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11.a. Types of solids
11.b. Theories of solids
11.c. Semiconductors
11.d. The laser
Types of bonds are: ionic, covalent, metalic, and Van der Waals.
Ohms Law: The diffrential form is:
J = s E,
where E is the electric field, J is the current density and s is the conductivity (inverse of the resistivity, r). The classical explanation of resistance start from the fact that there are free electrons in the metal. Normally electrons move around due to thermal motion as described by the classical Maxwell Boltzmann distribution. On top of that there is an extra velocity gained by the accceleration due to the electric field. Under the effect of this acceleration they would accelerate to infinity velocity if it was not for the ions. They collide with the ions lose their extra energy and start again. The net result is a relatively slow drift of constant average velocity in the direction of the electric field. The root mean squared thermal velocity is
vrms = (3kT/m)1/2.
Suppose the mean free path between collisions is L, then the time between collisions is
t = L/ vrms.
It is easy to calculate now the drift displacement in th edirection of the electric field after k collisions
s = (1/2) (eE/m) [ t12 + t22 + ... + tk2 ].
where the acceleration from Newton's law is a = eE/m. The average of this i
<s> = (1/2) (eE/m) k <t2>.
Using an exponential distribution (like in decay processes) as P(t)µ e-t/t we obtain <t2> = 2 t2, or
<s> = (eE/m) k t2.
Since for n collisons the total time is k t, <s> also equals to <s> = k t vd, where vd is the drift velocity. Comparing the last two equations we obtain an expression for the drift velocity:
vd = (eE/m) t = (eE/m) L/ vrms.
Now the drift velocity can easily be connected with the current density. In fact,
J = ne vd = ne (eE/m) L/ vrms,
where n is the density of electrons. This relation can be seen from the fact that the total amount of charge hitting unit surface area per unit time is contained in a column of unit cross section and length of vd . Now we can just read off the conductivity as
s = ne2 L/ mvrms = ne2 L / (3kTm)1/2.
Then the resistivity is
r =(3kTm)1/2/ ne2 L,
which has the wrong temperature dependence (experimentally it should be linearly rising) and wrong magnitude (even at room temperature it is 10 times as small as the experimental value).
So what's wrong? The reason is very simple. One needs to use Fermi-Dirac statistics instead of Maxwell-Boltzmann. We know already that at room temperature the metals represent a very degenerate electron gas, with a distribution close to what they have at absolute zero temperature. So how to use Fermi-Dirac statistics? One has to realize that deep inside the fermi sea (not near the fermi level) the electrons have nowhere to go under the influence of the electric field. They cannot accelerate because then they would take up velocities that other electrons already have (i.e. would go into an already occupied state). So which electrons can conduct? To see that it is necessary to consider the so-called fermi surface. This is a surface in momentum (or wavevector, k) space. The Fermi energy is a kinetic energy (remember we discussed it for free electrons) so we can define a sphere in momentum space as
px2 + py2 + pz2 < 2mEF.
This determines a sphere of radius pF = (
2mEF)1/2 in momentum space, where pF
is called the Fermi momentum. At absolute zero temperature all states
are inside the fermi sphere and none outside. At increased
temperature the occupancy becomes fuzzy near the Fermi surface.
Notice that we could just as well work in velocity v=p/m or k space,
k=p/h.
Now apply an external electric field. Each electron will gain extra momentum due to the acceleration of the electric field, so the net effect is that the Fermi sphere will shift slightly off center in the direction of the drift (electric field). The change is nothing else but we shave off a little from the Fermi sphere on on side and attach it on the other side. Since the momentum gained by the electric field is negligible compared to the fermi momentum, pF . In other words practically only particles at the Fermi surface participate in the drift. Consequently, the velocity in the formula for conductivity should be pF/m, rather than vrms used above. Then we obtain the expression for the conductivity as
s = ne2 L/ pF.
