The goals of this chapter are to understand · 2014. 3. 11. · Photonic Materials W-105 Example...

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

88616_online_ch18_ptg01.indd 100 20/01/14 3:14 PM

Photonic Materials

18

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.

Chapter


CHAPTER 18W-102 W-102

1019

1 nm

0.01 nm

400 nm

Gamma-raysFreq

uenc

y (H

z)

Wav

elen

gth

X-rays

Ultraviolet

Visible

Infra-red

MicrowavesUHF

VHF7-13

VHF2-6

FM

Radio, TV

Long-waves

Near IR

Thermal IR

Far IR

Radar

1000 MHz

500 MHz

100 MHz

50 MHzAM

500 nm

600 nm

700 nm

10 nm

100 nm

10 mm

100 mm

1000 mm1 mm{

1 cm

10 cm

1 m

10 m

100 m

1000 m

1018

1017

1016

1015

1014

1013

1012

1011

1010

109

108

107

106

{1000 nm1 mm

0.1 nm

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.

c 51

s�0 �0d

1y2 5 2.998 3 108 m/s 18.1


W-103Photonic Materials

If the wavelength of a photon is , then the photon energy (Ep ) in a vacuum or air is given by Equ- ation 18.2:

Ep 5hc

5 h�

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.

� 5c

Figure 18.1 gives the frequency ranges of various forms of EMR.

18.2

18.3

Example Problem 18.1

Confirm the value of the speed of light in a vacuum from Equation 18.1.

SolutionThe 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.

c 51

s�0 �0d

1y25

1fs8.85 3 10212 F/mds4� 3 1027 Wb/A ? mdg1/2 5

1s11.1 3 10218 s2/m2d1/2

c 51

3.33 3 1029 s/m5 3 3 108 m/s


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?

SolutionUse 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 5hc

5s6.63 3 10234 J ? sds3.00 3 108 m/sd

0.70 3 1026 m5 28.4 3 10220 J

Ep 5 28.4 3 10220 J 1 1 eV1.602 3 10219 J2 5 1.78 eV

Now make the calculation for violet.

Ep 5hc

5s6.63 3 10234 J ? sds3.00 3 108 m/sd

0.40 3 1026 m5 49.7 3 10220 J

Ep 5 49.7 3 10220 J1 1 eV1.602 3 10219 J2 5 3.10 eV


CHAPTER 18W-104

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.

s 51

s��d1y2

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.

n 5cs

5 sin �i

sin �t

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

18.4

18.5

Absorption

Refraction

Re�ected beamTransmitted beam It

I0

�i

�t

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.)




Calculate the speed of light that passes through silica glass.

SolutionThe speed of light in a material (s) with an index of refraction n is given by Equation 18.5.

n 5cs

5 1.46 53.00 3 108 m/s

s

Now solve for s.

s 5cn

53.00 3 108 m/s

1.465 2.05 3 108 m/s

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.

Table 18.1 The index of Refraction for Selected Materials, for Photons of Wavelength 589 nm

Index of Index ofMaterial Refraction (n) Material Refraction (n)

Air 1.00 Polystyrene 1.60

Ice 1.309 TiO2 1.74

Water 1.333 Sapphire (Al2O3) 1.8

SiO2 (glass) 1.46 Leaded glasses (crystal) 2.50

Polymethyl methacrylate 1.49 Rutile (TiO2) 2.6

Typical silicate glasses z1.50 Diamond 2.417

Polyethylene 1.52

Sodium chloride (NaCl) 1.54

SiO2 (quartz) 1.55

Epoxy 1.58

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.


CHAPTER 18W-106

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.

R 5 1n2 2 n1

n2 1 n12

2

The intensity of EMR reflected (Ir) is given by Equation 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:

R 5 1n 2 1n 1 12

2

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

18.6

18.7

18.8

18.9


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?

SolutionThe 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° sin �t

50.174 sin �t

sin �t 50.1741.46

5 0.119

�t 5 6.8°



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.

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* exps2�xd

where a is the linear absorption coefficient. Figure 18.3 shows the linear absorption coefficient for several semiconductor materials as a function of wavelength.

18.10

18.11


Calculate the reflectivity for visible light of wavelength 589 nm that is of normal incidence to the surface of silica glass in air.

SolutionThe 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.

R 5 1n 2 1n 1 12

2

5 11.46 2 11.46 1 12

2

5 10.462.462

2

5 s0.19d2 5 0.035

Thus 3.5% of the incident normal intensity of visible radiation is reflected from a polished surface of silica glass.

106

105

104

1000

Abs

orpt

ion

Coef

fici

ent [

cm2

1 ]

100

Si

InPGa

GaAs

InGaAs

100.3 0.6 0.9

Wavelength [mm]

1.2 1.5 1.8

Figure 18.3 The linear absorption coefficient for various semiconductors, as a function of wavelength. (Based on http://www.cleanroom.byu.edu/OpticalCalc.phtml )


CHAPTER 18W-108

18.3.1 fiber-optic cableAn 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 ultra-


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.

SolutionSince this radiation has already entered the silicon, we do not have to worry about reflection, because we are starting with I *

0 . The ratio of the transmitted to the initial intensity is 0.5, and it is given by

It

I *0

5 exps2�xd 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

(a)

(b)

Light

Light

n

n

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.



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.

n1

n2

5 sin �2

sin �1

5 sin �t

sin �i

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.

n1

n2

5 sin �t

sin �i

5 sin 90 sin �c

51

sin �c

18.12a

18.12b

Material 2 n5n2

Material 1 n5n1.n2

(a)

It

Io IrIo

Io

ItIr

Ir TIR

(b) (c)

�i �i

�t

�c �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).


