Homework Assignment 3
Suggested Problems – Chapter 4
HW#2 Problems for practice and grading are shown in the table below.
Chapter 4
1, 35, 11, 20, 29
KEY:
4.20
Let’s do a modified equation 4.9a (modified to return g/cm3 rather than Kg/m3) Thus:
CMo
16.4
g/ccMo 
 2.7627 g/ccMo
CW
16.4
83.6
 CMo




10.22
19.3
MO
W 



Now compute the number of moles of “Moly” per cc
2.7627/AMo = 2.7627/95.94 = 0.028796 moles
Therefore the Number of Atoms/cc =
#moles/cc*NA = .028796 * 6.023*1023 = 1.73*1022 atoms/cc
4.29
a.
b.
Surface energy must be higher than grain boundary energy. Along grain boundaries some
of the atoms will be able to bond across the boundary with other atoms thus G.B.’s will
be at somewhat lower energy than at a “free surface” where every atom will have
unfulfilled bonds.
For low angle grain boundaries the preceding argument is even more pronounced. That
is in low angle G.B.’s more atoms can find near (enough) neighbors to bond than is the
case with high angle G.B.’s.