Heat conductivity: Just like the electric current is driben by the voltage gradient (Ex=dV/dx), i.e.
J = s dV/dx,
thermal currents are driven by the change of the temperature.:
(1/A) dQ/dt = -K dT/dx,
where K is the thermal conductivity. The idea that these are very similar physical phenomena is fairly old. The random thermal motion is influenced by an increased thermal motion at one end of a thermal conductor and that extra energy is tranferred toward the other end. The net result is that one obtains a ratio of K/ s that is universal for all metals. This is called the Wiedemann-Franz law. It is valid assumingboth the classical and the Fermi-Dirac distributions, but the ratio is different.
Let us get an expression for K. Unfortunately the complete derivation is beyond what we learn in this course. First we deal with the derivation at the classical (Maxwell-Boltzmann) level. We use the following result from classical kinetic theory:
K = C vrms L/3= (3/2)k n vrms L/3 = k n vrms L/2,
where C is the heat capacity due to "classical" electrons, which is C = (3/2)k n.
Now K can also be calculated using the Fermi-Dirac statistics. The major change is that, as we seen earlier, C is proportional to the temperature and much smaller then the classical value at room temperature. We saw that it is proportional to T/TF and (3/2)k n. The exact result can be obtained by integrating the Fremi-Dirac distribution (this is the heat capacity per unit voulume, rather than per mole):
C = ( p2/3 ) ( T/TF ) (3/2)k n.
Then using the formula for K, except for substituting the Fermi-Dirac heat capacity and substituing vrms by pF/m, we obtain using TF= EF/k = pF2/2mk
K= ( p2/3 ) ( T/TF ) (3/2)k n ( pF/m) L/3 = ( p2/3 )(k2T/ pF) nL.
Wiedemann-Frantz law: Taking now the ratio of K/T s, which is called the Lorentz number, we obtain for the classical case (use <E> = 3m vrms2/2 = 3kT/2 )
K/T s= (3 k2/2e2) = 1.1x10-8 W W / K2
while in the quantum case we get
K/T s= ( p2 k2/3e2) = 2.45x10-8 W W / K2
Thought the fact that the Lorentz number is a universal constant is correctly predicted by the classical model, its value agrees reasonably well withthe Fermi Dirac calculation only.
Conductivity and mean free path revisited:
In spite of the success of the Wiedemann-Frantz law, the resistivity itself comes out to be too large even from the Fermi-Dirac model. The reason was understood later as the result of the wrong assumption concerning the mean free path. If we use the expression for s with measured conductivities then the mean free path comes out to be hundereds of the atomic distance. In fact, one can show that a perfect lattice tdoes not cause any resistance. There is no resistance just like there is no absorbtion of light in a perfectly transparent object. Unless there are no impurities, deffects (disclocations, voids, etc.) at absolute zero temperature there should be no resistance. Of course there is no matter without impurities (also there is no crystal or glass with no light absorbtion). At nonzero temperature the ions forming the lattice start to vibrate and they are out of their ideal, equilibrium positions. On the language of quantum mechanics we say that there are collective excitations, phonons present in the metal and the electrons scatter off the phonons, rather then the ions, and resistance is generated.
It i seasy to estimate the number of phonons as a function of temperature, at least at low temperature. The Bose Einstein statistics applies and we obtain
n µ 1 / [ ehf/kT - 1 ] µ 1 / [ 1 + hf/kT - 1 ] = kT/hf µ T
In other words teh number of phonons is proportional to the temperature. The probability of scattering off phonons and so the resistivity is proportional to the number of phonons, and so to T. Thus the resistivity has two parts one coming from impurities and largely independent of temperature and another coming from the scattering on phonon and linearly increasing with temperature.
