CHAPTER 5
5‐4
ATOM AND ION MOVEMENTS IN MATERIALS
Atoms are found to move from one lattice position to another at the rate of 5 of 5 x 105 jumps per second at 400°C when the activation energy for their movement is 30,000 cal/mol. Calculate the jump the jump rate at 750°C.
Solution:
5 1 0 30,000/1.987637 30,000/1.9871023 22 22.4 .434 34 14.7 14.759 59 510 5 exp‐7.6754.6410 ‐4 x
5 1 0 1.08 10 / 4.64 10 5‐5
The number of vacancies of vacancies in a material is related to temperature by an Arrhenius equation. If the fraction of lattice of lattice points containing vacancies is 8 x 10‐5 at 600°C, determine the fraction at 1000°C.
Solution: 8 1 0 /1.987 /1.987837 837
16,3 16,364 64 / /
/ 16 16,36 ,364/ 4/1.9 1.987 8712 1273 73 0.0015 0.001555 5‐6
The diffusion coefficient for Cr+3 in Cr2O3 is 6 x 10‐15 cm2/s at 727°C and is 1 x 10‐9 cm2/s at 1400°C. Calculate (a) the activation energy and (b) the constant Do.
Solution:
6 1 0 /1.9871000 1 1 0 /1.9871673 6 1 0 0 0.0 .000 0050 5033 0.00 0.0003 030 0 0. 0.00 0002 0203 03 12. 12.02 0244 0.0 0.000 0020 2033 59,2 59,230 30 / / 1 10 59 59,23 ,230/ 0/1.9 1.987 8716 1673 73 17.818 1 1 0 1.828 10 0.05 0.0555 / 5‐7
The diffusion coefficient for O‐2 in Cr2O3 is 4 x 10‐15 cm2/s at 1150°C and 6 x 10‐11 cm2/s at 1715°C. Calculate (a) the activation energy and (b) the constant Do.
Solution:
4 1 0 /1.9871423 6 1 0 /1.9871988 ©2009. Cengage Learning, Engineering. All Rights Reserved
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6.67 10 0.0001005 9.615 0.0001005 95,700 / 4 1 0 95,7000/1.9871423 2.02 10 1.98 / 5‐12
A 0.2 mm thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contains 1 Sb atom per 108 Si atoms and the other surface contains 500 Sb atoms per 108 Si atoms. The lattice parameter for Si is 5.407 Å (Appendix A). Calculate the concentration gradient in (a) atomic percent Sb per cm and (b) Sb atoms/cm3‐ cm.
Solution:
1/10 500/10 ∆/∆ 100% 0.02495 % / 0.02 5.4307 Å 160.16 10 8 /. . 1 /10 0.04995 10 / 160.16 10 /.. 8 /. . 500 /10 24.975 10 / 160.16 10 /.. 0.04995 24.975 10 ∆/∆ 1.246 10 / 0.02 5‐13
When a Cu‐Zn alloy solidifies, one portion of the structure contains 25 atomic percent zinc and another portion 0.025mm away contains 20 atomic percent zinc. If the lattice parameter for the FCC alloy is 3.63 x 10‐8cm, determine the concentration gradient in (a) atomic percent Zn per cm, (b) weight percent Zn per cm, and (c) Zn atoms/cm3. cm
Solution:
∆/∆
20% 25% 2000 % / 0.025 0.1 /
(b) We now need to determine the wt% of zinc in each portion:
%
2065.38 / 100 20.46 2065.38 8063.54
%
2565.38 / 100 25.54 2565.38 7563.54
∆/∆
20.46% 25.54% 2032 % / 0.0025 ©2009. Cengage Learning, Engineering. All Rights Reserved
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(c) Now find the number of atoms per cm3:
4 /0.2 0.0167 10 / 3.63 10
4 /0.25 0.0209 10 / 3.63 10
0.0167 10 0.0209 10 ∆/∆ 1.68 / 0.0025 5‐14
A 0.0025 cm BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 650°. 5 x 108 H atoms/cm3 are in equilibrium with the hot side of the foil, while 2 x 103 H atoms/cm3 are in equilibrium with the flux of hydrogen through the foil.
Solution:
2 1 0 5 1 0 ∆/∆ 1969 10 / 0.0025 ∆/∆ 0.0012 3600/1.9879231969 10 0.33 10 / 5‐15
A 1 mm sheet of FCC iron is used to contain nitrogen in a heat exchanger at 1200°C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. Determine the flux of nitrogen through the foil in atoms/cm2‐ s.
