Download The goals of this chapter are to understand

A photograph of an industrial laser
metal cutting facility, where light is
used to cut metal. Complex shapes
of metal parts can be cut with a
programmable laser cutter.
hxdbzxy / Shutterstock.com
The goals of this chapter are to understand
●●
The wave and photon character of electromagnetic radiation
●●
Reflection, refraction, absorption, and emission of electromagnetic radiation
●●
The effect of chemical bonding on absorption of electromagnetic radiation
●●
Color in materials
●●
X-ray absorption and emission spectra
●●
The operation of devices that depend upon the absorption or transmission of photons,
including fiber-optic cable, photoconductivity, infrared optics, and solar cells
●●
Spontaneous emission and light emitting diodes
●●
Stimulated emission and lasers
Chapter
18
Photonic Materials
18.1 Introduction
Electromagnetic radiation (EMR) is a wave that can be diffracted; it has energy, no mass, and
no electrical charge. As shown in Figure 18.1, EMR includes g-rays, X-rays, visible light,
infrared radiation, microwaves, radio, and TV waves. EMR travels in packets of energy,
called photons. A photon of sufficient energy can excite an electron out of a stable orbital.
Before the twentieth century, the only EMR properties studied and understood were the
optics related to lenses and the dispersion of visible radiation in materials, such as when
light passes through a glass prism and forms the color spectrum. Starting at the end of
the nineteenth century, the interaction of materials with all wavelengths of EMR shown
in Figure 18.1 was studied and developed. Figure 18.1 lists some of the inventions of the
twentieth century that relate to the interaction of EMR with materials, such as radio, TV,
microwave communications, and X-ray tomography. Other twentieth-century inventions
that depend upon the interaction of EMR with materials include lasers, photovoltaic solar
cells, fiber-optic cable, and infrared-vision (night-vision) optical systems.
W-102
Frequency (Hz)
Wavelength
CHAPTER 18
Gamma-rays
0.01 nm
19
10
0.1 nm
18
10
X-rays
1 nm
1017
400 nm
10 nm
1016
1015
14
10
Ultraviolet
Visible
100 nm
1000 nm
Near IR {
1 mm
Infra-red
Thermal IR
1012
Far IR
11
10
UHF
Microwaves
500 MHz
1010
600 nm
10 mm
1013
1000 MHz
500 nm
100 mm
700 nm
1000 mm
{ 1 mm
1 cm
Radar
10 cm
109
1m
VHF
7-13
100 MHz
108
Radio, TV
FM
10 m
VHF
2-6
107
100 m
50 MHz
6
10
AM
1000 m
Long-waves
Figure 18.1 Electromagnetic radiation showing the names, frequency, and wavelength. On the left the radio and
TV spectrum (channels 2–13) is expanded, and on the right the visible portion of the spectrum is expanded with
the wavelength and colors indicated. (Based on http://en.wikipedia.org/wiki/File:Electromagnetic-Spectrum.png)
18.2 Electromagnetic Radiation
in a Vacuum
All EMR in a vacuum, and for practical purposes in air, has the speed of light (c), and c is related to the
electrical permittivity (0) and the magnetic permeability (0) of a vacuum, as shown in Equation 18.1.
c5
1
5 2.998 3 108 m/s
s0 0d1y2
18.1
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Photonic Materials
If the wavelength of a photon is , then the photon energy (Ep ) in a vacuum or air is given by Equation 18.2:
Ep 5
hc
5 h
18.2
where h is Planck’s constant (6.63 3 10234 J ? s). The photon energy is inversely related to the wavelength,
as shown in Equation 18.2. The photon frequency () of oscillation in a vacuum or air is determined from
Equation 18.2 in Equation 18.3.
5
c
18.3
Figure 18.1 gives the frequency ranges of various forms of EMR.
Example Problem 18.1
Confirm the value of the speed of light in a vacuum from Equation 18.1.
Solution
The of the permittivity of a vacuum is equal to 8.85 3 10212 F/m, and the permeability of a vacuum is equal
to 4π 3 10212 Wb/A ? m in SI units. Substitute these values into Equation 18.1 and solve for the speed of light.
c5
1
1
1
5
5
s0 0d1y2 fs8.85 3 10212 F/mds4 3 1027 Wb/A ? mdg1/2 s11.1 3 10218 s2/m2d1/2
c5
1
5 3 3 108 m/s
3.33 3 1029 s/m
Example Problem 18.2
In Figure 18.1, the visible portion of the electromagnetic spectrum (EMS) has wavelengths from
0.70 3 1026 m (red) to 0.40 3 1026 m (violet). What is the energy range of these visible photons in
joules and electron-volts?
Solution
Use Equation 18.2 for photons to solve for the energy of red and violet light. Make the calculation first for red.
Convert this answer from SI units to eV, and then compare it with Table 18.1.
Ep 5
hc s6.63 3 10234 J ? sds3.00 3 108 m/sd
5
5 28.4 3 10220 J
0.70 3 10 26 m
Ep 5 28.4 3 10220 J
11.60213eV10 J2 5 1.78 eV
219
Now make the calculation for violet.
Ep 5
hc s6.63 3 10234 J ? sds3.00 3 108 m/sd
5
5 49.7 3 10220 J
0.40 3 10 26 m
Ep 5 49.7 3 10220 J
11.602 3 10 J2 5 3.10 eV
1 eV
219
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CHAPTER 18
18.3 Reflection, Refraction, and
Absorption in Materials
EMR that enters a material from a vacuum or air is reflected, refracted, and absorbed, as schematically
shown in Figure 18.2. A reflected beam has the same speed as the incident beam. The specular reflected
beam direction is at an angle equal to the angle of incidence (i ), but on the opposite side of the
normal to the surface. The incident and specular reflected beams and the normal to the surface all
lie in the same plane. Specular reflection is the reflection you see in a mirror. There is also a diffuse
reflected beam that is scattered in all directions. Diffuse reflection is due to scattering of photons from
imperfections in the surface, such as grain boundaries and scratches. When EMR enters a material, it
is refracted. The wave speed is reduced, and the direction of wave propagation is changed. Absorption
is the reduction in wave amplitude, as shown in Figure 18.2, that occurs with the distance the EMR
penetrates the material.
The speed (s) of EMR in a material is given by Equation 18.4, where is the electrical permittivity and is the magnetic permeability of the material.
s5
1
sd1y2
18.4
The speed of light in a material (s) can be substituted for the speed of light in a vacuum (c) in Equa­
tion 18.3, and this substitution relates s to and in a material. When EMR penetrates into a material
from air or a vacuum, the EMR frequency remains constant.
The change in speed from c to s when EMR enters a material from a vacuum or air results in a
change in the direction of the wave transmitted through the material to the angle t relative to the
incident direction i, where is measured relative to a normal to the surface, as shown in Figure 18.2.
If the medium surrounding the material is a vacuum or air, the index of refraction (n) is given by
Equation 18.5.
n5
sin i
c
5
s
sin t
18.5
Table 18.1 gives the index of refraction for a wavelength of 589 nm of some materials. The index of
refraction for air is 1.00; therefore, with three significant figures the speed of light in air and in vacuum
is the same. The index of refraction is a complex function of the wavelength of the photon. The index of
Refl
ecte
Absorption
d be
am
ted
it
nsm
Tra
m
bea
It
t
i
I0
Refraction
Figure 18.2 A schematic of the interaction of EMR with a material, showing the processes of reflection,
refraction, and absorption. (Based on Askeland, D. R., Fulay, P. P., and Wright, W. J., The Science and Engineering of Materials,
6th ed., Cengage Learning, Stamford, CT (2011), p. 802.)