Bands are a common phenomena in solids. There are energy ranges which are admissible for electrons. They are the collective modes equivalent to valence electrons. If there are widely separated atoms then their valence electrons all have exactly the same energy, i.e. we have kN electrons of the same energy, where k is the valence of the N atoms of the system. As we know from our investigation of molecules, if the atoms are brought close together so that their wavefunctions overlap then one needs to symmetrize or antisymmetrize the wavefunction of each electron in the two atoms creating the bonding and antibonding orbitals
y+(r) = y( r) + y(R-r) and y-(r) = y( r) - y(R-r).
These orbitals will correspond to different energies. In other words the energy levels split if we bring the atoms close together. If we bring N atoms close together then the levels split N ways. If N is very large then the levels will be so close together that in practice one obtains a continuous distribution of levels between a maximum and minimum energy states. This continuous spectrum is called a band. Inner electrons do not form bands because their wavefunction is shielded from each other and they do not overlap.
Metals are such elements that the valence band is half occupied. One electron is missing from every atom to complete the shell. A completed shell would generate a filled band. Since the band is half filled the Fermi level is in the middle of the band. Such a band is called a conduction band. The metal is a good conductor because a very small energy lifts electrons above the fermi level where at absolute zero temperature no electron levels are occupied. The valence band of insulators is completely filled (they have 2N electrons in an insolated s-shell) and the next band, the conduction band is too far that thermally electrons cannot be excited to get to this band where they would be able to conduct electricity. The energy gap of insulators is about 20 eV, such that the probabity of raising an electron to the conduction level is practically impossible (20eV -> 2.4x105 K). The probability of raising an electron to the conduction band at room temerature is approximately
P µ exp{ -(E-EF)/kT } =exp{ - 20eV / (1/40) eV } =exp{- 800} = 10-347 which is zero.
Semiconductors are insulators with a smaller energy gap (about 1eV or less). Intrinsic semiconductors (not doped by another element) conduct due to the effect that raising the temperature will raise the energy of some electrons to reach the conduction band. Say if the bandgap is 1eV then the probability that an electron reaches the conduction band is
P µ exp{ -(E-EF)/kT } =exp{ - 1eV / (1/40) eV } =exp{- 40} = 10-17.4. In one mole there are 6x1023 atoms or electrons in a single band, so altogether 6x1023x 10-17.4= 2.4x106 will reach the conduction band. This is of course not quite correct calculation because using the Maxwell Boltzmann statistics for electrons is allowed only at very high temperature. One needs to use Fermi statistics.
Since at absolute zero temperature the valence band of semiconductors is completely filled and the conduction band is completely empty, the Fermi energy is in the middle of the bandgap. At rising temperature the fermi distribution is not anymore rectangular, box-shaped, but rather goes smoothly from 1 to zero as one increases the energy. It will be smaller than one near the top of the valence band and nonzero at the bottom of the conduction band. In other words, the conduction band is not filled any more, some electrons are missing from it. These missing electrons are called holes. The same number of electrons appear in the conduction band. In other words, electron-hole pairs are created by thermal excitations. The holes also act as charge carriers, because they can hop from state to state inside valence band. Being positive they move under the influencre of external fields, contribute to electric conduction.
The reason for the existence of bands is the periodic nature of the lattice of positive ions in the metal. At some wavelengths (and in a "band" of wavelength around them) the wave can proceed unimpeded in the lattice, but at some other wavelengths (and in a band of wavelengths around them) the reflected waves and the incoming waves from subsequent ions interfere destructively and extended wavefunctions are impossible. Since wavelengths are related to momenta through the deBroglie relation, and momenta are related to energies, we get energy gaps.
The semiconductors we described above are intrinsic semiconductors. They usually have a very low conductivity, due to the difficulty of exciting an electron by ~1eV. Conductivity can be enhanced, and regulated if one adds trace elements to semiconductors. These elements are either called donors or acceptors. A donor atom has an electron at an energy level that is inside the bandgap but close to the conduction band. The atoms must be chosen so that it could replace an atom of the semiconductor in the lattice. Such an atom is called an impurity. The electron of the donor atom is localized, but being near to the conduction band can easily be thermally excited to go into the conduction band where it conducts electricity. This is called an n-type semiconductor, because the charge carriers are n(egative), electrons.
