Devin Gatherwright
09/14/2012
Materials Science Homework #2
1. What are Solid solutions? Explain interstitial and substitutional solid solutions (10points).
Answer: According to the material science textbook, a solid solution can be defined as “a homogeneous
crystalline phase that contains two or more chemical species, which makes both substitutional and
interstitial solid solutions possible.” Interstitial solid solutions are, according to the material science
textbook, “a solid solution where relatively small solute atoms occupy interstitial positions between the
solvent or host atoms.” Furthermore, a interstitial solid is formed when carbon is added to iron. In
contrast, substitutional solid solutions are, according to the material science textbook, a solid solution
where the solute atoms either replace or substitute for the host atoms.” Furthermore, a substitutional
solid is formed when copper is added to nickel in any quantity.
2. Explain in a paragraph each; (in detail along with schematic diagrams of the equipment) optical
microscopy, electron microscopy and scanning probe microscopy respectively? (10 points).
Answer: In optical microscopy, the microstructure is studied by a light microscope. The basic elements
that compose a light microscope are both optical and illumination systems. Light microscopes are used
to observe all metals and many ceramics and polymers due to the fact that these materials are opaque
to visible light (opaque meaning not impenetrable by light.) In order to observe these materials, the
microscope is set in reflection mode. Contrasts in the produced image are the result of the difference in
reflectivity of the microstructure’s various surface regions. This is known as metallographic
investigation. In order to be properly observed, the material must be both ground and polished to a flat,
mirror-like finish by the application of abrasive papers and powders that are tailor-made to this type of
work. The microstructure is then undergoes a surface treatment known as etching, where the surface of
the microstructure is treated with a chemical reagent. As a consequence of etching, small discernible
grooves form along the grain boundaries of the microstructure, which can be viewed under the light
microscope due to the fact that they reflect light.
The following diagram is that of a light microscope, which I found on vetmed.edu:
As you can see, a light microscope consists of a bulb to provide light, a mirror to reflect the light through
field diaphragm and lens, which then further reflects through the condenser and up through the stage
where the specimen sits. The specimen is then viewed using the lenses, and the light from the bulb is
further reflected through the lenses and up through the prism and into the eyepiece, so the viewer can
view the specimen. A focus knob is provided so the viewer can increase the resolution of the image
being viewed.
In electron microscopy, an electron microscope is used to observe structural elements that are either
too fine or too small to be observed by an optical light microscope. In electron microscopy, an image of
the small, fine structure is formed using beams of electrons instead of light radiation. A high-velocity
electron will become wave-like as a result, and will develop a wavelength that is inversely proportional
to the velocity of the electron. As a result of these electron beam wavelengths, the electron microscope
is capable of producing high magnifications with clear resolution. In order to form an image with the
electron microscope, magnetic lenses are used. Other than that, the components of the electron
microscope are relatively similar to that of a light microscope.
The following diagram of an electron microscope was taken from purdue.edu
As you can, the specimen being viewed sits on the stage and the electron gun shoots an electron beam
down through anode, magnetic lens, scanning coils, and the backscattered electron detector down to
the specimen. A TV scanner is hooked up to the scanning coils, the backscattered electron detector, and
a secondary electron detector, which allows the image to be viewed.
In scanning probe microscopy, a scanning probe microscope generates a type of topical map set to an
atomic scale that represents the surface features and characteristics of the specimen that is being
examined by the scanning probe microscope. Scanning probe microscopes are capable of magnifications
as high as 109 X (an incredibly small image to be sure,) as well as having an excellent resolution of the
image at this magnification as well. Scanning probe microscopes are also capable of generating 3
dimensional images of the topical area of a specimen, and lastly, the ability to examine a specimen in its
most suitable environment, no matter what that environment might detail.
The following diagram was taken from students.ycp.edu:
As you can see, the surface sample image is taken by the laser. The cantilever and tip is controlled by the
detector and feedback electronics, which moves the cantilever and tip over the portion desired for
viewing. The photodiode converts light energy and turns it into electrical energy.
3. What makes plain carbon steel harder than pure iron? (5 points)
Answer: Plain carbon steel is harder than pure iron because plain carbon steel is an alloy. According to
our materials science textbook, alloys are metals that contain impurity atoms that give the material
certain characteristics, such as increased mechanical strength for instance. Pure iron is soft because it
does not contain any impurity atoms like plain carbon steel does.
4. An ASTM grain size determination is being made from a photomicrograph of a metal at a
magnification of 100X. What is the ASTM grain-size number of the metal if there are 64 grains per
square inch? (10 points).
