(Solucionario) Ciencia e Ingenieria De Los Materiales - Callister - 6ed

=  1 ln  T 
l   R)2 
(1 

 

 1mmln  0.85  2.65 x 10-3 mm-1
20  5.33 x 102 =
 =  1  2 



Now, solving for l from Equation (21.19) when T = 0.75

l =  1 ln  T 
   R)2 
(1 

 

1  0.75 
x 103 5.33 x 102
 = ln  2 
2.65 mm1 
 1  


= 67.3 mm

21.22 (a) The characteristic color of a metal is determined by the distribution of wavelengths of the
nonabsorbed light radiation that is reflected.
(b) The characteristic color of a transparent nonmetal is determined by the distribution of
wavelengths of the nonabsorbed light radiation that is transmitted through the material.

21.23 For a transparent material that appears colorless, any absorption within its interior is the same for
all visible wavelengths. On the other hand, if there is any selective absorption of visible light (usually
by electron excitations), the material will appear colored, its color being dependent on the frequency
distribution of the transmitted light beam.

21.24 This problem calls for a calculation of the reflectivity between two quartz grains having different
orientations and different indices of refraction (1.544 and 1.553). We must employ Equation (21.12)
since the beam is normal to the grain boundary. Thus,

 R = 2
n2  n1
2
 n2  n1

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= (1.553  1.544)2 = 8.45 x 10-6
(1.553  1.544)2

21.25 Amorphous polymers are normally transparent because there will be no scattering of a light beam
within the material. However, for semicrystalline polymers, visible light will be scattered at
boundaries between amorphous and crystalline regions since they have different indices of
refraction. This leads to translucency or, for extensive scattering, opacity, except for semicrystalline
polymers having very small crystallites.

21.26 (a) The phenomenon of luminescence is described in Section 21.11.
(b) The feature that distinguishes fluorescence from phosphorescence is the magnitude of the time
interval between photon absorption and reemission events. Fluorescence is for delay times less
than a second; phosphorescence occurs for longer times.

21.27 (a) The phenomenon of photoconductivity is explained in Section 21.12.
(b) Zinc selenide, having a band gap of 2.58 eV, would be photoconductive. In order to be
photoconductive, electrons must be excited from the valence band into the conduction band by the
absorption of light radiation. According to Equation (21.16a), the maximum band gap energy for
which there may be absorption of visible light is 3.1 eV; since the band gap energy for ZnSe is less
than this value, photoinduced valence-band-to-conduction-band electron transitions will occur.

21.28 A photographic light meter is used to measure the intensity of incident light radiation. Each photon
of incident light induces a valence-band-to-conduction band electron transition in which both
electrons and holes are produced, as depicted in Figure 21.5(a). The magnitude of the
photoinduced current resulting from these transitions is registered, which is proportional to the
numbers of electrons and holes, and thus, the number of incident photons, or, equivalently, the
intensity of the incident light radiation.

21.29 Section 21.13 contains a description of the operation of a ruby laser.

21.30 This problem asks for the difference in energy between metastable and ground electron states for
a ruby laser. The wavelength of the radiation emitted by an electron transition from the metastable
to ground state is cited as 0.6943 m. The difference in energy between these states, E, may be
determined from using a combined form of Equations (21.6) and (21.3), as

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E = h = hc


  4.13 x 1015 eV - s 3 x 108 m/ s

= 6.943 x 107 m

= 1.78 eV

Design Problems

21.D1 This problem stipulates that GaAs and GaP have room-temperature band gap energies of 1.42
and 2.25 eV, respectively, that they form solid solutions in all proportions, that alloys of these two
semiconductors are used for light-emitting diodes wherein light is generated by conduction band-to-
valence band electron transitions, and that the band gap of a GaAs-GaP alloy increases
approximately linearly with GaP additions (in mol%). We are asked to determine the composition of
an alloy that will emit red light having a wavelength of 0.68 m. It first becomes necessary to
compute the band-gap energy corresponding to this wavelength of light using Equation (21.3) as

Eg = hc


=   4.13 x 1015 eV - s 3 x 108 m/ s = 1.82 eV

0.68 x 106 m

Realizing that at 0 mol% GaP, Eg = 1.42 eV, while at 100 mol% GaP, Eg = 2.25 eV, it is possible to
set up the relationship

100 mol%  CGaP = 2.25 eV  1.82 eV
100 mol%  0 mol% 2.25 eV  1.42 eV

Solving for CGaP, the composition of GaP, we get CGaP = 48.2 mol%.

21.D2 Relatively high densities of digital information may be stored on the CD-R (read only) compact
disc. For example, sound (i.e., music) may be stored and subsequently reproduced virtually free of
any interference. In essence, the CD-R is a laser-optical data-storage system, wherein a continuous
laser beam functions as the playback element. The input signal is stored digitally (as optical read-

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only memory or OROM) in the form of very small, microscopic surface pits that have been
embedded into the disc during the manufacturing process. The incident laser beam is reflected from
the surface of the disc, and modulation (i.e., variation of the phase) of this read or reflected beam is
achieved by optical interference that results from the depth of the pits.

These read-only discs consist of a substrate into which the datum pits have been replicated.
This substrate must be protected, which is accomplished by a thin and reflective layer of aluminum,
on top of which is coated an ultraviolet curable lacquer. Since the substrate is the key component of
the optical path, its properties are extremely important. Some of the substrate characteristics that
are critical are as follows: 1) it must be highly transparent; 2) it must be possible to economically
produce discs that are uniformly thick and extremely flat; 3) water absorption must be low so as to
avoid distortion; 4) high mechanical stability, good impact resistance, and high heat distortion
resistance; 5) good flow properties (while in a molten state) so as to avoid the establishment of
thermal stresses and subsequent optical nonuniformities (i.e., nonuniform birefringence); 6) the
material must be clean and defect-free in order to ensure error-free scanning; and 7) it must have a
long lifetime (on the order of 10 years).

The current material-of-choice for audio CD-Rs is a relatively low molecular weight
polycarbonate since it is the most economical material that best satisfies the above requirements.

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CHAPTER 22
ECONOMIC, ENVIRONMENTAL, AND SOCIETAL ISSUES IN MATERIALS SCIENCE AND

ENGINEERING
PROBLEM SOLUTION
22.D1W The three materials that are used for beverage containers are glass, aluminum, and the polymer
polyethylene terephthalate (designated as PET or sometimes PETE). Currently, the most commonly
used of these three materials is the PET. Its optical clarity is excellent, it is significantly lighter than
glass, PET has high burst and impact strengths and is shatter-proof, it is inexpensive to produce,
has high gas permeation resistance, is easily fabricated (by blow-molding), and PET containers are
safer (there is no breakage as with glass and no cuts result from pull-tabs as with the Al cans).
There are virtually no incineration and landfill problems with PET, although, PET is relatively
nondegradable. On the down side, PET containers are nonrefillable, but even so, they require less
energy to produce per filled unit volume than either aluminum or glass. Also, they can be recycled.
Glass containers are refillable and recyclable, are very impermeable to the passage of
gases and liquids, and are more expensive to produce and fabricate into bottles than is PET.
However, glass bottles are nonbiodegradable and can be dangerous when they break.
Aluminum beverage containers are nonrefillable and nonbiodegradable, but recyclable, and
are also light in weight. Again, they are more expensive to produce than are PET bottles.

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