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Essentially, we want to know the distribution of
photon energies in a piece of semiconductor with a certain volume V =
L3 (for simplicity) which is in thermodynamical
equilibrium at some temperature T. Don't confuse this
L with the diffusion length ! |
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We solved a similar problem already when we looked at the
distribution of electron energies in a
piece of semiconductor with a certain volume V =
L3; i.e. when we went through the free electron gas
model. |
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There is no reason why we should not follow the
free electron gas model. We do not have to solve the Schrödinger equation
because we already know that photons are waves described by some exp
(ik·r – wt) with w =
2pn |
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With that, we also know that the boundary
conditions imposed by the finite crystal will only allow wave vectors that fit
into the crystal and form standing waves. |
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All we have to do then is to figure out the
density of states at the energy hn, and the probability fph(hn) that these states are occupied . |
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For the density of states we obtain exactly as in the free
electron gas |
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and it has been taken into account that there are
two polarization states per photon for each
k. |
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Rewriting this for photon energies hn using k = 2pn ·nref
/c with nref = refractive index of the
semiconductor, yields |
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| Dph(hn)·dn |
= |
8 · p ·
nref3 · (hn)2
h3 · c3 |
· d(hn) |
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What is the probability that the states at some
energy hn are occupied? For electrons - which
are Fermions - this was given by the Fermi-Dirac
distribution which made sure that only one electron could occupy a given state. |
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Photons, however, are bosons and any number can share a given state. We therefore
must resort to the Bose-Einstein distribution given by |
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| fph(hn) |
= |
1
exp (hn/kT) – 1 |
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The density un · dn of
photons in the frequency interval n, n + dn (and, as always, per
volume unit) is then given by the product of the density of states and the
probability of occupation, it is |
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| undn |
= |
p · nref3 ·
(hn)2
h3 · c3 · exp (hn/kT) – 1 |
· d(hn) |
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This is Plancks formula for the number of photons per
volume element in the energy interval hn,
hn + d(hn) -
derived in just a few easy steps! |
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If you compare it with some text book formula, you
may find a different version - different by a factor hn! |
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This is because ususally it is not the
photon density, but the
energy density that is considered. The
total energy contained in a volume element between , hn + d(hn) is of course
simply the number of photons with that
energy interval times their energy
hn. |
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The energy density, if plotted, gives the well
known spectral intensity curves that made Planck famous. |
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If we want to know how many photons we have with
the band gap energy, we only have to insert
hn = Eg to get the
answer. |
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But how about hn =
Eg/2 or any other energy inside the band gap? After all,
photons with these energies can not be created in the semiconductor, while they
have a certain density according to Plancks formula. |
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Well, as in the free electron gas model (which does not have
band gaps after all) we have made approximations that do not quite apply to the
case of semiconductors. We have assumed a black
body that absorbs and emits at all frequencies - and this is not true for a "cool"
semiconductor. |
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Essentially, we should enter the proper density of states for photons, but this is
far beyond the scope of the course. |
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On the other hand, we can make a detailed
inspection of the the thermodynamic equilibrium just
for the frequencies corresponding to the band gap. |
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This was what Einstein
contributed to this field (should have been
his 3rd or 4th Nobelprize). It will yield expressions for the "Einstein coefficents" crucial for Lasers and
is demonstrated in
another advanced module |
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Finally, a few words to the reason why Plancks radiation law was so seminal and what
went wrong before it was found. |
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Before Planck, others had considered radiation equilibrium -
essentially people wanted to know why all
hot bodies "glowed" pretty much the same way, independent of their
composition. |
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Based on a the "final" knowledge of electromagnetism
as put down by Maxwell and a fully developed (classical) theory
of thermodynamics, Raleigh and Jeans showed - beyond a trace of doubt - that
the energy density hn · un · dn =
En · dn of electromagnetic radiation coming off a hot body
must be |
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| En · dn |
= |
8 · p ·
nref3 · (hn)2
h3 · c3 · kT |
· d(hn) |
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Big problem! The energy density increases forever, and there
should be tremendous amounts of UV and X-rays coming out of a hot
body. This is obviously not true, the term "ultraviolet catastrophe" was
coined. |
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But Raleigh and Jeans made no mistake - something
was fundamentally wrong with classical
physics as a discipline. The ultraviolet catastrophe, in fact, was one of the
many stumbling
boocks of classical physics at the close of the 19th century. Let's
see what went wrong: |
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Essentially , Raleigh and Jeans' formula says: |
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| Energy density En ·
dn |
= |
density of states D(E) · kT |
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This is nothing but the
equipartition
theorem that states that in a given system every energetic degree of
freedom is imbued with the same average
energy kT. Since the energy fluctuations - with kT
as the average value - can have any
value (just with different probabilities), all energies (= energy levels in modern parlor),
including the very large ones - can be reached and populated. |
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The correct formula of Planck now essentially
states that |
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Energy density Endn = density of
states D(E) times hn
times distribution function. |
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Energy now comes in quanta hn, and for finite temperatures there might not be any
to populate the high-up states. |
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Big difference! But not for low energies, where
the Raleigh-Jeans law is a good approximation of Plancks law. |
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© H. Föll
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