MLE 1101 Fall 2008 Tutorial II

Introductory Materials Science and Engineering

MLE 1101 Fall 2008 Tutorial II

C02(E3-06-07); D06(E1-06-04); A03(E3-06-07)Office hour: 4:30-5:30 pm on Tuesdays, E4-8-11

Assistant Professor Gengchiau [email protected]

Note: Solution will be provided on my website after class(http://www.ece.nus.edu.sg/stfpage/elelg/ )

mailto:[email protected]


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Tutorial 2 (Ch3)1. Show for the body-centered cubic crystal structure that the unit cell edge length a and the atomic radius R are related through a = 4R/√3.ANS: CH3/P.42

BCC


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Tutorial 2 (Ch3)2. Calculate the radius of a vanadium atom, given that V has a BCC crystal structure, a density of 5.96 g/cm3, and an atomic weight of 50.9 g/mol.

3364

34 33

R=R = VC ⎟⎟⎠

⎞⎜⎜⎝

⎛

AC

V

NVAn

=ρ

( ) ( ) ( )( ) ( ) ( )

1 3

1 3

3 23

3 364

3 3 2 / 50.9 /

64 5.96 / 6.023 10 /

V

A

n ARN

atoms unit cell g mol

g cm atoms mol

ρ⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

⎡ ⎤⎢ ⎥=⎢ ⎥×⎣ ⎦

•This problem asks for us to calculate the radius of a vanadium atom. For BCC, n = 2 atoms/unit cell, and

Since,

=1.32 × 10-8 cm =0.132 nm

and solving for R


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Tutorial 2 (Ch3)

ANS: CH3/P.43

3. Niobium has an atomic radius of 0.1430 nm and a density of 8.57 g/cm3. Determine whether it has an FCC or BCC crystal structure.

In order to determine whether Nb has an FCC or BCC crystal structure, we need to compute its density for each of the crystal structures. For FCC, n = 4, and a=2R√2. Also, from Figure 2.6 (As shown in Text book), its atomic weight is 92.91 g/mol. Thus, for FCC

For BCC, n = 2, and , thus

which is the value provided in the problem. Therefore, Nb has a BCC crystal structure.

Volume of a unit cell


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Tutorial 2 (Ch3)

ANS: CH3/P.43

4. Magnesium has an HCP crystal structure, a c/a ratio of 1.624, and a density of1.74 g/cm3. Compute the atomic radius for Mg.

This problem calls for us to compute the atomic radius for Mg. In order to do this we must use Equation (3.5) (as shown in text book), as well as the expression which relates the atomic radius to the unit cell volume for HCP,

,Where c = 1.624(2R)

By substituting it into eq. 3.5, A B

C

2R

R R

R

Area=(1/2)x2Rx√3R


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Crystallographic Directions

1. Vector repositioned (if necessary) to pass through origin.

2. Read off projections in terms of unit cell dimensions a, b, and c

3. Adjust to smallest integer values4. Enclose in square brackets, no commas

[uvw]

ex: 1, 0, ½ => 2, 0, 1 => [ 201 ]

-1, 1, 1

families of directions <uvw>

z

x

Algorithm

where overbar represents a negative index

[111 ]=>

y

Presenter

Presentation Notes

Lecture 2 ended here


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Crystallographic Planes• Miller Indices: Reciprocals of the (three) axial

intercepts for a plane, cleared of fractions & common multiples. All parallel planes have same Miller indices.

• Algorithm1. Read off intercepts of plane with axes in

terms of a, b, c2. Take reciprocals of intercepts3. Reduce to smallest integer values4. Enclose in parentheses, no

commas i.e., (hkl)


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z

x

ya b

c

4. Miller Indices (110)

example a b cz

x

ya b

c


1. Intercepts 1 1 ∞2. Reciprocals 1/1 1/1 1/∞

1 1 03. Reduction 1 1 0

1. Intercepts 1/2 ∞ ∞2. Reciprocals 1/½ 1/∞ 1/∞

2 0 03. Reduction 2 0 0

example a b c

Crystallographic Planes


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z

x

ya b

c•

••


example1. Intercepts 1/2 1 3/4

a b c

2. Reciprocals 1/½ 1/1 1/¾2 1 4/3

3. Reduction 6 3 4

(001)(010),

Family of Planes {hkl}

(100), (010),(001),Ex: {100} = (100),

Crystallographic Planes


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Crystallographic Planes (HCP)

• In hexagonal unit cells the same idea is used

example a1 a2 a3 c

4. Miller-Bravais Indices (1011)

1. Intercepts 1 ∞ -1 12. Reciprocals 1 1/∞

1 0 -1-1

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3. Reduction 1 0 -1 1a2

a3

a1

z

Adapted from Fig. 3.8(a), Callister 7e.