Answer:
𝑛=
𝐿𝑜𝑔 𝑁
+1
𝐿𝑜𝑔 2
𝑛=
𝐿𝑜𝑔 64
𝐿𝑜𝑔 2
+1 = 7
𝑛=7
5. Calculate the energy for vacancy formation in Tantalum, given that the equilibrium number of
vacancies at 6500C is 2.18 * 1024 m-3. The atomic weight and density (at 6500 C) for Tantalum are
180.95 g/mol and 16.19 g/cm3, respectively. (10 points)
Answer:
𝑁=
𝑁=
(6.022 ×1023
𝑁𝐴 𝑝
𝐴𝐶𝑢
𝑎𝑡𝑜𝑚𝑠
𝑔
𝑐𝑚3
)(16.19 3 )(106 3 )
𝑚𝑜𝑙
𝑐𝑚
𝑚
180.95 𝑔/𝑚𝑜𝑙
= 5.39 × 1029 𝑎𝑡𝑜𝑚𝑠/𝑚3
𝑁𝑣 = 𝑁 exp(−𝑄𝑣 /𝑘𝑇)
𝑁𝑣 = 5.39 × 1029 𝑎𝑡𝑜𝑚𝑠/𝑚3 exp(?/(8.62 × 10−5 𝑒𝑉/𝐾)(923.15 𝐾) = 2.18 × 1024 𝑚−3
2.18 × 1024 𝑚−3 = 5.39 × 1029
ln =
𝑎𝑡𝑜𝑚𝑠
𝑒𝑉
exp(−𝑄𝑣 /(8.62 × 10−5 )(923.15 𝐾)
3
𝑚
𝐾
2.18 ×1024 𝑚−3
5.39 ×1029 𝑎𝑡𝑜𝑚𝑠/𝑚3
= ln = (exp (−12.57 𝑄𝑣
𝑎𝑡𝑜𝑚𝑠
))
𝑒𝑉
ln = (4.04 × 10−6 ) ln = (exp(−12.57 𝑄𝑉 𝑎𝑡𝑜𝑚𝑠/𝑒𝑉
−12.4193 = −12.57 𝑄𝑉
12.4193
12.57 𝑄𝑣
= 0.99
𝑒𝑉
𝑎𝑡𝑜𝑚
𝑄𝑣 = 0.99 𝑒𝑉/𝑎𝑡𝑜𝑚
6. What is the composition, in weight %, of an alloy that consists of 32 atomic % Au and 68 atomic % Pb?
(10 points)
Answer:
𝐶1 =
𝐶′1 ×
𝐶′1 × 𝐴1
0.32 × 196.97
63.03
× 100 =
× 100 =
× 100
′
𝐴1 + 𝐶 2 × 𝐴2
0.32 × 196.97 + 0.68 × 207.2
203.93
= 30.91 𝑤𝑡%
𝐶 ′ 2 × 𝐴2
0.68 × 207.2
140.90
𝐶2 = ′
×
100
=
×
100
=
× 100
𝐶 1 × 𝐴1 + 𝐶 ′ 2 × 𝐴2
0.68 × 207.2 + 0.32 × 196.67
203.93
= 69.1 𝑤𝑡%
7. In structural applications, why do we use alloys rather than pure metals? (5 points)
Answer: We use metal alloys rather than pure metals in structural applications because alloys help
advance the metal’s mechanical strength, as well as help the metal material be more resistant to
corrosive elements found in nature. Metal alloys will allow a structure just as a bridge or building to be
sturdier and more resistant to corrosion, which will in turn make the structure last longer.
8. Pure copper is strengthened by addition of small concentration of Be. What mechanism of
strengthening is this related to? (5 points).
Answer: The strengthening of pure copper by the addition of a small concentration of beryllium is a
method of strengthening known as solid solution strengthening/alloying.
9. A slab of steel is transformed into a car chassis. Does this process involve plastic or elastic
deformation? (5 points)
Answer: It involves plastic deformation, which, according to the materials science textbook, is a type of
deformation that is “both permanent and non-recoverable” because the deformation of a metallic
element are permanent beyond a certain point, unlike elastic deformation, which is not permanent and
can be recovered.
10. Why would metals behave as brittle materials without dislocations? (5 points)
Answer: Metals would behave as brittle materials without dislocations because dislocations in metals
allow the metal material to have some ductility, which according to the materials science textbook,
“allows the material to undergo plastic deformation without fracture of the material.” Also, dislocations
allow for more toughness in metals, which helps combat both brittleness and fracture.