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Photonic Materials
Table 18.1 The index of Refraction for Selected Materials, for Photons of Wavelength 589 nm
Material
Index of
Refraction (n)
Material
Index of
Refraction (n)
Air
1.00Polystyrene
Ice
1.309TiO21.74
Water
1.333
Sapphire (Al2O3)1.8
SiO2 (glass)
1.46
Leaded glasses (crystal)
1.49
Rutile (TiO2)2.6
Polymethyl methacrylate
Typical silicate glasses
z1.50Diamond
1.60
2.50
2.417
Polyethylene1.52
Sodium chloride (NaCl)
1.54
SiO2 (quartz)
1.55
Epoxy1.58
Based on data from Askeland, D. R., Fulay, P. P., and Wright, W. J., The Science and Engineering of Materials, 6th ed., Cengage Learning,
Stamford, CT (2011), p. 803.
refraction approaches 1 for photons with very long wavelength, such as for radio and TV waves; and it
also approaches 1 at very short wavelengths in the X-ray and -ray regions. Refraction is not significant
if the index of refraction is close to 1. In between the very short and very long wavelengths, the index
of refraction goes through asymptotic maximums and minimums where there is resonance absorption
of the photons, as we will discuss in Section 18.4. In the visible portion of the EMS, if there is not
resonance absorption, the index of refraction decreases with increasing wavelength approaching 1 at very
long wavelengths. The change in index of refraction with wavelength is why the visible radiation forms a
rainbow when passing through water droplets in air. Each water droplet separates various wavelengths
of EMR because of the different angles of propagation of the different wavelengths of light in the water
droplets. Refraction is important in visible-light optics, where refraction results in the focusing of light
through lenses, and in fiber-optic cable for information transmission. We will discuss fiber-optic cable
later in this section.
Example Problem 18.3
Calculate the speed of light that passes through silica glass.
Solution
The speed of light in a material (s) with an index of refraction n is given by Equation 18.5.
n5
c
3.00 3 108 m/s
5 1.46 5
s
s
Now solve for s.
s5
c 3.00 3 108 m/s
5
5 2.05 3 108 m/s
n
1.46
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CHAPTER 18
Example Problem 18.4
Light with a wavelength of 589 nm enters a silica glass lens in an optical microscope, in air, at an
angle of 10° relative to normal incidence. What is the angle of the light transmitted inside the silica
glass lens relative to a surface normal?
Solution
The angle of the light is related to the index of refraction in Equation 18.5. The index of refraction of silica glass
is given in Table 18.1 as 1.46.
n 5 1.46 5
sin i
sin t
5
sin 10°
0.174
5
sin t
sin t
0.174
5 0.119
1.46
sin t 5
t 5 6.8°
When EMR passes from one material to another, part of the EMR is specular reflected, as shown in
Figure 18.2. The reflectivity (R) is the fraction of EMR intensity reflected from the interface between two
materials in a particular direction. The intensity of a beam of EMR is the average power transmitted
through a square meter oriented perpendicular to the direction of wave propagation, and the intensity
is proportional to the square of the wave amplitude. For a polished surface or a mirror the reflectivity is
greatest for the specular reflected EMR. If the EMR is perpendicular to the two material surfaces that
have an index of refraction n1 and n2, respectively, then the reflectivity (R) is given by Equation 18.6.
R5
1n 1 n 2
n2 2 n1
2
2
18.6
1
The intensity of EMR reflected (Ir) is given by Equation 18.7:
18.7
Ir 5 RI0
where I0 is the intensity of the incident beam in watts per square meter. If the radiation is not of normal
incidence, then the reflectivity depends upon the angle of incidence and the angle of measurement. The
reflectivity from an irregular dull surface where diffuse reflection dominates, such as amorphous carbon,
is nearly independent of angle. From Equation 18.6, the reflectivity of a material surface for EMR of
normal incidence in a vacuum or air is given by Equation 18.8:
R5
1
n21
n11
2
2
18.8
where n is the index of refraction of the material, and the index of refraction of a vacuum or air is equal
to 1. The intensity of EMR that enters the material (I *0 ) is then given by Equation 18.9.
I *0 5 I0 2 Ir 5 I0 2 I0 R 5 I0s1 2 Rd
18.9
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Photonic Materials
Example Problem 18.5
Calculate the reflectivity for visible light of wavelength 589 nm that is of normal incidence to the
surface of silica glass in air.
Solution
The reflectivity of a material in air is given by Equation 18.8. For silica glass, the index of refraction from
Table 18.1 is 1.46.
R5
1
n21
n11
2 1
2
5
1.46 2 1
1.46 1 1
2 1 2 5 s0.19d 5 0.035
2
5
2
0.46
2.46
2
Thus 3.5% of the incident normal intensity of visible radiation is reflected from a polished surface of silica glass.
As radiation passes through a material, some of the intensity is absorbed (Ia ), and the remainder of
the intensity is transmitted (It ), as shown in Figure 18.2. The sum of Ia and It equals the intensity of the
EMR that entered the material (I *0 ), as shown in Equation 18.10.
18.10
I *0 5 Ia 1 It
The transmitted intensity as a function of the penetration distance (x) is given by Equation 18.11:
It 5 I *0 exps2xd
18.11
where a is the linear absorption coefficient. Figure 18.3 shows the linear absorption coefficient for several
semiconductor materials as a function of wavelength.
Absorption Coefficient [cm21]
106
105
InGaAs
104
GaAs
InP
Ga
1000
Si
100
10
0.3
0.6
0.9
1.2
1.5
1.8
Wavelength [mm]
Figure 18.3 The linear absorption coefficient for various semiconductors, as a function of wavelength. (Based on
http://www.cleanroom.byu.edu/OpticalCalc.phtml )
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CHAPTER 18
Example Problem 18.6
Silicon is used in solar cells because it absorbs the solar spectrum. Calculate the distance over which
the intensity of a beam of orange light, the apparent color of the sun, of wavelength 600 nm that has
entered silicon at room temperature is reduced to half its initial intensity.
Solution
Since this radiation has already entered the silicon, we do not have to worry about reflection, because we are
starting with I0*. The ratio of the transmitted to the initial intensity is 0.5, and it is given by
It
I *0
5 exps2xd 5 0.5
From Figure 18.3, the linear absorption coefficient for photons with a 600 nm wavelength in silicon is 5.0 3
105 m21. Only the distance x is now unknown. Take the natural log of the ratio It yI *0 .
20.7 5 25.0 3 105 m21x
Now solve for x.
x 5 1.4 3 1026 m
18.3.1 Fiber-Optic Cable
An important application of the refraction and transmission of light in materials is the transmission of
data in the form of light pulses in fiber-optic cable. Fiber-optic cable has a glass fiber core that transmits
data over great distances without significant attenuation, and without loss of the photons, or information.
Figure 18.4 shows two designs for a fiber-optic cable. The core of fiber-optic cable is made of an ultran
Light
(a)
n
Light
(b)
Figure 18.4 Two designs for producing total internal reflection (TIR) for fiber-optic cable. (a) The core glass
fiber is coated with a glass of a lower index of refraction (n). The coating is the outer cylinder and the core glass is
the inner cylinder. The index of refraction (n) is plotted to the left of each figure. (b) Atoms are diffused into the
glass fiber, producing a gradual reduction in the index of refraction from the center to the surface.
W-109
Photonic Materials
Material 2 n5n2
It
t
It
c c
i i
Io
(a)
Ir
Io
Io
Ir
(b)
Material 1 n5n1.n2
Ir
TIR
(c)
Figure 18.5 (a) Light (I0) is incident on the material 122 interface. It is refracted and transmitted into material 2.
