mass.
So the large mass of the heavy fermions is only appropriate for questions about physics at long distances - much longer than the ordinary atom. If you look at very short distances - a would-be tiny atom with the heavy fermion - you cannot use the long-distance or low-energy effective approximations of condensed matter physics. You have to return to the more fundamental, short-distance or high-energy description which sees electrons again.
At any rate, you will find out that there can be no supertiny atoms created out of the effective particles such as heavy fermions. The validity of all such phenomenological effective theories - such as those with heavy fermions - is limited to phenomena at distances longer than a certain specific cutoff and highly sub-atomic distances surely violate this condition, so one must use a more accurate theory than this effective theory, and in those more effective theories, most of the fancy emergent condensed matter objects disappear.
Non-relativistic effective theories
Just a disclaimer for particle physicists: in this condensed matter setup, we are talking about non-relativistic theories so the maximum allowed energy $E$ of quasiparticles doesn't have to be $pc$ where $p$ is the maximum allowed momentum in the effective theory. In other words, we can't assume $v/c=O(1)$. Quite on the contrary, the validity of such effective theories in condensed matter physics typically depends on the velocities' being much smaller than the speed of light, too.
So the mass of the heavy fermions is much greater than $m_0$ which would make $m_e c^2$ much greater than $m_0 c^2$; however, the latter is not a relevant formula for energy in non-relativistic theories. Instead, $p^2/2m_e$, which is (for heavy fermions) much smaller than the kinetic energy of electrons, is relevant. The maximum allowed $p$ of these quasiparticles is much larger than $\hbar/r_{\rm Bohr}$ - the de Broglie wavelength must be longer than the Bohr radius. That makes $p^2/2m_e$ really tiny relatively to the Hydrogen ionization energy.
In classical mechanics, you can describe a crystal (in some approximation) by a Hamiltonian that is a quadratic form in coordinates and momenta of atoms. After you diagonalize this quadratic form, you obtain a Hamiltonian of a set of independent, rather than coupled, oscillators (modes). Then you can quantize this system and you do get zero-point energy for each independent mode (by the way, this energy is $\frac{1}{2}$ℏω, not ℏω). However, not all independent modes are purely longitudinal or transversal. In other words, longitudinal modes are often coupled with transversal modes and, therefore, they are not independent modes that you get as a result of diagonalization. In other words, longitudinal and transverse modes are some linear superpositions of independent modes (which are also called eigen modes, or characteristic modes, or normal modes:-) ) .
Best Answer
Neumann's rule can actually predict the form of these anisotropic efective mass tensors and the existence of the longitudinal and transverse masses, if you provide it with the right information. Therefore, I will first make explicit some preliminaries, so that the final calculation resulting in this prediction makes sense.
Silicon's crystal structure has the $O_h$ (octahedral) point group, therefore any quantity associated with the crystal itself should transform under this point group. Specifically, the symmetry elements $\mathbf{A},\mathbf{B}\in O_h,$ can be specified by matrices that act on vectors $\mathbf{v}\in\mathbb{R}^3$ by $$(\mathbf{A}\mathbf{v})_i=A_{ij}v_j,$$ act on each other by $$(\mathbf{A}\mathbf{B})_{ij}=A_{ik}B_{kj},$$ and as a consequence must act on rank-2 tensors ${\boldsymbol\sigma}\in\mathbb{R}^3\otimes\mathbb{R}^3$ by $$(\mathbf{A}\boldsymbol\sigma)_{ij}=A_{ii'}A_{jj'}\sigma_{i'j'}.$$
The group $O_h$ can be generated by three reflections, which in our presentation (choosing Cartesian coordinates) can be written $$A_{ij}=\begin{bmatrix}1&&\\&1&\\&&-1\end{bmatrix}\quad B_{ij}=\begin{bmatrix}1&&\\&&1\\&1&\end{bmatrix}\quad C_{ij}=\begin{bmatrix}&1&\\1&&\\&&1\end{bmatrix}.$$
Neumann's rule is that any rank-2 tensor-valued property of the crystal must be invariant under all of $O_h$, and because $\mathbf{A},\mathbf{B},\mathbf{C}$ generate $O_h$, it boils down to equations like $\sigma_{ij}=A_{ii'}A_{jj'}\sigma_{i'j'}$ (resp. for $B_{ij}$ and $C_{ij}$). If you calculate this out, you indeed find $\sigma_{ij}=\sigma\delta_{ij}$—any rank-2 tensor (such as effective mass) should be diagonal and thus isotropic (same in every direction), since it acts like just a scalar $\sigma$.