The n-type semiconductor should be distinguished from a 10-17.4 in which the charge carriers are holes, i.e. positive. In them the doping element has an unfilled shell that is ready to accept an electron the energy of which is also in the gap, but only slightly below the top of the valence band. Then an electron from the valence band can easily be thermally excited to fill the state in the acceptor. This will ;leave a hole in the valence band that is free to move around and thus conduct electricity. In p-type semiconductors the valence band is the real conduction band. The energy separating the donor or acceptor levels from the conduction or valence bands, respectively can be as small as 0.05 eV, a small fraction of the bandgap, enhancing conductivity tremendously.
One can get ane extriemely useful device if one joins an n-type and a p-type semiconductor, i.e. forms a p-n junction. The p-n junction can be imagined as an interface of a p and an n junction. Near the interface the donor atoms on the n-type side readily relinquish their electrons that move to the p-type semiconductor side, where they get attached to the acceptor atoms. The process is energetically favorable. This process happens in a narrow region around the interface, called the depletion region. Then, first of all, a very strong electric field is created that prevents the diffusion of more electrons into the p side.
The n-p junction can be imagined as a dipole layer (the depletion region) between neutral regions. The total charge in the depletion region is zero, so there will be no electric field in the non-depleted regions. Inside the depletion region, however, there will be a linearly decreasing negative electric field as we apporach the interface from the p side and a linearly increasing negative electric field on the n side. Integrating the electric field we obtain that the neutral part of the n-type superconductor will be at a higher potential than the p-type. This generates a barrier against electrons (the charge carriers in the n-type superconductor) to move across the barrier towards the p-type semiconductor. If we apply a forward bias, i.e negative potential to the n side then the valence electrons from the n side will be swept over the barrier and a current is generated. The magnitude of the current depends on the number of avilable carriers that is given by the inverse of the Boltzmann factor eeV/kT, where V is the forward voltage. If we apply a reverse bias (positive voltage on the n-type conductor, then we increase the potential barrrier and no carriers will be able to move from one side to the other. The reverse current will go into saturation at a very low level current, I0, which is of the order of tens of micro amps. The current-voltage relation for the junction is then
I = I0( eeV/kT-1).
This is a characteristic behavior of a diode that allows currents to flow in one direction but not in the other direction. Diodes are very important in electronic circuits.
A slightly more compicated device is the pnp or npn transistor. This device can be used for the amplification of signals. The pnp transistor consists of two junctions, an n-type semiconductor sandwiched between two p-type semiconductors. The two p type regions are called the emitter and the collector, while the n-type region is called the base. When we apply negative voltage to the collector then the charge carries (holes) of the p-type emitter can move through the barrier into the collector and generate a current. The base is very narrow and holes accumulated in its donors are not sufficient stop the current completely. If we draw the positive charge off the base by applying a negative voltage to it then the barrier is decreased and the flow of holes from emitter to collector increases tremendously. If you apply a variable electric signal to the base that will be amplified and modulate the current flowing through the transistor. The relation between the current on the base and the current at the emitter has the form
Ic = b Ib
where b is the current gain, which is between 10 and 100.
Integrated circuits are etched on silicon wafers. They are complicated electronic circuits combined from a large number of diodes, transistors, resistors and capacitors. The number of elements in integrated circuits are usually in the millions. They are much more reliable than circuits formed from individual transistors, diodes, etc, that require a lot of soldering.
Suppose a collection of identical atoms, let it be a solid, liquid, or gas, is such that the lowest excitation energy is DE = E2 - E1 Then there are three different processes that photons of energy DE can participate in. These are:
Now the transition rates (probability of transition per unit time per atom) for these processes are 1. A21, B12 u(f,T), and B21 u(f,T), respectively, where u(f,T) is the density of the electromagnetic energy at frequency f, where DE=hf, and temperature T. A21= 1/ t, where t is the lifetime of the excited state.