Ir is reflected from the material 122 interface back into material 1. Material 2 has a lower index of refraction than
does material 1. (b) Materials 1 and 2 have indices of refraction, resulting in t 5 90° and i 5 c. (c) i is greater
than c, and t is greater than 90°. In (c) the light wave is totally internally reflected (TIR).
high-purity silica glass where impurity elements and defects, such as bubbles, have been eliminated to
minimize absorption and scattering of the light signal. The signal attenuation in the core glass is typically
1 millionth that of ordinary glass. If ocean water was this clear, you could see the ocean bottom at its
deepest location (11 km deep). In Figure 18.4a, the core is surrounded by a cladding designed such
that the light incident upon the core-cladding interface remains in the fiber-optic cable by total internal
reflection (TIR). In Figure 18.4b, the glass is modified by diffusing or implanting atoms into the glass
to reduce the index of refraction in a gradual manner, thereby producing TIR. Figure 18.5a shows the
conditions where light is transmitted from material 1 to a different material 2. The light transmitted into
material 2 is refracted at an angle t that is measured from the normal to the surface. Figure 18.5b shows
the conditions where material 2 has an index of refraction such that the refracted beam in material 2
is at an angle t of 90°. If the incident-beam angle is greater than the critical angle (c ), then the light does
not enter material 2. All of the light is reflected back into material 1 for TIR. TIR occurs as long as i is
greater than c. From Snel’s law of refraction, the ratio of the indices of refraction for light going from
material 1 into material 2 (n1/n2) is given by Equation 18.12a.
sin t
n1
sin 2
5
5
n2
sin 1
sin i
18.12a
Note that if air or vacuum is material 1, then Equation 18.12a reduces to Equation 18.5. If t is 90° and i
is the critical angle for total internal reflection (c ), Equation 18.12a results in Equation 18.12b.
sin t
n1
sin 90
1
5
5
5
n2
sin i
sin c
sin c
18.12b
Example Problem 18.7
Calculate the minimum angle of incidence for a laser beam to have total internal reflection in fiberoptic cable made from dense optical flint glass, with an index of refraction of 1.65, that is coated
with silica glass.
W-110
CHAPTER 18
Solution
If the incident angle of the light from the normal to the surface of the fiber-optic cable in the core is at the
critical angle (c ), then the angle of the refracted light from the normal to the surface of the fiber-optic cable
is 90°, and the light does not enter the coating. The light reflects back into the flint glass at the angle 1 5 c as
shown in Figure 18.5b. The critical angle (c ) is calculated from Equation 18.12b.
n1
n2
5
sin t sin 90
1.65
1
5 1.31 5
5
5
1.46
sin i
sin c
sin c
sin c 5 0.88
c 5 62°
18.4 Absorption and Chemical
Bonding
The index of absorption (K ) is given by Equation 18.13.
K5
4n
18.13
Figure 18.6 presents a schematic of the index of absorption for metals, semiconductors, and dielectrics,
as a function of the logarithm of the wavelength (top scale). The wavelength scale is logarithmic and
decreases in the positive x direction.
Absorption of visible, UV, and X-ray EMR occurs when a photon has sufficient energy (h) to excite
an electron from a low energy (E1), shown in Figure 18.7a, into an unfilled higher energy level E2. -Rays
are of such high energy that they can be absorbed by the nucleus in addition to absorption by electrons.
Index of absorption, (K)
10
4
2
log10 (wavelength, , m) 22
24
Metal
8
6
Semiconductor
4
Dielectric
Visible spectrum
2
Infrared
Radio
Optical
Ultraviolet
X–rays
–rays
Figure 18.6 A schematic of the index of absorption for a dielectric, metal, and semiconductor as a function
of the logarithm of the wavelength of EMR from radio waves to -rays. (Based on W. D. Kingery et al. Introduction to
Ceramics, 2nd ed. John Wiley & Sons, New York (1976), p. 647.)
W-111
Photonic Materials
E2
h
E2
E2
h
h
h
h
E1
(a)
E1
E1
(b)
(c)
Figure 18.7 (a) Absorption of a photon of energy Ep 5 h and excitation of an electron from a low-energy E1
to a higher-energy E2. (b) Spontaneous emission of a photon of energy Ep 5 h when an electron goes from the
higher-energy E2 to the lower-energy E1. (c) Stimulated emission of multiple photons of energy Ep 5 h when an
incident photon of energy Ep 5 h stimulates many electrons at the excited energy level E2 to make the transition to
the energy level E1. The incident photon is not absorbed.
We will discuss the processes of emission and simulated emission in Figures 18.7b and 18.7c in this
section (Section 18.4).
Emission is the reverse process of absorption. Assume that the atomic energy level E1 is occupied by
electrons and the atomic energy level E2 is unoccupied. An electron in the atomic energy level E1 is excited
into the unoccupied energy level (E2 ) by photons of sufficient energy. If the incident photon energy (Ep )
is less than E2 2 E1, there is no absorption of EMR, as shown in Equation 18.14a and schematically
in Figure 18.8a, where the absorption coefficient () is plotted as a function of energy for free atoms of
an element, such as sodium.
Ep 5 h < E2 2 E1 transmission
18.14a
Ep 5 h $ E2 2 E1 absorption
18.14b
The penetration of photons is not like the penetration of a bullet, where a higher kinetic energy increases
the penetration distance.
Equation 18.14b, with the equality, gives the threshold absorption energy; this is the minimum-energy
photon that can excite an electron in the atom. An absorption edge is the sharp increase in absorption of
EMR that occurs when the energy of the photon is sufficient to excite an electron from one energy level
to another. The Laporte selection rule states that only transitions between electron orbitals that change
the angular-momentum quantum number by ±1 are allowed for absorption or emission. Because the
energies for the absorption edges of EMR in a material are dependent upon the energy levels in the atom,
the energy of the absorption edge is used to identify the material in chemical analysis. A discussion of
absorption spectroscopy techniques for chemical analysis is presented in Chapter 15.
Example Problem 18.8
Calculate the threshold absorption energy in joules and electron-volts for sodium vapor knowing
that the lowest-energy transition in a free sodium atom is from 3s to 3p. This transition corresponds
to a wavelength of 589.6 nm.
W-112
CHAPTER 18
Solution
The threshold absorption energy is the lowest-energy photon capable of exciting an electron from a filled electron
energy level to an unfilled level. It is possible for electrons to be excited from the filled 2p level into the unfilled orbital
of the 3s, but this takes more energy than excitations from the 3s orbital to the unfilled 3p. The energy of the electron
is primarily determined by the principal quantum number (n) and to a much lesser extent by the angular-momentum
quantum number (l ). The photon energy for the wavelength of 589.6 nm is calculated with Equation 18.2.
Ep 5
hc s6.626 3 10234 J ? sds2.998 3 108 m/sd
5
5 3.369 3 10219 J
5.896 3 10 27 m
Ep 5
3.369 3 10219 J
5 2.103 eV
1.602 3 10219 J/eV
a
3s23p
Energy (E)
E
(a)
a
Energy (E)
E
(b)
a
E
Eg
Energy (E)
(c)
Figure 18.8 A schematic of the linear absorption coefficient () as a function of energy for (a) a free atom, such
as sodium; (b) a metal, such as sodium; and (c) a transparent dielectric material, such as sodium chloride. The
absorption scales in (a), (b), and (c) are not necessarily the same.
Photonic Materials
The fraction of intensity absorbed decreases as energy increases above the absorption edge if there
are no other transitions excited, as shown in Figures 18.6 and 18.8. High-energy transmission of EMR is
demonstrated by -rays that pass through thick sections of all forms of material. -Rays are used for the
inspection of containers for illegal items, such as nuclear materials or weapons.
The effect of chemical bonding upon the absorption of EMR is demonstrated in Figure 18.8, where
a schematic of the absorption of EMR in an element, such as sodium, is shown as a function of photon
energy for free atoms in a vapor, a metal, and a dielectric compound. In a free atom of sodium, there
is no absorption of EMR until photons have sufficient energy to excite the 3s valence electron into the
3p level, as shown in Figure 18.8a. The absorption spectrum of a metal is very different from that of
the free atom. The index of absorption for the metal shown in Figure 18.6 increases off the scale at
low energies (long wavelengths), and the schematic of the absorption coefficient of the metal in Figure
18.8b shows absorption of EMR down to the lowest of energies. In sodium metal the 3s electrons form
a conduction band of free electrons, with the energy levels filled up to the Fermi energy (EF ) at 0 K. The
filled electron energy levels of the conduction band in sodium are the energy level E1 in Figure 18.7a.
There are many unfilled energy levels (E2 in Figure 18.7a) above the Fermi energy in a metal, such as
sodium, that can be filled with excited electrons. The unfilled electron energy levels are at increments
on the order of 10214 eV above the filled energy levels in a metal, such as sodium. Metals can absorb the
smallest-energy photons by transitions from filled states to unfilled states within the conduction band.
This is why metals are utilized as antennae to detect radio and TV waves that have photon energies of
the order of 1025 to 10211 eV. If the photon is absorbed by the antennae, then electrons are excited in the
metal and are detected by the radio or TV electronics, and subsequently amplified to produce a signal.