Calculate the minimum angle of incidence for a laser beam to have total internal reflection in fiber-optic cable made from dense optical flint glass, with an index of refraction of 1.65, that is coated with silica glass.


CHAPTER 18W-110

18.4 absorptIon and chEmIcal bondIng

The index of absorption (K ) is given by Equation 18.13.

K 5�

4�n

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.

18.13

2

4

6

8

104 2

Metal

log10 (wavelength, , �m)

Semiconductor

Visible spectrumDielectric

Infrared

Radio Optical

Ultra-violet

X–rays �–rays

Inde

x of

abs

orpt

ion,

(K)

22 24

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.)

SolutionIf 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

51.651.46

5 1.31 5 sin �t

sin �i

5sin 90sin �c

51

sin �c

sin �c 5 0.88

�c 5 62°



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

Ep 5 h� $ E2 2 E1 absorption

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.

18.14a

18.14b

E2

E1

(a)

h�

E2

E1

(b)

h�

E2

E1

(c)

h�

h�

h�

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.


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.


CHAPTER 18W-112

E

3s23p

E

E

(c)

(b)

Eg

a

a

a

Energy (E)

Energy (E)

Energy (E)

(a)

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.

SolutionThe 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 5hc

5s6.626 3 10234 J ? sds2.998 3 108 m/sd

5.896 3 1027 m5 3.369 3 10219 J

Ep 53.369 3 10219 J

1.602 3 10219 J/eV5 2.103 eV



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,


CHAPTER 18W-114

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, light- emitting 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.

Eg

VB

h� > EgCB–

+

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.


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?

Solutiona) 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



18.4.1 colorThe 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.


What should be the color of sodium vapor when exposed to white light? Refer to Example Problem 18.8.

SolutionFor sodium atoms in their ground state, there are electrons in the 3s level, but no electrons in any of the higher-energy 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.

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.

5hcEp

5s6.626 3 10234 J ? sds2.998 3 108 m/sd

s0.057 eVds1.602 3 10219 JyeVd5

20 3 10226 m9.13 3 10221 5 2.19 3 1025 m

b) From Figure 18.1, we see that this wavelength is in the infrared portion of the spectrum.


CHAPTER 18W-116

18.4.2 X-ray absorptionX-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 higher-energy 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 state, n = 1 electron removed

L state, n = 2 electron removed

M state, n = 3 electron removed

N state, n = 4 electron removed

valence electron removedneutral atom

Ener

gy o

f Ato

m

K ex

cita

tion

L ex

cita

tion

K� K emission

L�

M�M

N

0

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.)



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 absorption-edge 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.


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.

SolutionWavelengths 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.

5hcEp

56.63 3 10234 J ? ss3 3 108 m/sd

1.1 eVs1.602 3 10219 J/eVd5 11 3 1027 m

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.


CHAPTER 18W-118

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 cost-competitive with conventional power generation systems.

(a)

p-typen-type

Light hv > Eg

– –– – –

–

Eg

(b)

Load

I

Light

I

n-type

p-type

+ + + + ++

Load

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.



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 power-conversion efficiency of 20%.

+

I

–

M1(ITO)

Low workfunction metal 2

Fermi energy

Free Space

CB

(a)

Eg1 2

F1F2

VB

Polymer

–

+ + + + +

– – – –

(b)

Light

PolymerM2

Glass

High workfunction metal 1

Load

Load

hv > Eg

I

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).


CHAPTER 18W-120

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� 5hc

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.

18.15


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.

SolutionThe 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�

5hc

5s6.626 3 10234 J ? sds2.998 3 108 m/sd

0.154178 3 1029 m5 128.8 3 10217 J

EK�

51.288 3 10215 J

1.602 3 10219 J/eV5 8043 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.



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 lasErsSeveral 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� 5hc

18.16

1 2

Eg

Va

P nElectron energy

(a) (b)

EgEcn

Evn

Ecp

Evp

p n

– – – – – – – –

– – –– – – – – – – –

+ + + + ++

+++++

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.


CHAPTER 18W-122

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 high-temperature 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

(a) (b)

Eghv

CB

Polymer

M1

M2

EF1

F1

F2

EF2

1

1

2

2

2

e(Vc2Va)

VB

Light emission

Va

Va

M2

Glass substrate

+++

+

+

––––

M1 (transparent)Polymer emissive layer

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.



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 fiber-optic 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

Polished faces

Laser output

2

1

p-typen-ty

pe

Figure 18.15 A schematic of the configuration of a pn-junction laser with an applied forward bias.


CHAPTER 18W-124

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.

Ruby rod

Xenon �ashtube

Re�ecting cavity

Laseroutput

Mirror

Partialmirror

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)

Nonradiative

Radiative694.3 nm

E2

E1

E0

Pumping

Figure 18.17 Energy levels, electron transitions, and wavelength of a ruby laser.



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.


CHAPTER 18W-126

● 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 AuthorsFull references are listed at the end of the book.

General: Askeland, Fulay, and Wright

X-ray di�raction, 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 _____.



Engineer in Training–Style Questions1. 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


CHAPTER 18W-128

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?



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?


The goals of this chapter are to understand · 2014. 3. 11. · Photonic Materials W-105 Example...

Documents

Transcript of The goals of this chapter are to understand · 2014. 3. 11. · Photonic Materials W-105 Example...