Light absorption in solid matter:
First experiment on external photoelectric effect
Photoelectric cell (~1890)
In photoelectric cell, electrons are liberated from the
surface of a metallic conductor (cathode) by
absorbing energy from light shining on the metal's
surface. Free electrons drift to anode and form the
current through the cell.
The number of electrons released from the metal can be determined by
measuring the current. The kinetic energy of electrons could be measured by
the force needed to stop them (the charge applied to the metal)
As the light shining on a metal becomes increasingly intense, the classical
wave (electromagnetic) theory of light suggests that the electrons that absorb
the light will be liberated from the metal with more and more energy.
In 1902, Phillip Lenard studied the
photoelectric effect and discovered a number of
key features:
1. The greater the intensity of the incoming light, the greater the number of electrons that
were released. After all, when bigger, more powerful waves start hitting an ocean beach,
they dislodge more sand than smaller, weaker waves. That observation made sense.
2. Next discovery did not make sense at all in the world of classical physics. He
discovered that the kinetic energy contained in each of the escaping electrons did not
increase when the intensity of the light increased. Where did the extra energy go? No
answer.
3. Next discovery did not make sense at all in the world of classical physics. While
increasing the intensity of light didn’t increase the energy of the released electrons,
increasing the frequency of light (light color) did. Weaker waves with higher frequency
supplied more energy to electrons than stronger waves with lower frequency.
4. Next discovery did not make sense at all in the world of classical physics. Light
with low enough frequency could not release electrons from the metal no matter how
strong it was. The photo-effect occurred only if the light frequency was above certain
threshold value.
Quantum nature of light
Albert Einstein (1879-1955), probably the best-known scientist of the
twentieth century, did not talk until he was three. However, even as a youth,
he showed great mathematical ability.
Einstein dropped out of school for a year when he was fifteen. He
eventually attended the Swiss Federal Institute of Technology, but often cut
classes; he passed by studying a classmate's notes. His professors refused to
recommend him for an academic position— which is why he was working
in the patent office in Bern, Switzerland, in 1905.
Quantum nature of light
In 1905, Einstein published a theory of the
photoelectric effect.
He believed that light and other forms of
radiations are made up of discrete bundles of
energy (particles, or quasi-particles), and that
the energy of each bundle depends on the
light’s frequency.
This idea was soon supported by Max Planck’s theory, stating that
the energy E of a single particle of light (photon) is given by:
E = h ν, where h is a Plank constant and ν is the light frequency.
In 1921, Albert Einstein received the Nobel Prize in Physics "for
his services to Theoretical Physics, and especially for his
discovery of the law of the photoelectric effect."
Absorption
Related electrical process:
electron - hole pair generation
The photon with the energy exceeding the bandgap energy of
semiconductor can be absorbed.
The photon disappears; the photon energy excites the electron
from the valence band into the conduction band.
As a result, one e-h pair is being created
Photon
absorption
E = h ν = Εg
Radiation
Related electrical process:
electron - hole pair recombination
When the excited electron meets the hole in the valence band, it
may occupy that place. As a result the e-h pair disappear; this
process is called recombination.
During recombination, the electron energy is released as a
photon with the energy closed to the bandgap energy of the
semiconductor.
Photon
Photon
emission
absorption
E = h ν = Εg
E = h ν = Εg
Photon energy – wavelength relation
E ph = h × ν = h ×
c
λ
Here h = 6.626 × 10-34 J×s; v = 3×108 m/s;
Substituting the constants into the equation for Eph , we get:
1.24
Ε PH [eV ] =
λ [ µm]
Here, we have used: 1 eV = 1.6×10-19 J; 1 µm = 10-6 m
Example: find the energy of the photon corresponding to the light wavelength of
λ = 0.63 µm (RED LIGHT)
Substituting the value of λ into the above expression for the photon energy, we find:
Eph ≈ 1.97 eV;
Photon energies, wavelengths and semiconductor bandgaps
Eph ≈ 1.97 eV;
What semiconductor material can absorb and emit the photon of such energy?
Example problem
The beam of red photons (λ=0.63 µm) creates an optical power
density of 1 W/cm3 in a GaAs sample).
The electron-hole life time is GaAs 5 µs.
Find the concentration of photo generated electron-hole pairs
Solution
Optical power density =
Photon energy × Number of photons per second per unit volume:
Note: Number of photons per second per unit volume absorbed in
the material, is the generation rate G
Popt = Eph× G;
G = Popt/Eph; Eph = 1.97 eV;
G = (1W/cm-3) /(1.97*1.6e-19)J = 3.17 x 1018 cm-3 s-1.
G = R = ∆n/τ; > ∆n = G x τ;
∆n = 3.17 x 1018 cm-3 s-1 x 5x10-6 s = 1.58x1013 cm-3
The energy and momentum conservation laws require
certain conditions to be satisfied to make this absorption possible.
Electron
--
Photon
Hole
For the direct photon into e-h pair transformation,
Eph = Eg;
pph = | pe - ph|,
where Eph = h× v is the photon energy, p is the momentum
of the photon, of the electron or of the hole.
The energy and momentum conservation laws require
certain conditions to be satisfied to make this absorption possible.
Eph = Eg;
pph = | pe - ph|,
The mass of the photon is negligibly small:
pph ≈ 0;
(because p = m×v)
Therefore after the absorption the e and the h must have equal
momentums:
pe ≈ ph;
The momentum of the quantum-mechanical particle of the mass
m and velocity v is given by
p = m×v = = k
where
k=
2π
λ
is the wave vector of the particle.
Band diagram of
direct-band semiconductor
material and
absorption/emission processes
in such material:
1 – absorption
2 - emission
Comments.
A simple way to understand the semiconductor band diagram in energy
– k coordinates is as follows.
The kinetic energy of a free particle using a classic mechanics approach
is
2
mv
E=
2
Since m×v = = k , we expect that in quantum mechanical
approximation,
=2 k 2
E=
2m
i.e. the E(k) dependence in the simplest case of free or quasi-free particle
must be parabolic. This indeed is the case for some semiconductors (as
shown above for the materials like GaAs, InP, GaN etc.)
For other materials, for instance, Si,
the dependence is more complicated:
For the photon emission (top) or
absorption (bottom), the THIRD particle
IS NEEDED to meet the momentum
conservation.
This third particle is usually a PHONON
(the crystal lattice vibration).
Donor - Acceptor and Impurity-band Absorption
Band – donor (a) and acceptor - band (b) absorption
Low-energy (a) donor-band and (b) acceptor-band absorption
Radiation in semiconductors
In general, the reverse of all the absorption processes considered
above can occur to produce radiation.
1) Band - to -band recombination (radiation)
2) Donor- acceptor recombination
3) Impurity band recombination.
Not ANY recombination produces the RADIATION
In case of indirect bands, the recombination produces phonons rather than photons.
This is particularly the case of the most
important semiconductor material, Si.
Log-scale
Absorption spectra