A general note on answering assignment questions: your answer should never contain more significant digits than were supplied in the data given. For example, if the question says that a sample was annealed for 10 hours, you can't assume that this figure is accurate to the closest millisecond. Even assuming it's accurate to within five minutes may be over-optimistic. So your answer should not be given to six significant figures -- this claims a degree of precision which you can't justify, even if your calculations are faultless.
Question 4.2
Calculate the number of vacancies per cubic metre in gold at 900 C. The energy for vacancy formation is 0.98 eV/atom. Furthermore, the density and atomic weight for gold are 19320 kg/cubic metre and 196.9 gm/mol respectively.
Answer 4.2
This involves evaluating the equation
Nv=N exp(Qv/kT))
We first find N, using the equation
N = NAp/A = 6.023*1023 * 193200 / 196.9 = 5.9* 1028 atoms/m3.
Then substitute this into the former equation to get:
Nv= 5.9*1028*exp(-0.98/(8.62*10-5* 1173)) = 3.65 * 1024 vacancies/cubic metre.
Question 4.7
What is the composition, in atom percent, of an alloy that consists of 92.5% silver and 7.5% copper, by weight?
Answer 4.7
(There are various formulas offered in Callister for calculating this kind of thing, but the best approach is to figure it out from first principles:)
Suppose we have 1 kg of the alloy. Then we have 0.925 kg of silver and 0.075 kg of copper. So we have 0.925/As moles of silver, and 0.075/ Ac moles of copper, where As and Ac are the atomic weights of silver and copper respectively. So the total number of moles present are 0.925/As + 0.075/Ac. So the fraction, in atom percent, made up of silver must be
(0.925/As)(0.925/As + 0.075/Ac) = 0.925/(0.925 + 0.075As/Ac)
As = 107.87 and Ac = 63.55, so this evaluates to 87.9% silver and 12.1% copper.
Question 4.21
Gold forms a substitutional solid solution with silver. Compute the weight percent of gold that must be added to silver to yield an alloy that contains 5.5 x 1027 gold atoms per cubic meter. (The densities of pure gold and silver are 19320 and 10490 kg/cubic meter respectively.)
Answer 4.21
First off, consider a cubic meter of pure silver. This contains 10490 kg of silver, which is 10490 * Na / As atoms, where As is the atomic weight of silver, 108, and Na is Avogadro's number, 6.023 * 1026 atoms/kg-mole. So this evaluates to 58.5* 1027 atoms. Now, turn 5.5*1027 of these atoms to gold. Then total mass of gold = 5.5*1027*Ag/Na kg, where Ag is the atomic weight of gold, 197 amu. So there is a total of 179.9 kg of gold in the alloy. And there is (58.5-5.5)*1027* As/Na = 951.1 kg silver. So the fraction of gold by weight is 179.9/(179.9+951.1) = 15.9% by weight of gold.
Question 4.24
Cite the relative Burgers vector dislocation line for edge, screw, and mixed dislocations.
Answer 4.24
For screw dislocations, the Burger's vector is parallel to the dislocation line. For edge vectors, it is perpendicular; and for mixed dislocations, it is neither perpendicular nor parallel.
Question 5.7
A sheet of metal 1.5 mm thick has nitrogen atmospheres on both sides at 1200 C, and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6 x 10-11 m2/s, and the diffusion flux is found to be 1.2x10-7 kg/m2-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2 kg/m3? (Assume a linear concentration profile.)
Answer 5.7
To solve this, we need to work out the concentration of nitrogen in the steel on the low-pressure side; call this N. If we assume a linear concentration profile, we can do this from Fick's first law:
J = -D dC/dx
Hence 1.2 * 10-7 = 6 * 10-11 * (4 - N)/0.0015
This can be solved to get N = 1 kg/m3. So the concentration gradient is 1 kg/m^3 per 0.0005 m, so the concentration drops from 4 to 2 in 0.001 m, which is 1 mm.
Second Thoughts on Answer 5.7
Come to think of it, we don't need to know the concentration of nitrogen on the low pressure side at all. We can just apply Fick's Law to the portion of the sheet between the high-pressure side and the (unknown) cross-section at which the concentration is 2 kg/m3. If we suppose that this cross-section lies at a distance x from the high-pressure surface, we can write:
1.2 * 10-7 = 6 * 10-11 * (4 - 2)/(x - 0)
from which x can be found directly.
Question 5.13
Nitrogen from a gaseous phase is to be diffused into pure iron at 700 C. If the surface concentration is to be maintained at 0.1 wt% N, what will be the concentration 1 mm from the surface after 10 h? The diffusion coefficient for nitrogen in iron at 700 C is 2.5 x 10-11 m2/s.
Answer 5.13
This is a time-varying problem, so we need Fick's second law. Since the question involves diffusion into a semi-infinite slab, we can use the error function solution,
(Cx - C0)/ (Cs-C0)=1-erf(x/(2 sqrt(Dt)))
The question says `pure iron', so C0 = 0. Cs = 0.1%, and Cx is what we have to find.
The square root of Dt is 0.000949 m, so x/(2sqrt(Dt)) = 0.5269. Looking this up in Callister's table of error function values and interpolating, I get that erf(0.5269) = 0.544
So Cx/Cs = 1 - 0.544 = 0.456
So Cx = 0.456 * 0.1 = 0.0456% by weight nitrogen.
Question 5.31
An FCC iron-carbon alloy initially containing 0.1wt% carbon is carburised at an elevated temperature and in an atmosphere in which the surface carbon concentration is maintained at 1.1 wt%. If after 48 hours the concentration of carbon is 0.3 wt% at a position 3.5 mm below the surface, determine the temperature at which the treatment was carried out.
Answer 5.31
This uses the error function solution again. This time, we have
(Cx - C0)/ (Cs - C0)= (0.3 - 0.1) / (1.1 - 0.1) = 0.2.
So 1-erf(x/(2sqrt(Dt))) = 0.2. So erf(x/(2sqrt(Dt))) = 0.8, and, from Callister's table,
x/(2 sqrt(Dt)) = 0.9
0.0035/sqrt(D * 48 * 3600) = 2 * 0.9
(0.0035/1.8)2 = D * 48 * 3600
From which we get that D = 2.188 * 10-11 m2.s. However, the question doesn't ask for D, but for the temperature. So we will need to use the diffusion data Callister provides in Table 5.2. We note that for carbon diffusing through FCC iron, D0 = 2.3 * 10-5 m2.s, and Qd = 148000J/mol. So
D0exp(-Qd/RT) = 2.3 * 10-5exp(148000/(8.31*T)) = 2.188 * 10-11=D
So
exp(-148000/(8.31T)) = 0.9517 * 10-6
-148000/8.31T = - 13.86
8.31T = 148000/13.86
T = 148000/(13.86 * 8.31) = 1284 K = 1011 C.
(This is hot, but not unbelievably hot.)