KMU 396

MATERIALS SCIENCE AND TECHNOLOGY I

 

Midterm Examination

Thursday, May 8, 2008

 

Solutions

 

Question 1. (18 pts)

a) (14 pts.)

§     For brass, the bonding is metallic since it is a metal alloy.

§     For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)

§     For BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of Ba and S in the periodic table.

§     For solid xenon, the bonding is van der Waals since xenon is an inert gas.

§     For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin).

§     For nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.)

§     For AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of Al and P in the periodic table.

 

b) (4 pts) Since the hydrogen bond is stronger than van der Waals, HF will have a higher

melting temperature ) (19.4 vs. -85ºC)

 

 

Question 2. (10 pts.)

Beryllium and magnesium, both in the 2A column of the periodic table, are lightweight

metals. Which would you expect to have the higher modulus of elasticity?

Explain, considering binding energy and atom radii and using appropriate sketches

of force versus interatomic spacing.

Solution: 4 Be 1s22s2  E=42x106 psi rBe =1.143 Å

12 Mg 1s22s22p63s2 E =6x106 psi rMg =1.604 Å

Be has higher melting and boiling T than Mg as shown on the periodic table on pg.32 meaning Be has higher binding energy. In addition, Be atoms are smaller than Mg atoms. Therefore smaller Be electrons are held closer to the core therefore held more tightly, giving a higher binding energy.

                                                         

                          Be

 

 

 

 

Question 3. (12 pts.)

In this problem we are asked to cite which of the elements listed form with Cu the three possible solid solution types.  For complete substitutional solubility the following criteria must be met:  1) the difference in atomic radii between Ni and the other element (ΔR%) must be less than ±15%, 2) the crystal structures must be the same, 3) the electronegativities must be similar, and 4) the valences should be the same, or nearly the same.  Below are tabulated, for the various elements, these criteria.

 

                                                             Crystal           ΔElectro-                

                 Element         ΔR%             Structure          negativity           Valence

 

                 Cu                                         FCC                                          2+

                 C                    -44

                 H                    -64

                 O                    -53

                 Ag                  +13                 FCC                   0                     1+

                 Al                   +12                 FCC                 -0.4                   3+

                 Co                   -2                   HCP                 -0.1                   2+

                 Cr                    -2                  BCC                 -0.3                   3+

                 Fe                    -3                  BCC                 -0.1                   2+

                 Ni                    -3                   FCC                 -0.1                   2+

                 Pd                   +8                  FCC                 +0.3                  2+

                 Pt                    +9                  FCC                 +0.3                  2+

                 Zn                   +4                  HCP                 -0.3                   2+

 

       (a)  Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete solubility.

       (b)  Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility.  All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Ni are greater than ±15%, and/or have a valence different than 2+.

       (c)  C, H, and O form interstitial solid solutions.  These elements have atomic radii that are significantly smaller than the atomic radius of Cu.

 

 

Question 4. (20 pts.)

a) A material in which atomic bonding is predominantly ionic in nature is less likely to form a noncrystalline solid upon solidification than a covalent material because covalent bonds are directional whereas ionic bonds are nondirectional;  it is more difficult for the atoms in a covalent material to assume positions giving rise to an ordered structure.

 

b) A crystal structure is described by both the geometry of, and atomic arrangements within, the unit cell, whereas a crystal system is described only in terms of the unit cell geometry.  For example, face-centered cubic and body-centered cubic are crystal structures that belong to the cubic crystal system.

 

c) The vacancy concentration in a crystal structure increases with __temperature_.

d) The surface energy of a single crystal depends on crystallographic orientation because the atomic packing is different for the various crystallographic planes, and, therefore, the number of unsatisfied bonds will vary from plane to plane.

 

e)  The surface energy will be greater for an FCC (100) plane than for a (111) plane because the (111) plane is more densely packed (i.e., has more nearest neighbor atoms in the plane); as a consequence, more atomic bonds will be satisfied for the (111) plane, giving rise to a lower surface energy.

 

 

Question 5. (20 pts.)

 

Peak No

2θ (degrees)

sin2θ

sin2θ/0.1391

h2+k2+l2

(hkl)

dhkl (nm)

1

43.8

0.1391

1

3

(111)

0.2067

2

50.8

0.1840

1.3227

4

(200)

0.1797

3

74.4

0.3655

2.6279

8

(220)

0.1275

4

90.4

0.5035

3.6196

11

(311)

0.1087

a)  For each peak, in order to calculate the interplanar spacing we must employ Bragg’s law.  For the first peak which occurs at 43.8o

 

b) The first peak results by diffraction from (111) planes.

 

c) The h2+k2+l2 order of 3,4,8,11 is for an FCC crystal structure in which only those peaks for which h, k, and l are all either odd or even will appear. Consequently, the material could be Cu (Fe has BCC and Zn has HCP structure)

 

d)  Computation of atomic radius R for FCC crystal structure is given by:

    a=d111(1+1+1)1/2=0.2067(3)1/2=0.3580 nm

    a=d200(4+0+0)1/2=0.1797(2)=0.3594 nm

 

       Similar computations are made for the other peaks which results are tabulated below:

 

                   Peak Index           2q                 dhkl(nm)              R (nm)

                   200                     50.8                0.1797              0.1271

                   220                     74.4                0.1275              0.1275

                   311                     90.4                0.1087              0.1274

 

 

Question 6 (20 pts.)

a)      CRS, HRS, Brass, Al. Al has fcc, Brass has hcp, CRS and HRS have bcc structure. As CRS is cold worked its hardness is enhanced.

 

b)      CRS and HRS have ductile to brittle transition as they have bcc crystal structure. There are no closed packed planes, and thus, no slip planes in the bcc structure whereas fcc has many and hcp has some (three) slip planes along which dislocations can move. At very low temperatures, atoms are frozen with very small internal energy and, therefore, cannot take up the impact energy and disperse it in vibrational, rotational, and other forms. If there are no slip planes dislocations cannot move and increase the internal stresses resulting in breakage. As fcc and hcp structures have slip places, they do not present such a transition.

 

c)      The hardest material in the list would be the best choice for a knife-blade. Therefore, cold rolled steel is the best choice.

 

d)      The toughest material, hot rolled steel, would be best for a construction nail as nails should be able to resist high impact.