Solution:
0.00005 0.00044 /3.589 10 ∆/∆ 1 0.1 / 3.03 10 / ∆/∆ 0.0034 34,6000/1.98714733.03 10 7.57 10 / 5‐16
A 4 cm‐diameter, 0.5 mm‐thick spherical container made of BCC iron holds nitrogen at 700°C. The concentration at eh inner surface is 0.05 atomic percent and at the outer surface is 0.002 atomic percent. Calculate the number of grams of nitrogen that are lost from the container per hour.
Solution:
∆/∆
0.00002 0.00052 //2.866 10 0.5 0.1 / ©2009. Cengage Learning, Engineering. All Rights Reserved
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8.16 10 / 0.0047 18,300/1.9879738.16 10 2.97 10 / 4 42 50.27
3600 /
/ 2.97 10 50.273600 5.37 10 / 5.37 10 14.007 / 1.245 10 / 6.02 10 / 5‐17
A BCC iron structure is to be manufactured that will allow no more than 50 g of hydrogen to be lost per year through each square centimeter of the iron at 400°C. If the concentration of hydrogen at one surface is 0.05 H atom per unit cell and is 0.001 H atom per unit cell at the second surface, determine the minimum thickness of the iron.
Solution: 0.05 /2.866 10 212.4 10 /
0.001 /2.866 10 4.25 10 / 4.25 10 212.4 10 2.08 10 ∆/∆ ∆ ∆ 50 / 6.02 10 / 1.00797 /31.536 10 / 9.47 10 / 2.08 10/∆0.00123600/1.987637 ∆ 0.179 5‐18
Determine the maximum allowable temperature that will produce a flux of less than 2000 H atoms/cm2‐s through a BCC iron foil when the concentration gradient is ‐5 x 1016 atoms/cm3‐ cm. (Note the negative sign for the flux).
Solution: 2000 / 0.0012 3600/1.9875 10 /
3.33 10 3600/1.987 3600/24.121.987 75 198 5‐19
As mentioned before in example 5‐6, the diffusion of yttrium ions in chromium oxide (Cr2O3) has been studied Lesage and co‐workers (Ref. J.Li, M.K.Loudjani, B.Lesage, A.M.Huntz: Philosophical Magazine A, 1997, 76[4], 857‐69). In addition to the measurement of diffusion of yttrium ion in bulk chromia scale grown on a Ni‐Cr alloy, these researchers also measured the diffusion of yttrium along the grain boundaries. These data are for grain boundary diffusivities are shown below. ©2009. Cengage Learning, Engineering. All Rights Reserved
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Temperature (C)
Grain Boundary Diffusion Coefficient (D) (cm 2/s) 1.2 × 10-13 5.4× 10-13 6.7× 10-13 1.8 × 10-12 4.6 × 10-12
800 850 900 950 1000
(a) From these data show that the activation energy for grain boundary diffusion of yttrium in chromia oxide scale on nickel‐chromium alloy is 190 kJ/cal. (b) What is the value of the prexponetial term D0 in cm2/s? (c) What is the relationship between D and 1/T for the grain boundary diffusivity in this temperature range? (c) At any given temperature, the diffusivity of chromium along grain boundaries is several orders of magnitude higher than that for within the bulk (See Example 5‐4). Is this to be expected? Explain. Solution:
ln D
=
ln D 0 −
Q RT
Temperature C
800
850
900
Temp(K)
1/T (K‐1)
1073 0.000932
1123
0.00089
1173 0.000853
D cm2/s
ln D
1.22E‐ 13
‐ 29.7348
5.4E‐13
‐ 28.2472
6.7E‐13
‐ 28.0315
950
1223 0.000818
1.8E‐12
‐ 27.0432
1000
1273 0.000786
4.6E‐12
‐26.105
Fitting the data of ln D vs. 1/T gives us a slope of ‐23132.9, this means the activation energy is ~ 46 kcal/mole or 190 kJ/mol.
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The intercept for data of ln D vs. 1/T is ‐8.0390, this is the value of ln(D0 ). This corresponds to value of D0 = 0.000322612 or 3.23 × 10 4 cm2 /s. ‐
One way to write this relationship is: D = 3.23 × 10 4 cm2 /s exp ( ‐190,000 Joules/RT) ‐
Yes, in general at grain boundaries there is more disorder and hence diffusion of atoms or ions is faster compared to that in bulk.