11. What is added to an alumina crystal to turn it into a ruby crystal? (5 points)
Answer: According to the Christi Jordan’s article “How Do Rubies Form?” on ehow.com, alumina crystal
turns into a ruby crystal from a process that is known as “isomorphous” which creates corundum
(Jordan). According to Jordan, isomorphous occurs when aluminum ions are replaced with chromium
(Jordan). Furthermore, according to Jordan, elements such as chromium, iron, titanium, vanadium, or
rutile can form a ruby with mixed with aluminum oxide (Jordan). So, either chromium, iron, titanium,
vanadium, or rutile can be added to alumina crystal to create a ruby crystal.
12. Sterling silver consists of approximately 93 wt% silver and 7 wt% copper. Discuss all the ways by
which the addition of 7 wt% copper is beneficial. (5 points)
Answer: Sterling Silver is an metal alloy. As previously stated, alloys allow metals to have certain
characteristics that they would not achieve otherwise. According to the material science textbook, pure
silver in its natural environment is corrosive resistance, but is “very soft.” So, when pure silver is alloyed
with copper, the mechanical strength of the silver is greatly increased, making it harder. Also, the
addition of the copper does not hinder the corrosive resistance of the silver in the slightest according to
the materials science textbook.
13. Describe the structure of a grain boundary. Why are grain boundaries favorable sites for the
nucleation and growth of precipitates? (5 points)
Answer: According to the materials science textbook, a grain boundary is defined as “the interface that
separates two contiguous grains that have different crystallographic orientations.” Now the structure of
the boundary region between the grains is no more than several atoms distances wide according to the
textbook. Also, according to the textbook, there is always an atomic disparity in the transition from the
crystalline orientation. Grain boundaries are favorable sites for nucleation and the growth of
precipitates due to the relatively high grain boundary energy, which, according to the textbook is a
result of “the atoms being bonded less regularly along the grain boundary.” This high grain boundary
energy makes the grain boundaries more chemically reactive than the grains themselves, therefore
making the boundary an idea place for both nucleation and growth of precipitates to occur.
14. What are edge and screw dislocations? (5 points)
Answer: According to the materials science textbook, an edge location is “a linear crystalline defect that
is related to the lattice distortion created in the area of the end of an extra half-plane of atoms within
the crystal.” Furthermore, in the case of an edge location, the Burgers vector is perpendicular to the
dislocation line.
In contrast, a screw location is, according to the textbook, is “a linear crystalline defect that is related
with the lattice distortion that is created when normally parallel planes are joined together in order to
generate a helical ramp.” Furthermore, in the case of a screw location, the Burgers vector is parallel to
the dislocation line.
15. What are a solute and a solvent? Give 5 examples of a solute and a solvent from our daily lives? (5
points)
Answer: According to the materials science textbook, a solute can be defined as “one component or
element of a solution that is in a minor concentration, and is dissolved by the solvent.” Five examples of
solutes are sugars, salts, dissolvable powders such as laundry detergent and protein powder, flour, food
color dye, and nail polish.
According to the materials science textbook, a solution can be defined as “the compound of the solution
that is present in the greatest amount, as well as being the compound that dissolves the solute.” Five
examples of solutions are water (dissolves sugars, salts, dissolvable powders, flour, and food color dye);
milk (dissolves sugars, salts, dissolvable powders such as protein powder and chocolate powder, and
food color dye); acetone (dissolves nail polish and glue); ethanol (dissolves fats); and turpentine
(dissolves paint.)
Work Cited
Image of light microscope was taken from:
http://www.vetmed.vt.edu/education/curriculum/vm8054/labs/lab2/IMAGES/BRIGHTFIELD%20SCOPE
%20DIAGRAM.JPG
Image of electron microscope was taken from: http://www.purdue.edu/rem/rs/graphics/sem2.gif
Image of scanning probe microscope was taken from
http://www.google.com/imgres?imgurl=http://students.ycp.edu/~danders7/Instrument%2520Portfolio
s/SPM%2520Block%2520Diagram.jpg&imgrefurl=http://students.ycp.edu/~danders7/Instrument%2520
Portfolios/SPM%2520Portfolio.htm&h=315&w=350&sz=13&tbnid=Xkndd5kAiO6T0M:&tbnh=95&tbnw=
106&zoom=1&usg=__VKHE2CGTnrW5fWCy6_X7x3MaBH0=&docid=Yp1_5BB0QRhNFM&sa=X&ei=PuBV
UK_SOeqe2QXk7YAg&ved=0CEMQ9QEwAw&dur=262
Jordan, Christi. “How Do Rubies Form?” Ehow.com. taken 9/14/2012, from:
http://www.ehow.com/how-does_4963260_rubies-form.html