Solutions Midterm #1 CEMM 260/261 Spring 2005
1 A. Al = 4 x 27 amu/a3 A = 2 x 21/2 x 0.143 nm = 0.404 nm
Al = 1632 amu/nm3
1632 amu/nm3 /(6.02 x 1023 amu/g x (10-9 m/nm)3) = 2.71x 106 g/m3 = 2.71 g/cm3
B. Planar density in (100) = 2 atoms/a2 = 12.22 atoms/nm2
C. Linear density in [111] = 1 atom/31/2 a = 1.43 atoms/nm
D. No. The crystal structure and the atomic dimensions match, as required by the
Hume-Rothery rules, but the electonegativities of the atoms differ, so
compounds will form rather than a solid solution.
2. ln D500 = ln 4 x 10-18 = ln Do –Q/(R x 773)
-40.06 = ln Do –189,000/(8.314 x 773)
-40.06 = ln Do –29.41
ln Do = -10.65
Do = 2.37 x 10-5
D400 = 2.37 x 10-5 x exp (-189,000/(8.314 x 573)) = 5.06 x 10-20 m2/s
3. E = 1200MPa/0.012 = 105 MPa = 100 GPa +/- 50%
Y.S = 860 MPa +/- 50 MPa
T.S. = 1,000 MPa +/- 50 MPa
%Elongation = 11.2% +/- 1% All read from graph
4. A. Simple cubic structures are not close packed. The non-directional
bonds in metallic materials tend to pack atoms densely, favoring fcc or
bcc structures.
B. These are both close packed directions, that is, the atoms are touching one
another along these directions.
C.  =  cos  cos 
MPa =  cos 60 cos 35 =  x 0.409,  = 2.44 MPa
D. Metallic bonds contain mobile electrons that can move in response to
electrical fields and can carry thermal energy. Polymers contain only
covalent and secondary bonds, so all of the electrons are localized with the
atoms and cannot move in response to an electrical field.
E. The ratio will be higher at 600oC than at 700oC because the activation
energy for grain boundary diffusion is smaller than that for lattice
diffusion. Therefore, when the temperature is reduced by 100oC, the
lattice diffusion coefficient will decrease by more than the grain boundary
diffusion coefficient does and the ratio between these two quantities will
increase.
F. Annealing heat treatments allow heavily cold worked grains to recrystallize and remove the effects of cold work, reducing the yield strength
of the metal.