5‐20
Certain ceramic materials such as those based on oxides of yttrium, barium, and copper have been shown to be superconductors near liquid nitrogen temperature (~ 77‐110 K). Since ceramics are brittle it has been proposed to make long wires of these materials by encasing them in a silver tube. In this work, researchers investigated the diffusion of oxygen in a compound YBa2Cu3O7 (D.K.Aswal, S.K.Gupta, P.K.Mishra, V.C.Sahni: Superconductor Science and Technology, 1998, 11[7], 631‐6.) The data are in the temperature range 550 to 750 C are shown below for undoped (i.e. silver free) samples. Diffusion Coefficient (D) (cm 2/s)
Temperature (C)
2.77 × 10-6 5.2 × 10-6 9.24 × 10-6
500 600 650
Assume that these data are sufficient to make a straight line fit for the relationship between ln (D) and 1/T and calculate the values of the activation energy for diffusion of oxygen in YBa2Cu3O7 containing no silver. Solution: The data for ln (D) vs. 1/T are shown below. Temperature C 500 600 650
Temp(K) 773 873 923
1/T (K-1) 0.001294 0.001145 0.001083
D cm2/s 0.00000277 0.0000052 0.00000924
ln D -12.7967 -12.1669 -11.592
When fitted to straight line the data for ln (D) vs. 1/T gives a slope of ‐5460.33, this corresponds to an activation energy value of 10.84 kCal/mol or ~ 45 kJ/mol. 5‐21
Diffusion of oxygen in YBa2Cu3O7 doped with silver was also measured (D.K.Aswal, S.K.Gupta, P.K.Mishra, V.C.Sahni: Superconductor Science and Technology, 1998, 11[7], 631‐6.) It was seen that the diffusion of oxygen was slowed down by silver doping as shown in the data below.
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Temperature (C)
Diffusion Coefficient (D) (cm2/s) 2.89 × 10‐7 8.03 × 10‐7 3.07× 10‐6
650 700 750
Ideally more data points would be better. However, assume that these data are sufficient to make a straight line fit for the relationship between ln (D) and 1/T and calculate the values of the activation energy for diffusion of oxygen in YBa2Cu3O7 containing silver. Solution: The data for ln D vs. 1/T are as shown below. Temperature C 650 700 750
Temp(K) 1/T (K‐1) 923 0.001083 973 0.001028 1023 0.000978
D cm2/s 2.89E‐07 8.03E‐07 8.03E‐06
ln D ‐15.0568 ‐14.0349 ‐11.7323
The slope of a straight line fitted to ln (D) vs. 1/t is – 31156.5. This corresponds to an activation energy (Q) of ~ 62 kCal/mol or 259 kJ/mol. 5‐22
Zinc oxide (ZnO) ceramics are used in a variety of applications such as surge protection devices. The diffusion of oxygen in single crystals of ZnO was studied by Tomlins and co‐ workers (G.W.Tomlins, J.L.Routbort, T.O.Mason: Journal of the American Ceramic Society, 1998, 81[4], 869‐76). These data are shown in the table below.
Temperature (C)
Diffusion Coefficient (D) (cm2/s) 2.73 × 10‐17 8.20 ×10‐17 2.62 ×10‐15 2.21 ×10‐15 5.48 × 10‐15 4.20 ×10‐15 6.16 × 10‐15 1.31 ×10‐14 1.97 × 10‐14 3.50 × 10‐14
850 925 995 1000 1040 1095 1100 1150 1175 1200
(a) Using these data calculate the activation energy for the diffusion of oxygen in ZnO. What is the value of D 0 in cm2/s.
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Solution: These data are shown below. Temperature C 850 925 995 1000 1040 1095 1100 1150 1175 1200
Temp(K) 1123 1198 1268 1273 1313 1368 1373 1423 1448 1473
1/T (K-1) 0.000890472 0.000834725 0.000788644 0.000785546 0.000761615 0.000730994 0.000728332 0.000702741 0.000690608 0.000678887
D cm2/s 2.73E-17 8.2E-17 2.62E-15 2.21E-15 5.48E-15 4.2E-15 6.16E-15 1.31E-14 1.97E-14 3.5E-14
ln D -38.1396 -37.0398 -33.5756 -33.7458 -32.8377 -33.1037 -32.7207 -31.9662 -31.5582 -30.9834
These values of ln(D) vs. 1/t when fitted to straight ‐line give a value of activation energy ~ 275 ‐ kJ/mol or 65.6 kcal/mol. The value of D0 is 2.05 × 10 4 cm2 /s.