Sodium is present in the dielectric sodium chloride, which is table salt. The index of absorption of
a dielectric is shown in Figure 18.6, and a schematic of the absorption spectrum of sodium chloride
is shown in Figure 18.8c. We know that the absorption spectrum of sodium chloride differs from that
of sodium metal, because sodium metal is opaque to visible light, but a crystal of sodium chloride is
transparent to visible light. This difference is shown in Figure 18.6 by the low index of absorption for
dielectric sodium chloride in the visible region, and by the low absorption coefficient at low energy in
Figure 18.8c. Because electrons are bound to Na1 and Cl2 ions in sodium chloride, which has ionic
bonds, there are no free conduction electrons. The low energy state E1 in Figure 18.7a is when the two
ions (Na1 and Cl2) are present. For photons to be absorbed, they must have sufficient energy to remove
an electron from the Na1 or Cl2 ion and to excite the electron into the conduction band that is E2 in
Figure 18.7a. In NaCl, the energy level of the conduction band (E2) is equal to the energy gap (Eg 5
7.8 eV ). An absorption edge energy of 7.8 eV is in the ultraviolet portion of the EMS. Visible light is
transmitted through sodium chloride, because visible light has insufficient energy to excite electrons in
NaCl into the conduction band. Dielectric materials are highly transparent to photons of energy just
below the value of the energy gap (Eg ). Photon energies greater than Eg are absorbed when they excite
electrons from filled chemical-bond states to the unfilled conduction band.
In sodium chloride and other ionic dielectric materials, there is also an absorption in the infrared
region, as shown in Figure 18.6 and at low energy in Figure 18.8c. This absorption results from ionic
polarization, when infrared radiation displaces the positive ions in the crystal in one direction and the
negative ions in the opposite direction.
In pure semiconductors, EMR is absorbed if photons have energies greater than the magnitude of
the energy gap (Eg ) between the valence band that is filled with electrons and the conduction band
where there are unfilled quantum states, as shown in Figure 18.9. For the semiconductor in Figure 18.6,
the index of absorption is small for low-energy, long-wavelength EMR such as infrared, because the
low-energy photons have insufficient energy to excite valence band electrons into the unfilled quantum
states in the conduction band. For example, diamond has a large-energy band gap (Eg) of 5.5 eV.
The visible portion of the EMS extends from 0.7 mm (1.8 eV) to 0.4 mm (3.1 eV). A photon with 3.1 eV
of energy cannot excite an electron from the chemical bonds of the valence band into the conduction
band of diamond. Since the photon cannot excite the valence band electron into the conduction band,
W-113
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CHAPTER 18
h > Eg
–
CB
Eg
+
VB
Figure 18.9 A schematic of the absorption of a photon with energy h in a semiconductor, with an energy gap
Eg by excitation of an electron from the valence band (VB) to the conduction band (CB), resulting in a hole (yellow
circle with 1 sign) in the VB and a free electron (red circle with 2 sign) in the CB.
the photon passes through diamond without absorption. Diamond is transparent to all visible radiation,
and it is transparent to photons in the ultraviolet region of the spectrum up to 5.5 eV. If a diamond
has color, it is due to defect energy levels in the energy gap. The absorption and transmission of light
in semiconductors leads to interesting device applications, such as lasers, photovoltaic solar cells, lightemitting diodes, and night vision optics.
At low temperatures, n-type semiconductors have electrons at the donor energy level (Ed ), and in p-type
semiconductors there are empty energy states at the acceptor energy level (Ea ). In n-type semiconductors,
if the photon energy is greater than Ec 2 Ed , shown in Figure 16.12, a photon is absorbed by exciting
an electron from the donor level into the conduction band. In p-type semiconductors, if the photon
energy is greater than Ea, shown in Figure 16.15, a photon is absorbed by exciting an electron from the
valence band into the acceptor level. At high temperatures in n-type semiconductors, the donor atoms
are all ionized, and in p-type semiconductors the acceptor energy states are all filled. At these higher
temperatures the light absorption at the donor atoms and acceptor atoms does not occur.
Polymers absorb EMR that has sufficient energy to excite an electron out of a covalent bond to become
an electron free of the covalent bond. Polymers are mixed with other elements, compounds, or polymers
that absorb different wavelengths of EMR and change the color. We will discuss color in the next section.
The absorption of photons by excitation of electrons from a filled quantum state to an unfilled state is
the primary mechanism of absorption of visible, UV, X-ray, and low-energy -rays in materials. However,
there are other ways that EMR is absorbed in materials. Very-high-energy -rays are absorbed by the
nucleus of the atom. Infrared radiation is absorbed by atomic and molecular vibrations, and by lattice
vibrations in crystals. Microwaves are absorbed by exciting molecular vibrations. Microwave ovens,
for example, emit radiation that is tuned to the vibrations of the H2O molecule. Low-energy radio and
TV waves are absorbed when the conduction electrons in metals are excited into unfilled energy states
in the conduction band. All low-energy EMR is absorbed by the excitation of conduction electrons in
metals. This low-energy photon absorption is why metals are used as radio and TV antennas.
Example Problem 18.9
Silicon doped with aluminum produces a p-type semiconductor with an acceptor level (Ea) at
0.057 eV in an energy gap Eg of 1.1 eV. (a) What is the wavelength of the lowest-energy absorption
edge for this semiconductor? (b) To what region of the EMS does this wavelength correspond?
Solution
a) The lowest-energy absorption edge corresponds to excitation of valence-band electrons into Ea at 0.057 eV.
The excitation from the valence band to the conduction band requires a minimum photon energy equal to
Photonic Materials
the energy gap (Eg) of 1.1 eV, and 1.1 eV is a much higher energy than Ea. Use Equation 18.2 to solve for the
wavelength of valence electrons excited to the acceptor level.
5
hc s6.626 3 10234 J ? sds2.998 3 108 m/sd 20 3 10226 m
5
5
5 2.19 3 1025 m
Ep
s0.057 eVds1.602 3 10219 JyeVd
9.13 3 10221
b) From Figure 18.1, we see that this wavelength is in the infrared portion of the spectrum.
18.4.1 Color
The color of a material that transmits light, such as a ruby, results from the visible radiation that is not
absorbed and is transmitted through the material. A ruby is made from a single crystal of Al2O3 doped
with approximately 5% chromium (Cr) atoms. A pure crystal of Al2O3 is transparent; it is called a white
sapphire. In ruby there is an absorption edge at 550 nm. The ruby appears to be red when exposed
to white light, because wavelengths shorter than 550 nm in the orange, yellow, green, blue, and violet
portions of the EMS are absorbed, and the longer-wavelength red portion of the EMS is transmitted. A
color center is the absorption of radiation in materials at defects in the structure, such as the absorption
of 550 nm light at the Cr ions in ruby.
The photons are absorbed at the substitutional Cr atoms by excitation of electrons that on an isolated
Cr atom would be 3d electrons into unfilled 3d orbitals. This excitation would appear to be a violation of the
Laporte selection rule that Δl 5 ±1. When a Cr atom is substituted for Al in Al2O3 there is a combination of
covalent and ionic bonding between the Cr atom and O atoms in Al2O3. The bonding results in the formation
of molecular orbitals that are a linear combination of the 3d and 4p orbitals from the Cr atom and the
2p orbitals from the surrounding O atoms. Electronic transitions are allowed between these molecular
orbitals because they now have both p and d character, and the Δl 5 11 rule is obeyed for these transitions.
Doping Al2O3 with impurity atoms of iron (Fe) and titanium (Ti) produces a blue sapphire. The
absorption in sapphire is due to the excitation of an electron from Fe21 to Ti41, where both iron and
titanium are substitutional in Al2O3. The absorption process is Fe21 1 Ti41 1 Ep → Fe31 1 Ti31, which
results in absorption in the red part of the EMS, and blue is transmitted.
Colored surfaces result in a similar way. A red surface results when the green, blue, and violet portions
of the EMS are absorbed at a surface, and where the red portion of the EMS is reflected.
Example Problem 18.10
What should be the color of sodium vapor when exposed to white light? Refer to Example Problem 18.8.