This means silicon crystals cannot have any "one" (i.e. in some sense preferred or unique) anisotropic effective mass. However, silicon's conduction band actually has, in addition to one minimum centered at wavevector $\mathbf{k}=\mathbf{0}$, 6 lower minima centered at $\mathbf{k}_{+x}=\begin{bmatrix}k\\0\\0\end{bmatrix},\mathbf{k}_{-x}=\begin{bmatrix}-k\\0\\0\end{bmatrix},$ etc. for $\mathbf{k}_{\pm y},\mathbf{k}_{\pm z}.$ Note that each of these "preferred directions" alone is not invariant under the symmetry of the crystal, and so you might think Neumann's rule excludes their existence, but actually they transform into each other, so they can exist when they are all together. Similarly, the effective mass around each of these points can be anisotropic, but together you find that the effective mass around one point transforms under the crystal's symmetries into the effective mass around another, and so the entire system is symmetric enough to exist in the crystal. (Note that the central minimum's effective mass is isotropic as predicted.)
Explicitly, we can define the effective mass as a function $\sigma_{ij}(\mathbf{k})$ of which "well" we're in (where $\mathbf{k}$ may only be the $\mathbf{k}_{\pm x},\mathbf{k}_{\pm y},\mathbf{k}_{\pm z}$ from above, and optionally $\mathbf{0}$). Neumann's rule then predicts that this function must satisfy $$\sigma_{ij}(\mathbf{A}\mathbf{k})=A_{ii'}A_{jj'}\sigma_{i'j'}(\mathbf{k})$$ for all $\mathbf{A}\in O_h$ in order to be allowed. If you compute this out for each of the generators above you will again come to some set of requirements for $\sigma_{ij}(\mathbf{k}).$
Doing the calculation, I get $$\sigma_{ij}(\mathbf{0})=m_0\delta_{ij}\\\sigma_{ij}(\mathbf{k}_{+x})=\sigma_{ij}(\mathbf{k}_{-x})=\begin{bmatrix}m_l&&\\&m_t&\\&&m_t\end{bmatrix}\\\sigma_{ij}(\mathbf{k}_{+y})=\sigma_{ij}(\mathbf{k}_{-y})=\begin{bmatrix}m_t&&\\&m_l&\\&&m_t\end{bmatrix}\\\sigma_{ij}(\mathbf{k}_{+z})=\sigma_{ij}(\mathbf{k}_{-z})=\begin{bmatrix}m_t&&\\&m_t&\\&&m_l\end{bmatrix},$$ where $m_{0,l,t}$ are the remaining degrees of freedom after all equations have been eliminated. They correspond respectively to the isotropic (central minimum) and longitudinal and transverse (anisotropic minima) effective masses. You can see how Neumann's rule is actually rather predictive (42 DOF have become 3!): I only inputted that there was some symmetric set of privileged directions each with an associated effective mass tensor and it suggested the existence of longitudinal and transverse masses along/perpendicular to those directions, while eliminating cross terms. Further, these masses must be the same in every well.
To get really abstract about what's going on here, note that Neumann's principle only applies to those quantities that truly belong to the crystal, without anything external breaking the symmetry. If I just gave you a piece of silicon, you could determine and write down as a 2-tensor the effective electron mass in the central minimum of the conduction band without breaking the crystal's symmetry, but you cannot do that with the effective mass in the anisotropic minima. You'd have to pick one minimum to do the measurement around, and then you've broken the symmetry of the crystal and Neumann's principle can't be blamed if the tensor you get is not symmetric enough. However, you can preserve the symmetry by realizing that "the effective mass around the anisotropic minima" is a quantity on its own right, which contains all 6 anisotropic effective masses in a symmetric container. This symmetric quantity is no longer "contaminated" by your choice of minimum to measure it at, so it indeed "belongs" to the crystal alone and you can apply Neumann's principle to this larger object (which is the function ${\boldsymbol\sigma}(\mathbf{k})$) to inspect the structure of all of the anisotropic effective masses at once. More generally, $\boldsymbol\sigma$ is a function of all of (continuous) $\mathbf{k}$-space anyway, so the principle ${\boldsymbol\sigma}(\mathbf{A}\mathbf{k})=\mathbf{A}{\boldsymbol\sigma}(\mathbf{k})$ comes up that way too.
Partial Reference