5‐23
Amorphous PET is more permeable to CO2 the PET that contains microcrystallites. Explain why.
Solution: The microcrystallites in PET provide a more compact structure compared to amorphous PET. 5‐24
Explain why a polymer balloon filled with helium gas deflates over time.
Solution: Helium atoms diffuse through the chains of the polymer material due to the small size of the helium atoms and the ease at which they diffuse between the loosely ‐ packed chains. 5‐26
Compare the diffusion coefficients of carbon in BCC and FCC iron at the allotropic transformation temperature of 912°C and explain the difference.
Solution: 0.011 20,900/1.9871185 1.51 10 /
0.23 32,900/1.9871185 1.92 10 / Packing factor of the BCC lattice (0.68) is less than that of the FCC lattice; consequently atoms are expected to be able to diffuse more rapidly in the BCC iron.
5‐28
A carburizing process is carried out on a 0.10% C steel by introducing 1.0% C at the surface at 980°C, where the iron is FCC. Calculate the carbon content at 0.01 cm, 0.05 cm, and 0.10 cm beneath the surface after 1 h.
Solution:
0.23 32,900/1.9871253 42 10 / 1 /2√42 10 3600 /0.778 10.1 ©2009. Cengage Learning, Engineering. All Rights Reserved
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5‐29
Iron containing 0.05% C i heated to 912°C in an atmosphere that produces .20% C at the surface and is held for 24 H. Calculate the carbon content at 0.05 cm beneath the surface if (a) the iron is B C and (b) the iron is FCC. Explain the ifference.
Solution:
Faster diffusion occurs in the looser p cked BCC str ucture, leadi g to the hig er carbon content at point “x”. 5‐30
What temp rature is required to obtain 0.50% C at a distance of 0.5 mm beneath the sur ace of a 0.20%
steel in 2 h.
hen 1.10% C is present t the surface? Assume th t the iron is FCC. ©2009. Cengag Learning, Engineering. All Rights Reserved
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Solution:
1.1 0.5 0.667 0.05/2√ 1.1 0.2 0.05/2√ 0.685 √ 0.0365 0.00133 23600 / 7200 0.00133/7200 1.85 10 0.23 32,900/1.987 16,558/ 8.043 10 1180 907 5‐31
A 0.15% C steel is to be carburized at 1100°C, giving 0.35% C at a distance of 1 mm beneath the surface. If the surface composition is maintained at 0.90% C, what time is required.?
Solution:
0.9 0.35 0.733 0.1/2√ 0.9 0.15 0.1/2√ 0.786 √ 0.0636 0.00405 0.23 32,900/1.9871373 1.332 10 / 0.00405/1.332 10 3040 51 5‐32
A 0.02% C steel is to be carburized at 1200°C in 4 h, with a point 0.6 mm beneath the surface reaching 0.45% C. Calculate the carbon content required at the surface of the steel.
Solution:
0.45 0.06/2√ 0.02 0.23 32,900/1.9871473 3.019 10 / 4 3600 14,400 √ √3.019 1014,400 0.2085 0.06/20.2085 0.144 0.161 0.45 0.161 0.53% 0.02 5‐33
A 1.2% C tool steel held at 1150°C is exposed to oxygen for 48 h. The carbon content at the steel surface is zero. To what depth will the steel be decarburized to less than 0.20% C?
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Solution:
00.2 0.1667 /2√ 0.149 01.2 0.23 32,900/1.9871423 2.034 10 / 48 3600 / 17.28 10 √ 0.5929 Then from above,
5‐34
0.14920.5929 0.177
A 0.80% C steel must operate at 950°C in an oxidizing environment, where the carbon content at the steel surface is zero. Only the outermost 0.02 cm of the steel part can fall below 0.75% C. What is the maximum time that the steel part can operate?