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)100(),010(),001()001(),010(),100(Equivalent:

The family: {100}

10

0

0

=∞=∞=

cba

∞===

0

0

0

11

cba

111

0

0

0

===

cba

(001)

(110)

(111)


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Tutorial 2 (Ch3)

ANS: CH3/P.49-535. Within a cubic unit cell, sketch the following directions:

6. Sketch the (0111) and (2110) planes in a hexagonal unit cell.ANS: CH3/P.54

(2110) ½, -1, -1, ∞1st Step

2nd Step

½, -1, -1, ∞


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Tutorial 2 (Ch3)

ANS: CH3/P.49-537. Sketch within a cubic unit cell the following planes:


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Tutorial 2 (Ch3)8. (a)Derive the linear density expression for BCC [110] direction in terms of atomic radius R(b)Derive the planar density expression for BCC (110) plane in terms of atomic radius R

(a) In the figure below is shown a [110] direction within a BCC unit cell.

For this [110] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalence of 1 atom that is centered on the direction vector. The length of this direction vector is denoted by x in this figure, which is equal to

where y is the unit cell edge length, which, from Equation 3.3 is equal to 4R/sqrt(3)

...


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Furthermore, z is the length of the unit cell diagonal, which is equal to 4R Thus, using the above equation, the length x may be calculated as follows:

Therefore, the expression for the linear density of this direction is


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Thus, in terms of R, the area of this (110) plane is just

And, finally, the planar density for this (110) plane is just

(b)


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In chemistry, the term transition metal (sometimes also called a transition element) has two possible meanings:

• It commonly refers to any element in the d-block of the periodic table, including zinc, cadmium and mercury. This corresponds to groups 3 to 12 on the periodic table.

• More strictly, IUPAC defines a transition metal as "an element whose atom has an incomplete d sub-shell, or which can give rise to cations with an incomplete d sub-shell." By this definition, zinc, cadmium, and mercury are excluded from the transition metals, as they have a d10 configuration. Only a few transient species of these elements that leave ions with a partly filled d subshell have been formed, and mercury(I) only occurs as Hg2

2+, which does not strictly form a lone ion with a partly filled subshell, and hence these three elements are inconsistent with the latter definition.[1] They do form ions with a 2+ oxidation state, but these retain the 4d10 configuration. Element 112 may also be excluded although its oxidation properties are unlikely to be observed due to its radioactive nature. This definition corresponds to groups 3 to 11 on the periodic table.

In class, the first one is used to explain transition metal. However, the text book uses the 2nd definition. Therefore, let’s follow it.

http://en.wikipedia.org/wiki/Transition_metal#cite_note-0


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In group A03, I said that H is not metal (alkali metal) based on the old definition. In general, it is considered to only exist as the gas phase in nature. However, the recent discovery shows that the solid H can be found under extremely high pressure environment and persist the alkali metal properties. Therefore, it is also cataloged as Metal


Questions

1. why are there some covalent characters in BaS ? Similarly, why are there some ionic bondings in AlP?

2. Can I say fluctuating bonding and permanent bonding all belong to Van der Waal? For hydrogenal bonding is a particular example of permanent, hydrogenal bonding also belongs to Van der Waal, is that correct?


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Tutorial 2 (Ch3)9. The metal niobium has a BCC crystal structure. If the angle of diffraction for the (211) set of planes occurs at 75.99º (first order reflection) when monochromatic x-radiation having a wavelength of 0.1659nm is used, compute (a) the interplanar spacing for this set of planes and (b) the atomic radius for the niobium atom.

(a) From the data given in the problem, and realizing that 75.99° = 2θ, the interplanar spacing for the (211) set of planes for Nb may be computed using Equation 3.13 as follows:


Diffraction Phenomena:

Bragg’s Law:

θλ sin2 ⋅⋅= hkld

( )222 lkh

ad o

hkl++

=θθ sinsin ⋅=⇒= dMLd

ML

λθλθθ

λ

⋅=⋅⋅⋅=⋅+⋅

⋅=+

ndndd

nLNML

sin2sinsin

θ

M L

K

d

cubic


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(b) In order to compute the atomic radius we must first determine the lattice parameter, a, using Equation 3.14, and then R from Equation 3.3 since Nb has a BCC crystal structure. Therefore,

And from equation 3.3,

Introductory Materials Science and EngineeringIntroductory Materials Science and Engineering

• APF for a face-centered cubic structure = 0.74

ATOMIC PACKING FACTOR: FCC

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MLE 1101 Fall 2008 Tutorial II