HW KEY
MATL2100 HW Set 6
4/20/16
7.20 Briefly describe why small angle grain boundaries are not as effective in interfering
with the slip process as are high angle grain boundaries.
Small angle grain boundaries are not as effective at interfering with sli
HW KEY
MATL2100 HW Set 2
4/20/16
3.2 If the atomic radius of aluminum is 0.143 nm, calculate the
volume of its unit cell in cubic meters.
r = 0.143 nm (0.143 x 10-9) m
Aluminum is FCC, 4 atoms per unit cell.
(2)(a) = (4)(r) 0.572 nm
a = (.572 nm)(cos45*)
CHAPTER 6
MECHANICAL PROPERTIES OF METALS
PROBLEM SOLUTIONS
6.1
This problem asks that we derive Equations (6.4a) and (6.4b), using mechanics of materials principles. In Figure (a) below is shown a block element of material of cross-sectional are
CHAPTER 2
ATOMIC STRUCTURE AND INTERATOMIC BONDING
PROBLEM SOLUTIONS
2.1 (a) When two or more atoms of an element have different atomic masses, each is termed an isotope. (b) The atomic weights of the elements ordinarily are not integers because:
CHAPTER 3
THE STRUCTURE OF CRYSTALLINE SOLIDS
PROBLEM SOLUTIONS
3.1 Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the
CHAPTER 5
DIFFUSION
PROBLEM SOLUTIONS
5.1 Self-diffusion is atomic migration in pure metals-i.e., when all atoms exchanging positions are of the same type. Interdiffusion is diffusion of atoms of one metal into another metal.
5.2 Self-diffusion m
CHAPTER 4
IMPERFECTIONS IN SOLIDS
PROBLEM SOLUTIONS
4.1 In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation (4.1). As stated in the problem, Q = 0.55 eV/atom. Thus, v
N
N
V = exp
Q
V
kT
=