At thermal equilibrium the Boltzmann relation (that is valid except at very low temperature) for the occupancy of states tells us that the artio of numbers of atoms in the ground and excited states is
N2/N1 = exp{-( E2 - E1 )/kBT }
The thermal equilibrium implies the number of transitions 2 ->1 is the same as the number of transitions 1 ->2.
N2( A21 + B21 u(f,T) ) = N1 B12 u(f,T), or
exp{-( E2 - E1 )/kBT }( A21 + B21 u(f,T) )= N1 B12 u(f,T)
Solving for u(f,T) and comparing with the known Planck distribution we obtain
(8p h f3/ c3) 1 / (ehf/kT - 1) = A21 / (B12 exp{( E2 - E1 )/kBT }- B21)
Since the constants A21, B12, and B21 are independent of the temperature, we obtain that
B21 = B12, and A21 = B12 (8p h f3/ c3) .
In other words, the probability of induced emission and absrorbtion is the same. Since u(f,T) decreases exponentially with f at high f, the transitions are dominated by spontaneous emission, the term A21 is much larger than the term B21 u(f,T).
If one wishes to generate a large number of coherent photons, that can only achieved by stimulated (induced) emission. Under normal circumstances the number stimulated emissions is very low. It is necessary to achieve a so-called population inversion to have a lasing action. Population inversion means that there are more excited state atoms than ground state atoms. Since B21 = B12, N2/N1 will determine whether the there will be more stimulated emission or absorbtion processes Since the first increases the number of photons by one, while the second decreases it by one if N2/N1>1 then the number of coherent photons increases if N2/N1<1 the number of coherent photons decreases. If we can achieve N2/N1>1 somehow then the number of photons will increasew, in principle, without any bound.
A population inversion is achieved if we "pump" the excited state somehow, with a process that is nothing to do with the excitation with photons of energy hf. This can be done if there is a third state, 3, of the system, having energy E3 > E2, such that the 3 -> 2 decay is admissible. Then we produce a lot of particles in the excited state 2 and reverse the population (have N2/N1>1). This reversion cannot happen unless the number of 3 -> 2 transitions is larger than the number of 2 -> 1 (spontaneous decay) transitions. That is possible only if the lifetime of the 3 state is shorter than the lifetime of the 2 state.
In gas lasers an electrical discharge produces electrons that collide with atoms of Neon and excite them into a state of small lifetime that decay into a metastable state. In ruby lasers a broad band optical light pulse produces the same effect.
In lasers to keep the coherent light to leak out from the laser until a sufficient intensity is achieved mirrors are used to reflect light back at both ends. A standing wave is produced, which is even stronger if the length of the laser tube is a multiple of the half wavelength. If one of the end mirrors is slightly transparent than the laser beam can come out of the laser. The intensity, monochromacity, and the collimation of laser beams is over a hundred times stronger than anything one can achieve with ordinary light sources.
In semiconductor lasers the population inversion is achieved by intense radiation or electrical current transmission. The top of the valence band is emptied and the bottom of the conduction band is fully populated. Then again more and more coherent photons (their energy is exactly the bandgap) are produced because the number of atoms in the conduction band is more than the those in the valence band and the probabilities for induced emission and absorbtion are the same. The situation is even slightly better than in conventional lasers because the absorbtion energy is eventually significantly larger then the emission energy, because the conduction band is filled up to level X and the valence band is depleted by energy Y. Thus the emission energy is Eg, while the absorbtion energy is Eg + X + Y. The photons produced in stimulated emeission will not have sufficient energy to be absorbed by the remaining electrons in the valence band.
The advantage of solid state lasers is their small size, their reliability and their great effectiveness (large percentage of the energy output is turned into laser light).