Solution
For sodium atoms in their ground state, there are electrons in the 3s level, but no electrons in any of the higherenergy states. If sodium vapor is exposed to white light that can excite electrons, the lowest-energy transition for the
3s electrons is to the 3p quantum state. Absorption is the reverse of the emission. If the photon wavelength is greater
than 589.6 nm, there is no absorption, and the photons are transmitted. From Figure 18.1, the orange and red part
of the spectrum is transmitted. Wavelengths less than 589.6 nm are absorbed. From Figure 18.1, the wavelengths
that are absorbed are yellow, green, blue, and violet. Sodium vapor should have a red-orange color in transmission
when exposed to white light.
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CHAPTER 18
18.4.2 X-Ray Absorption
X-rays are absorbed if they excite 1s, 2s, 2p, and so on electrons from occupied orbitals on atoms
into unoccupied states or out of the atom. The index of absorption for X-rays is shown on the right
side of Figure 18.6. The absorption-edge energy resulting from the excitation of electrons from these
orbitals is dependent upon the atomic number of the element. X-ray absorption is present in all forms
of matter: free atoms, metals, dielectrics, and semiconductors. The absorption of radiation by the
core electrons of the atom has important applications in the fields of X-ray diffraction, spectroscopy,
and radiography. Figure 18.10 shows the ground state and excited electron energy levels for an atom.
The ground state is the unexcited atom at the bottom of Figure 18.10, at zero energy. The higherenergy states, such as the K state, correspond to the energy of the atom with an electron of principal
quantum number equal to 1 removed from an atom. The energy absorbed to create the K excited state
is designated by the up arrow. The designation of the excited states is K, L, M, and N, if the principal
quantum number of the electron missing from the atom is, respectively, 1, 2, 3, and 4. In the L state a
2s or 2p electron is excited out of the atom. The lowest energy of the L edge corresponds to exciting a
2p electron from the atom.
K
K emission
Energy of Atom
K excitation
K state, n = 1 electron removed
L excitation
L state, n = 2 electron removed
L
M state, n = 3 electron removed
M
M
N
0
N state, n = 4 electron removed
valence electron removed
neutral atom
Figure 18.10 A schematic of the energy of the neutral atom at zero energy, excited energy states resulting from
removal of an electron with principal quantum number (n), and the identification of the excited atom states K, L,
M, and N. The up arrows indicate the energy required to remove an electron with principal quantum number n
to free space. The down arrows indicate the energy of the photon emitted when an electron of a higher principal
quantum number fills a vacant electron orbital with a smaller principal quantum number. (Based on Cullity, B. D.,
Elements of X-Ray Diffraction, Addison-Wesley Pub. Co., Inc., Reading, MA (1956), p. 14.)
Photonic Materials
If a photon has an energy equal to or greater than the energy of the K absorption energy, it can
excite the 1s electrons from its orbital, and the photon energy is absorbed. This absorption results in
a large index of absorption, as shown on the right of Figure 18.6 at the wavelength of approximately
1 3 1024 mm. If the energy of the incident photon is less than the K excitation energy (longer wavelength),
the 1s electron is not excited. The high index of absorption for energies above the K excitation energy
(shorter wavelengths) produces a K absorption edge. Table 15.1 presents the wavelengths for the
K absorption edge for a number of elements. This table also shows emission lines, which we will discuss in
Section 18.6.4.2. The absorption edge of an atom is characteristic of that atom type, and the absorptionedge energy is used as a chemical analysis tool to identify the presence of an atom type, as we discussed
in Section 15.3.2.
18.4.3 Photon Absorption and Devices
Silicon provides an interesting example of how absorption properties are utilized to produce unique
optical devices, such as infrared-vision optics, light meters, and solar cells. Figure 18.9 shows electron
transitions that absorb visible-light photons in a semiconductor, such as silicon. In silicon, the energy
gap is 1.1 eV. The entire visible spectrum from 1.78 eV (0.4 3 1026 m) to 3.1 eV (0.7 3 1026 m) has
sufficient energy to excite electrons from the valence band of silicon into the conduction band. The
result is that the visible portion of the EMS is absorbed, and silicon is opaque with a gray metallic
color. The absorption of visible radiation makes silicon a good photoconductor.
Photoconductivity is conductivity that results from incident photons. In a semiconductor or insulator,
valence-band electrons are excited into the conduction band by incident photons of sufficient energy,
creating conduction-band electrons and valence-band holes. The light meter in a camera is an example
of a photoconductor. Photoconductivity increases the conductivity of a semiconductor, such as silicon,
by a factor as large as 1010.
Silicon is transparent to infrared radiation of energy less than 1.1 eV and wavelengths greater than
1.1 3 1026 m. Therefore, silicon is utilized for such products as infrared optical (night-vision) lenses. The
far-infrared radiation is transmitted, and all of the visible wavelengths are absorbed. Electronic systems
are necessary to convert the infrared image to an image visible to the human eye.
Example Problem 18.11
Calculate the shortest wavelength transmitted through pure silicon with nearly 100% transmission,
given that the energy gap between the valence band and the conduction band in silicon is 1.1 eV.
Solution
Wavelengths shorter than that corresponding to a photon energy of 1.1 eV are absorbed by silicon. The
wavelength corresponding to a photon of energy 1.1 eV is calculated from Equation 18.2.
5
hc 6.63 3 10234 J ? ss3 3 108 m/sd
5 11 3 1027 m
5
Ep
1.1 eVs1.602 3 10219 J/eVd
Wavelengths longer than this (lower energy) are transmitted in pure silicon. This wavelength is in the infrared
region of the EMS, as shown in Figure 18.1.
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CHAPTER 18
p-type
Light hv > Eg
–
n-type
Eg
+
+ ++ +
– – – – –
+
I
Load
(a)
Light
I
n-type
Load
p-type
(b)
Figure 18.11 (a) The electron-band structure of a pn-junction irradiated with visible light of energy hv.
Electrons in the valence band of the junction region are excited to the conduction band, resulting in conduction
electrons (red circle with 2 sign) and additional holes (yellow circle with 1 sign) in the valence band. The
conduction electrons flow to the n-type region, charging it negative, and holes migrate to the p-type region,
charging it positive creating a solar battery. (b) The configuration of a solar photovoltaic cell with a positive
current I passing through the external load.
A photovoltaic solar cell generates a potential difference (a voltage) from light. One type of photovoltaic
solar cell is a pn-junction, which produces electrical energy from incident light. Figure 18.11a is a
simplified version of the pn-junction in Figure 16.18b showing only the charges that contribute to the
generation of a voltage. Light of sufficient energy excites valence-band electrons into the conduction
band, resulting in conduction electrons and holes in the valence band. The electrons in the conduction
band lower their energy by migrating to the n-type semiconductor, charging it negative. The holes in the
valence band migrate to a higher energy to reduce the energy of electrons in the valence band of the
p-type semiconductor, charging it positive and creating a solar battery. If the solar cell is connected to an
external circuit, as shown in Figure 18.11b, a positive current I passes through the circuit from the p-type
to the n-type material. The incident light provides a continuous supply of conduction electrons and holes
for the current.
The pn-junction for a photovoltaic solar cell is produced by diffusing or implanting donor atoms
into a p-type semiconductor creating a thin layer of n-type material on the surface. The thickness
of the n-type material is controlled so that light incident on the surface is able to penetrate to the
junction region.
Solar cells are utilized for generation of electrical power in spacecraft and at remote locations where
conventional power is not available. Photovoltaic solar cells are used for conventional power generation;
however, the use is limited because of the high cost of making pn-junctions from single silicon crystals.
Developments in materials may allow for less-expensive materials, such as polycrystalline or polymer
semiconductors. Less-expensive solar cells could result in solar generation of electricity being costcompetitive with conventional power generation systems.
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Photonic Materials
Free Space
hv > Eg
F1
CB
– –
– –
–
F2
2
1
Eg
Fermi energy
Low work
function metal 2
High work
function metal 1 + + +
+ +
VB
Polymer
I
Load
(a)
Light
+
Glass
M1(ITO)
–
Polymer
M2
Load
I
(b)
Figure 18.12 (a) A schematic of the energy levels and the operation of a metal-polymer-metal (MPM) solar cell.