Solution:
00.75 0.9375 /2√ /2√ 1.384 00.8 0.02/2√ 1.384 √ 0.007226 5.22 10 0.23 32,900/1.9871223 3.03 10 / 5.22 10 /3.03 10 172 2.9 5‐35
A steel with BCC crystal structure containing 0.001% N is nitride at 550°C for 5 h. If the nitrogen content at the steel surface is 0.08%, determine the nitrogen content at 0.25 mm from the surface.
Solution:
0.08 0.025/2√ 0.08 0.001 5 3600 / 1.8 10 0.0047 18,300/1.987823 6.488 10 / √ 0.0342 0.025/20.0342 0.3655 0.394 0.08 0.394 0.049% 0.079 5‐36
What time is required to nitride a 0.002% N steel to obtain 0.12% N at a distance of 0.005 cm beneath the surface at 625°C? The nitrogen content at the surface is 0.15%.
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Solution:
0.15 0.12 0.2027 /2√ /2√ 0.2256 0.15 0.002 0.0047 18,300/1.987898 1.65 10 / 0.005 0.005 0.2256 2√1.65 10 1.267 10 1.267 10 /1.65 10 768 12.8 5‐37
We can successfully perform a carburizing heat treatment at 1200°C in 1 h. In an effort to reduce the cost of the brick lining in our furnace, we propose to reduce the carburizing temperature to 950°C. What time will be required to give us a similar carburizing treatment?
Solution:
0.23 32,900/1.9871473 3.019 10 / 0.23 32,900/1.9871223 3.034 10 / 1 / 5‐39
3.019 101 9.95 3.034 10
During freezing of a Cu‐Zn alloy, we find that the composition is nonuniform. By heating the alloy to 600°C for 3 hours, diffusion of zinc helps to make the composition more uniform. What temperature would be required if we wished to perform this homogenization treatment in 30 minutes?
Solution:
0.78 43,900/1.987873 7.9636 10 3 0.5 / 7.9636 10 3/0.5 4.778 10 0.78 43,900/1.987 6.1258 10 23.516 43,900/1.987 940 667 5‐40
A ceramic part made of MgO is sintered successfully at 1700°C in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to 1500°C. Which will limit the rate at which sintering can be done – diffusion of the magnesium ions or diffusion of oxygen ions? What time will be required at the lower temperature? ©2009. Cengage Learning, Engineering. All Rights Reserved
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Solution: Diffusion of oxygen is the slower of the two, due to the larger ionic radium of the oxygen.
0.000043 82,100/1.9871973 3.455 10 / 0.000043 82,100/1.9871773 3.255 10 / 5‐41
3.455 1090 / 955 15.9 3.255 10
A Cu‐Zn alloy has an initial grain diameter of 0.01 mm. The alloy is then heated to various temperatures, permitting grain growth to occur. The times required for the grains to grow to a diameter of 0.30 mm are Temperature (°C) 500 600 700 800 850
Time (min) 80,000 3,000 120 10 3
Determine the activation energy for grain growth. Does this correlate with the diffusion of zinc in copper? (Hint – note that rate is the reciprocal of time.) Solution: Temperature (°C) (K) 500 773 600 873 700 973 800 1073 850 1123
1/T (K‐1) 0.00129 0.00115 0.001028 0.000932 0.000890
Time (min) 80,000 3,000 120 10 3
Rate (min‐1) 1.25 x 10‐5 3.33 x 10‐4 8.33 x 10‐3 0.100 0.333
From the graph, we find that Q = 51,286 cal/mol, which does correlate with the activation energy for diffusion of zinc in copper.
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5‐42
A sheet of g ld is diffusion‐bonded to a sheet of silver in 1 h at 00°C. At 500°C, 440 h are required to btain the same degree o bonding, and at 300°C, bonding requires 1530 yea s. What is the activation energy for the iffusion bonding process Does it app ar that diffusion of gold or di fusion of silver controls t e bonding r te? (Hint – note that rate is the recipr cal of time.)
Solution: Temperature (°C) 7 0 5 0 3 0
(K) 973 773 573
1/T ‐ (K 1) 0.001007 0.001294 0.001745
Time (s) 36 0 1.584 x 106 4.825 x 1010
Rate ‐ (sec 1) ‐ 0.278 x 10 3 ‐ 0.631 x 10 6 ‐ 0.207 x 0 10
The activati n energy for the diffusion of gold in solver is 45,50 cal/mol; thus the diffusion of gold appear to control t e bonding r te.
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