In textbooks the characteristic length scale of an exciton, or an electron bound to dopant atom, in silicon is calculated by analogy to the vacuum case.

Bohr radius in vacuum:

$$a_0 = \frac{4 \pi \varepsilon_0 \hbar^2}{m_0 e^2} = 0.53\ {\rm Å}$$

where $\varepsilon_0$ is the vacuum permittivity, $m_0$ is the electron mass, and $e$ is the electron charge.

Exciton bohr radius in material:

$$a = \frac{4 \pi \varepsilon_{\mathrm{r}} \varepsilon_0 \hbar^2}{m_e e^2} = a_0 \varepsilon_{\mathrm{r}} \frac{m_0}{m_e}$$

where $\varepsilon_{\mathrm{r}}$ is the relative permittivity, and $m_e$ is the effective mass of the electron in the material.

For silicon, these can be looked up in a properties table and are:

$$\varepsilon_{\mathrm{r}} = 11.7$$

$$m_e = 0.26 m_0$$

Which gives:

$$a = \frac{11.7}{0.26} 0.53\ {\rm Å} = 23.85\ {\rm Å}$$

However, experimental evidence suggests an exciton radius of $4.3\ \mathrm{nm}$ in silicon (according to Yoffe, AD. Advances in Physics, 42(2) pg 173, 1993). This seems quite a bit off to me.

Is the idea of effective mass not entirely appropriate for this analogy?
And more specifically, what about in heavy fermion systems where the effective mass of the electron can be even larger than a muon!? This would give an incredibly small scale for the exciton or defect bound radius.

  • 3
    $\begingroup$ Interesting discovery with \AA. I've never actually used that macro myself; I guess the MathJaX developers thought it was obscure enough to leave out. \overset{\circ}{\mathrm{A}} is a decent substitute. $\endgroup$
    – David Z
    Apr 11, 2011 at 7:33
  • $\begingroup$ I have replaced this contrived sequence by the actual Unicode character, Å ... You may also get it as ... & A r i n g ; ... in HTML, without the dots and spaces between characters. $\endgroup$ Apr 11, 2011 at 7:50

1 Answer 1


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.


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).


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.


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.

  • $\begingroup$ Oops, forgot to use reduced mass! So the exciton case would have a larger bohr radius ~ 4 nm, while the binding to a defect/dopant atom would give the smaller bohr radius ~ 2 nm. Am I understanding that alright? Now what about in the second part of the question regarding heavy fermion systems... it predicts such a small bohr radius can it really be correct!? $\endgroup$
    – John
    Apr 11, 2011 at 8:12
  • $\begingroup$ Yes, I think that the bound state to a dopant would have a radius near your 2 nm, one half of this exciton one. I will add a comment on heavy fermions. $\endgroup$ Apr 11, 2011 at 8:19

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