Photons of energy hv excite electrons from the valence band (VB) into the conduction band (CB) of the polymer
across the energy gap Eg. The electrons (red circles with a 2 sign) in the conduction band migrate down in energy
to metal 2 (M2), charging it negative, and holes (yellow circles with a 1 sign) in the valence band migrate up in
energy to metal 1 (M1), charging it positive. (b) A schematic of a possible MPM junction solar cell with M2, the
semiconducting polymer, and M1 of indium tin oxide (ITO).
Photovoltaic cells have also been produced from metal-polymer-metal (MPM) junctions. We discussed
MPM junctions in Section 16.7.5, and Figure 16.21 shows the energy-level diagram of an MPM junction.
Figure 18.12a shows a schematic of the operation of an MPM solar cell. One contact (M1) has a large
work function FM1. For solar cells, M1 is usually indium-tin-oxide (ITO), because ITO thin films are
relatively transparent to visible radiation. ITO is an n-type semiconducting oxide that has an electrical
resistivity comparable to that of a metal, of approximately 1026 Ω ? m. The high conductivity results from
a donor energy level 0.03 eV below the conduction band. The second metal contact (M2), which is made
from aluminum, silver, or magnesium, has a small work function FM2. The polymer is a semiconductor
with an energy gap (Eg). We discussed semiconducting polymers in Section 16.6. Photons absorbed in
the polymer create conduction electrons and holes. The electrons migrate to lower energies and give M2
a negative charge. Holes migrate to a higher energy to reduce the energy of electrons in the valence band
of the polymer, and they give M1 a positive charge. The charging of the two contacts as negative and
positive creates a voltage. If a circuit is connected to the two terminals, a positive current flows from M1
to M2. Figure 18.12b shows a possible configuration for a polymer-based photovoltaic cell. Polymer
solar cells have been developed with power-conversion efficiencies of 6 to 8%. The cost of MPM junction
solar cells is potentially less than that of single-crystal silicon systems, which typically have a powerconversion efficiency of 20%.
W-120
CHAPTER 18
18.5 Photon Emission
For an excited-state atom that has a vacant inner electron orbital, the atom decays to the lower-energy
state (E1), when an electron from the energy level E2 fills the vacant orbital. The electron energy change
results in the spontaneous emission of a photon of energy (Ep) equal to the energy difference between the
energy level E2 and the state E1, as shown in Figure 18.7b and in Equation 18.15.
Ep 5 E2 2 E1 5 h 5
hc
18.15
In Equation 18.15, it is assumed that the photon is in a vacuum or air, and the speed of the photon is c. A
material spontaneously emits photons in all directions. The energies of photon emission are characteristic
of the element, because the emission results from the difference in the energy levels of the atom. The
energy of the emitted photon is used to identify the atom in chemical analysis. Section 15.3 presented
a discussion of emission spectroscopy techniques for chemical analysis. The energy of characteristic
emitted photons ranges from low energies, such as visible EMR, to high-energy X-rays.
Example Problem 18.12
Using the data from Table 15.1, calculate the energy in joules and eV of the photons emitted by
electrons in the 2p states that decay into an unfilled 1s ground state in copper.
Solution
The electron transition from the 2p level to an unfilled 1s electron orbital results in the K X-ray whose wavelength
is 0.154178 nm. The energy of this wavelength is calculated from Equation 18.15.
EK 5
hc s6.626 3 10234 J ? sds2.998 3 108 m/sd
5
5 128.8 3 10217 J
0.154178 3 10 29 m
EK 5
1.288 3 10215 J
5 8043 eV
1.602 3 10219 J/eV
Emitted photons are broadened into an energy band if the energy E2 in Figure 18.7b is a band of
energies. Emission spectra are used to determine the energy width of the conduction band of a metal by
determining the width of the emission band resulting from electrons going from the conduction band of
a metal to a defined energy level (E1).
Luminescence is the emission of EMR that is not due to thermal vibrations. The decay time is the time
between the creation of the excited state of the atom and the decay to the ground state with emission of
a photon. Emission is fluorescence if the decay time is less than 1028 s. Although the term fluorescence
is commonly used for emission in the visible range, as in fluorescent lighting, it also applies to other
energy ranges. For example, we discuss X-ray fluorescence spectroscopy in Section 15.3.1 and will cover
it in Section 18.5.1. Phosphorescence is the emission of visible radiation with a decay time of more than
1028 s. In general, the higher the energy of the excited state or of the emitted radiation, the shorter is the
decay time. The decay time for the emission of high-energy X-rays is as short as 10217 s. If the energy of
the excited state is not large, and if the transition probability to a lower-energy state is low, the decay time
can be quite large, which makes possible the production of the lasers that we discuss in Section 18.5.2.
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Photonic Materials
18.5.1 X-Ray Emission
X-rays have energies from approximately 100 eV (soft X-rays) to approximately 105 eV (hard X-rays). Soft
X-rays are absorbed in air and must be studied in a vacuum. Hard X-rays are transmitted through air.
Figure 18.10 shows the designation of X-ray emission lines, and we discuss characteristic X-ray emission
by atoms in Section 15.2.1. The wavelength of photons emitted from some atoms is presented in Table 15.1.
The wavelength of the X-ray emission is characteristic of the atoms, and it is used to identify the presence
of that type of atom in chemical analysis. In X-ray fluorescence spectroscopy, the relative intensity of the
X-ray emission line is used for quantitative chemical analysis, as we discussed in Section 15.3.1.
18.5.2 Emission of Electromagnetic Radiation
and Devices: LEDs, OLEDs, and LASERs
Several devices, such as the laser and the light-emitting diode (LED), are based upon photon emission. A
laser uses light amplification by stimulated emission of radiation (laser) to create a low divergence beam
of photons that are in phase with each other. LEDs produce photons by the spontaneous emission of
photons resulting from the application of a voltage. LEDs are utilized in electronic systems for display
readout, and they are now becoming a general light source because of their high efficiency.
Inorganic LEDs are made from pn-junctions. Figure 18.13a shows a schematic of the band structure
of a pn-junction with no applied voltage. By the application of a forward bias voltage (Va ), as shown
in Figure 18.13b, electrons from the conduction band of the n-type material are pumped up into the
conduction band of the p-type material, and holes are pumped from the p-type material into the
n-type material. If a conduction electron in the junction region recombines with a hole, this energy
difference (Eg ) is emitted as a photon of energy Ep, given by Equation 18.16.
Ep 5 Eg 5 h 5
hc
18.16
Electron energy
p
Ecp
+
+
– – –
n
– – – – – – – –
Eg
Evp
n
P
+
Ecn
Eg
+
+
– – – – – – – –
+
+
+
+
+
+
Evn
1
(a)
Va
2
(b)
Figure 18.13 (a) Energy bands of a pn-junction diode with no bias voltage. (b) Energy bands of a pn-junction diode
with a forward bias voltage Va that pumps electrons into the conduction band (red circles with – sign) of the p-type
semiconductor and pumps holes (yellow circles with 1 sign) into the n-type semiconductor. Photons are emitted with
energy equal to the energy gap when electrons recombine with holes in the transition region of the pn-junction.
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CHAPTER 18
The recombination of electrons with holes occurs primarily in the junction region between the p-type
and n-type materials, because this is the region where both conduction electrons and holes exist when
the forward bias is applied. In an LED with an applied forward bias, the emission is spontaneous and
in all directions. Silicon is not an efficient emitting material for LEDs. Inorganic LEDs are made from
materials such as GaAs, GaN, and GaP. LED crystals are also doped with elements such as sulfur and
cadmium to obtain different emission colors.
Organic light-emitting diodes (OLEDs) are made with organic semiconducting polymers. A
metal-polymer-metal (MPM) junction can be made into an OLED. In an MPM junction OLED, a
semiconducting polymer with an energy gap (Eg ) has two metal contacts M1 and M2, as shown in the
energy-level diagram of Figure 18.14a. One metal contact (M1), usually transparent indium-tin oxide,
has a large work function. The metal contact M2, such as aluminum, silver, or magnesium, has a small
work function. A forward bias voltage (Va ) is applied, and M1 is charged positive and M2 negative. With
the forward bias, holes are created in the valence band of the polymer at M1, and at M2 electrons are
injected into the conduction band of the polymer. The injected holes and electrons diffuse toward the
center of the polymer, where they combine to create a photon of energy Ep, as given by Equation 18.16.
Figure 18.14b is a schematic of a possible polymer-based OLED. OLEDs have been developed that have
nearly 100% efficiency. OLEDs are mechanically more flexible than inorganic LEDs. OLEDs can be
produced with inexpensive methods similar to printing, whereas inorganic LEDs require expensive hightemperature vacuum techniques. OLEDs are more suitable for large displays than inorganic LEDs are,
because of the ease of fabrication of MPM junctions.
An inorganic diode laser is made from a pn-junction, as shown in Figure 18.15. The energy-band
structure and the bias is similar to that of an LED. The difference between an LED and a diode laser
is that in an LED the recombination of the electron in the conduction band with the hole in the valence
e(Vc2Va)
F2
– – –
CB –
F1
2
hv
M2
Eg
EF1
EF2
+
+ +
+ +
M2
VB
M1
Polymer
1
(a)
2
Polymer emissive layer
M1 (transparent)
Glass substrate
2
Va
Va
1
Light emission
(b)
Figure 18.14 (a) A schematic of the energy levels and the operation of an MPM-junction organic LED with
an applied forward bias voltage Va. The applied forward bias shifts the energy levels of metal 2 (M2) up relative
to metal 1 (M1). Electrons (red circles with a 2 sign) that enter the conduction band (CB) of the semiconducting
polymer migrate down in energy, and holes (yellow circles with a 1 sign) that enter the valence band (VB) of the
semiconducting polymer migrate up in energy. When the electrons and holes recombine, photons are emitted from
the polymer. (b) The configuration of a possible LED based on a MPM-junction with a semiconducting polymer.
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Photonic Materials
Polished faces
2
1
pe
pe
ty
p-
ty
n-
Laser output
Figure 18.15 A schematic of the configuration of a pn-junction laser with an applied forward bias.
band happens spontaneously without stimulation, as shown in Figure 18.7b. In a laser, the electron
recombination with the hole is stimulated by photons from other emissions, as shown in Figure 18.7c. In
a diode laser, many electrons are pumped into the conduction band of the junction region by exceeding a
threshold current density with a forward bias. It is also necessary to have photons in the junction region
to stimulate the emission of other photons. Keeping photons in the junction region is accomplished by
polishing the ends of the laser materials so that emitted photons are reflected back into the junction.
Also, highly doped semiconductors have a lower index of refraction than do pure semiconductors. The
emitting pn-junction is sandwiched between highly doped semiconductors with a low index of refraction
so that emitted photons are refracted back toward the emitting junction, in a manner similar to how fiberoptic cable works, which we discussed in Section 18.3.1. The stimulated photons are in phase with the
stimulating photon, as shown in Figure 18.7c. Lasers utilized as pointers or for surveying are pn-junctions
made from GaAs with an energy gap of 1.42 eV resulting in photons in the red region of the EMS. Other
semiconductor materials that are used for diode lasers include InP, GaSb, and GaN. Silicon has not been
been an effective laser material; however, research is being conducted to produce silicon lasers.
A laser can also be made out of a ruby crystal, as shown in Figure 18.16. In a ruby laser, a xenon flash
lamp surrounds the ruby laser crystal and emits photons that excite (pumps) electrons in the ruby from
their ground state (E0) into an excited state (E2), as shown in the energy-level diagram in Figure 18.17.
The energy level E2 is a band of energies to take advantage of the spectrum of energies coming from the
xenon flash tube. The radiation from the xenon flash tube must have a wavelength of less than 550 nm to
excite electrons from E0 to E2. Electrons decay rapidly from E2 to E1 without the emission of radiation.
However, the decay time from E1 to E0 is longer than from E2 to E1, and E1 becomes highly populated
with electrons. The eventual spontaneous transition of electrons from E1 to E0 produces the ruby red laser
light, with a wavelength of 694.3 nm. This radiation stimulates the transition of other electrons in energy
state E1 to make a transition to E0, emitting a photon of wavelength 694.3 nm that is in phase with the
photon that stimulated the emission.
A laser crystal, whether ruby or pn-junction, is elongated so that photons can travel back and forth
many times, stimulating other emissions with the eventual small divergence laser beam. In a diode laser
the pn-junction is parallel to the long axis of the crystal, as shown in Figure 18.15. Initially emission
occurs in all directions due to both spontaneous and stimulated emission. One end of the laser is highly
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CHAPTER 18
Reflecting cavity
Partial
mirror
Xenon flash
tube
Mirror
Ruby rod
Laser
output
Figure 18.16 A schematic of the configuration of a ruby laser consisting of a xenon flash tube and a ruby rod.
There is a polished mirror surface on one end of the ruby rod to reflect light and a partial mirror on the other end
to both reflect and transmit light. (Based on http://en.wikipedia.org/wiki/File:Ruby_laser.jpg)
reflecting, and the other end is partially reflecting and partially transparent. Photons that are emitted
parallel to the long axis of the laser reflect from the crystal ends and move back and forth through the
laser many times, stimulating the emission of other photons in phase with themselves with a propagation
direction parallel to the long axis of the laser. A coherent (in phase) beam is produced that is parallel to
the laser long axis that emerges from the partially reflecting end. Photons that are emitted in a direction
off the long axis exit the laser without stimulating a significant number of emissions.
Lasers have been produced from other materials, including Y3Al5O12 (YAG) doped with neodymium
and CO2 gas lasers. Polymer lasers have been produced from MPM-junctions. However, polymer lasers
have not found commercial applications, because it is necessary to stimulate the laser emission in the
polymer with another laser.
Lasers have many technological uses, including cutting, ablating, and welding of materials; reading
and writing data on compact disks; surveying, guiding, and targeting; and generating signals for
communications through fiber-optic cable. A laser beam with a small divergence can send information to
satellites or spacecraft without significant loss of the signal intensity, due to the small beam divergence.
Pulsed laser deposition (PLD) utilizes a high-intensity pulsed laser beam to vaporize material from
a target. The vaporized material is then deposited onto a surface to produce a desired coating. Laser
medical applications include eye surgery, the removal of skin blemishes, tissue removal, and the removal
of dental cavities.
E2
Nonradiative
E1
Pumping
Radiative
694.3 nm
E0
Figure 18.17 Energy levels, electron transitions, and wavelength of a ruby laser.
Photonic Materials
Summary
●●
●●
●●
●●
●●
●●
●●
●●
●●
●●
Electromagnetic radiation (EMR) is a wave that can be diffracted. It has energy, no mass, and
no electronic charge. EMR includes -rays, X-rays, visible light, infrared radiation, microwaves,
radio, and TV waves. EMR behaves like it travels in packets of energy, called photons. A photon
of sufficient energy can excite an electron out of a stable orbital.
All EMR in a vacuum, and for most practical purposes in air, has the speed of light (c) equal to
2.998 3 108 m/s.
When EMR enters a material from a vacuum or air, the beam is reflected, refracted, and absorbed.
The reflected beam has the same speed as the incident beam, but the intensity is a small fraction
of the incident intensity, and the direction is at an angle equal to the angle of incidence on the
opposite side of the normal to the surface. When EMR enters a material the wave speed is
reduced; it is refracted and absorbed. Refraction is the change in direction of wave propagation.
Absorption is the reduction in amplitude and intensity that occurs with the distance penetrated
into a material.
An important application of the refraction and transmission of light in materials is the
transmission of data in the form of light pulses in fiber-optic cable. The core of fiber-optic
cable is made of ultra-high-purity silica glass, for which impurity elements and defects, such
as bubbles, have been eliminated to minimize absorption and scattering of the light signal. The
signal attenuation in the core glass is typically 1 millionth that of ordinary glass. The core is
surrounded by a material designed so that the light incident upon the core-material interface
remains in the fiber-optic cable by total internal reflection.
Absorption of EMR in a material occurs when a photon has sufficient energy to cause excitation
of the material. The absorption mechanism of EMR depends upon the EMR energy and
the material. Visible, UV, X-ray, and low-energy -ray radiation is primarily absorbed by the
excitation of electrons on atoms. High-energy -rays are absorbed by the nucleus. Infrared
radiation is primarily absorbed by atomic and molecular vibrations and lattice vibrations. Radio
and TV waves are absorbed by excitation of electrons in the conduction band of metals.
Applications that depend upon absorption or transmission of EMR include photoconductivity,
infrared optics, X-ray absorption-edge spectroscopy, solar cells, and radio and TV antennas.
Photovoltaic solar cells are made from pn-junctions and from metal-polymer-metal (MPM) junctions.
Absorption is affected by chemical bonding. Metals absorb all EMR down to the lowest of
energies. Semiconductors and dielectrics absorb EMR when the photon energy is greater than
the energy gap, and when the photon energy can excite electrons from or to a defect energy level.
Polymers absorb EMR that has sufficient energy to excite an electron out of a covalent bond to
become an electron free of the covalent bond.
The color of a transparent material, such as colored glass, results from the visible radiation that
is not absorbed when passing through the material.
Photon emission occurs when an electron in an excited state decays to a lower-energy state. The
photon energy is equal to the energy difference between the excited state and the lower-energy
state. The wavelength of the emission is characteristic of the atoms, and emission lines are used
to identify the amount of an atom type present in chemical analysis.
Inorganic light-emitting diodes (LEDs) are made from semiconductor pn-junctions of materials
such as GaAs, InP, GaSb, and GaN. Organic light-emitting diodes (OLEDs) are made from
MPM-junctions. Light emission results when a sufficient forward bias is applied to these devices.
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The difference between an LED and a laser is that an LED has spontaneous emission of
photons, and a laser has stimulated emission of photons. Lasers require an excited electron
state with a relatively long lifetime that can be stimulated to decay. Lasers have been made from
semiconductor pn-junctions, ruby, Y3Al5O12 (YAG) doped with neodymium, and CO2. Polymer
lasers have been produced from MPM-junctions. However, polymer lasers have not yet been
used in commercial applications, because it is necessary to stimulate the emission in the polymer
laser with another laser.
Supplemental Reading: Subjects and Authors
Full references are listed at the end of the book.
General:
Askeland, Fulay, and Wright
X-ray diffraction, spectroscopy, and absorption:
Barrett and Massalski; Cullity; Massa
Optoelectronics and Photonics:
Kasap; Kwok
Photonic polymers:
Hadziioannou and van Hutten; Heeger, Sariciftci, and Namdas
Homework
Concept Questions
1. Because electromagnetic radiation (EMR) of sufficient energy can excite electrons from stable orbitals,
EMR is considered to travel in packets of energy called _____________.
2. Photon energies and wavelengths are ______________ related to each other.
3. The speed of light divided by the wavelength of light is the wave ________________.
4. The speed of light in a vacuum divided by the speed of light in a material is equal to the index of
__________________ for the material.
5. Fiber-optic cable relies upon __________ _________ _____________ to keep all of the light in the cable.
6. For an electron transition between orbitals to occur, the angular-momentum quantum number must
change by plus or minus _____________.
7. An absorption ___________ is the sharp increase in absorption that occurs when the energy of a photon
is sufficient to excite an electron from one energy level to another.
8. ______________________ is the increase in conductivity that results when photons excite electrons in a
semiconductor or insulator into the conduction band.
9. Emission is called fluorescence if the decay time is less than __________ seconds.
10. The difference between a LED and a laser is that a LED has spontaneous photon emission, and a laser
has _____________ photon emission.
11.A K X-ray corresponds to the vacant electron orbital on an atom going from 1s to _____.
Photonic Materials
Engineer in Training–Style Questions
1. If the energy gap in a semiconductor is 1.0 eV, which of the following energy photons will have the lowest
absorption coefficient in the semiconductor?
(a) 0.2 eV
(b) 0.8 eV
(c) 1.2 eV
(d) 2 eV
2. If the index of refraction for a silica glass fiber is 1.46, which of the following refraction indices for a
coating would produce total internal reflection?
(a)1.30
(b)1.46
(c)1.50
(d)1.60
3. Which of the following electron orbital transitions is not allowed?
(a)3p to 3s
(b)4d to 3p
(c)5d to 4s
(d)5p to 4d
4. Which of the following material types absorb all wavelengths of photons?
(a)Metals
(b)Semiconductors
(c)Dielectrics
(d) Free atoms
5. The primary mechanism for the absorption of infrared radiation is by excitation of:
(a) The core atom electrons
(b) Atomic vibrations
(c) The nucleus
(d) Valence-band electrons
6. If X-rays excite a 2s electron from an atom, this corresponds to which of the following excited atomic states?
(a) K
(b) L
(c) M
(d) N
7. An inorganic photovoltaic solar cell is made from which of the following?
(a) pnp-junction
(b)MPM-junction
(c) Pure silicon
(d) pn-junction
8. A photon with energy of 104 eV is in what part of the electromagnetic spectrum?
(a) -ray
(b)X-ray
(c)UV
(d)Visible
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CHAPTER 18
9. An organic light-emitting diode (OLED) can be made from which of the following?
(a) pnp-junction
(b)PMP-junction
(c)MPM-junction
(d) pn-junction
Problems
Problem 18.1 The light that comes from a ruby laser has a wavelength of 694.3 nm.
(a) What is the color of the laser light?
(b) What is the energy of a photon from a ruby laser in air, in joules and in electron volts?
Problem 18.2 Silicon (Si) is used for night-vision infrared optics. Si is transparent to infrared radiation
because the energy gap of Si is 1.1 eV. What is the longest-wavelength photon that has sufficient
energy to excite a valence electron across the energy gap?
Problem 18.3 UHF radio and TV broadcast frequencies are from 300 MHz to 3 GHz. What are the photon
energies in joules and eV and the wavelengths of the extremes of UHF waves in air?
Problem 18.4 Diamond is transparent to the visible spectrum. Table 18.1 gives the index of refraction at
589 nm. Calculate the speed of 589 nm light in diamond.
Problem 18.5 Light from the sun is illuminating a fish in the water. The sun is at 45° relative to a perpendicular
to the water surface. What is the angle of the sun’s rays in the water relative to a perpendicular
to the water surface?
Problem 18.6 Compare the relative reflected intensity in air from the surface of soda-lime silica glass with an
index of refraction of 1.51, and dense optical flint glass with an index of refraction of 1.65.
Problem 18.7 (a) Calculate the distance required to reduce a beam of photons of energy 1.0 eV that have
entered germanium to 50% of the initial intensity.
(b) What is the wavelength of photons of energy 1.0 eV?
(c) Photons of 1.0 eV correspond to what type of electromagnetic radiation?
(d) Comment on using germanium for the lenses of an optical system to operate with
1.0 eV photons.
Problem 18.8 In a solar cell the light initially passes through the n-type silicon to reach the pn-junction,
where it is desired that the light absorption occur. If we want at least 90% of the light entering
a silicon solar cell to reach the pn-junction region, what is the maximum thickness of the n-type
material for 0.6 mm light assuming pure silicon?
Problem 18.9 (a) Calculate the shortest wavelength EMR that has high transmission through CdS. The
energy gap in CdS is 2.25 eV.
(b) Is a photon with energy 2.26 eV transmitted or absorbed?
(c) Is a photon of energy 2.24 eV transmitted or absorbed?
Photonic Materials
Problem 18.10 Calculate the critical angle for total internal reflection in a fiber-optic cable made from quartz
glass with a coating of polyethylene.
Problem 18.11 Silicon doped with arsenic produces an n-type semiconductor with a donor level at 0.049 eV
below the conduction band in an energy gap of 1.1 eV.
(a) What is the wavelength of the lowest-energy absorption edge for this semiconductor?
(b) To what region of the electromagnetic spectrum does this wavelength correspond?
Problem 18.12 Red-colored diode lasers for pointers and surveying are made from pn-junctions of the
compound semiconductor GaAs that has an energy gap of 1.42 eV. What is the wavelength of
the light from a GaAs laser?
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