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Unformatted text preview: Solutions problem set #2, 2011 2.1 1) For each of the following alloys, state the factor (electrochemical, size, or electron concentration) that you expect to dominate in formation of intermediate phases and explain briefly how you made your selection. a) Cu-Zn • Size difference is small (8%) • Difference in electronegativity is small (0.25) • Cu has 1 valence electron and Zn has 2 valence electrons ⇒ Intermediate phases should be electron compounds b) Fe-C • Size difference is large, rFe = 1.260 and rC = 0.914 • The difference in electronegativity is moderate (1.83 versus 2.55) • Fe has 2 valence electrons and C has 4 ⇒ Size difference dominates c) Mg2Sn • Size difference is small (1%) • Difference in electronegativity is large (0.65) • Mg has 2 valence electrons and Sn has 4 ⇒ Electrochemical effects dominate In v e s tig a tin g t h e s y m m e try a n d l a t ti c e s tru c t ure o f b o t h c ry s t a ls . Solutions problem set #2, 2011 2.2 Principles 2) The figure on the right illustrates the experimental setup for taking a Laue image of a single crystal. On a separate page below a Laue diagram of a NaCl crystal oriented along a high-symmetry axis is shown. The distance between the film and the single F i g . 1 S c h e m e o f t h e s e t u p f o r t a k i n g a Laue d i a g r a m o f a m o n o crystal is 15 25 mm. c ry s t a l a X - r a y tidentify which axis 〈hkl〉 of the a) From the symmetry of the Laue diagram,m autb e b C o lli or c C ry s t a l NaCl crystal is parallel to the incoming x-ray beam. d X - r a y f il m • The Laue diagram shows 4-fold symmetry. 1 9 1 Max t • Hence the NaCl crystal is orientedIsnu p p o 2e d w a v e c h aincominga yx-rayiifdrea c tviodne n tc eparale.such, thatvon Laue tp rroo foX -erd ts b yod fbeam a c rfyosrt ahs the r a c e p s o p r v e i is s l Friedrich a n d Knipping t o o k u p h is p ro p o s a l a n d e x p o s e d a lel to the 〈001〉 axis F ig . 2 tiv e in t e rf e r e n c m ulti p l e o f t h e i n a ll t h r e e s p a t c r y s t a l t o a c o lli m a t e d r a y f r o m a n X - r a y t u b e . O n a p h o t o g r a p hi c p l a t e b e hin d t h e c ry s t a l t h e y o b s e rv e d – a s e x p e c t e d – d i s c r e t e r e f l e c t i o n s . F o r t h e f ir s t t i m e t h e y a l s o c o n f ir m e d t h e 16 abrightestsdiffractionyspots. s u b s t a n c e s w it h t his s p t i a l l a t t i c e t r u c t u r e o f c r s t a lli n e e x p e ri m e n t . In a c u b i c c ry s t s p a t i a l d ir e c t i o n t a n c e a0 b e t w e c o n d iti o n f o r c o 1 0808-Ste b) Measure the coordinates of the • First identify the center of the Laue image using the symmetry of the diffracLaue condition: tion spots. I n h i s i n t e r p r e t a t i o n o f t h e s e fi n d i n g s , von Laue c o n s i d e r e d t h e c r y a l t o b e diffraction o f t h r e o u p s • Then measure the x-y position of thes tbright a l a t ti c e b uilt u pspotseingrmm.o f o n e - d im e n s i o n a l e q u i d i s t a n t ro w s o f p o i n t s . W h e n a n X - r a y i s d if fr a c t e d • Make a table with the coordinates.a t a ro w o f p oin t s , Number a0 " c o s 1 ! a0 a0 " c o s 1 ! a0 a0 " c o s 1 ! a0 w it h in t e g e r v a l H e re 1 , 1 a n d a n d t h e ro w s o b e t w e e n t h e d if ti e s h, k, l a re c a r e s m a ll i n t e g f u l f ill e d f o r a r b ( “ a p p r o p ri a t e ” ) = 1 ! 2 = a0 " c o s 1 ! a0 " c o s 2 (I) y a0 : xi s t a n c e b e t w e e n t h e p o i n t s d 1 : a n g l e b e t w e e n t h e i n c o m i n g X - r a y a n d t h e ro w o f p o i n t s 12.6 e d iff 2 : a n g l e b e t w e e n t h -0.4 r a c t e d X - r a y a n d t h e r o w o f p o i n t s i s 0.2 d if f e r e n c e o f p a t h o f p a rt i a l r a y s s c a t t e r e d a t t w o n e i g h th e 14.5 b o u ri n g p o i n t s ( l a t t i c e e l e m e n t s , s e e F i g . 2 ) . T h e r e i s c o n s t r u c - 2 3 -13.2 0.0 9.0 -11.0 -9.0 9.0 11.5 4.2 -4.3 -12.2 -11.2 -0.1 -11.2 9.2 11.0 -9.2 -9.0 3.6 12.4 13.0 4.0 -3.9 Tw o - d i m enc e of p w hi c h a r 1 Solutions problem set #2, 2011 2.3 Number x y -3.3 3.2 10.5 -10.0 -10.0 -3.9 c) Index the 16 brightest spots in the Laue image. You can try coming up with a method yourself or use the following derivation. • We derive a relationship between the Miller indices (hkl) of the lattice planes and the x-y position of the diffraction spot of the x-ray beam reflected off these (hkl) lattice planes. • From the x-y coordinates of the diffraction spot and the distance between the crystal and the film, we can determine the following relationships for the diffraction angle θ ￿ x2 + y 2 sin(2θ) = ￿ x2 + y 2 + L2 L cos(2θ) = ￿ x2 + y 2 + L2 ￿ x2 + y 2 tan(2θ) = L • Using the trigonometric relationship tan(θ) = yields tan(θ) = ￿ 1 − cos θ sin θ x2 + y 2 + L2 − L ￿ x2 + y 2 • Assuming that the x-ray beam is parallel to the (001) axis of the crystal, the angle θ between the incident x-ray beam and a lattice plane with Miller indi- T h e b u il t - i n p r o t e c t i o n a n d s c r e e n i n g m e a s u r e s r e d u c e t h e lo c a l d o s e r a t e o u t si d e o f t h e X -r a y a p p a r a t u s t o l e s s t h a n 1 S v / h , a v a lu e w hi c h is o n t h e o rd e r o f m a g nit u d e o f t h e n a tSolutions g ro u n d r aseta#2,n2011 u r a l b a c k problem d i t i o . B e f o re p u t tin g t h e X -r a y a p p a r a t u s in t o o p e r a tio n , in s p e c t it f o r d a m a g e a n d m a k e s ure t h a t t h e hi g h v o lt a g e i s s h u t o f f w h e n t h e s li d i n g d o o r s a r e o p e n e d ( s e e cio n (hkl) as a y a p p a r a t u s) i n s t r u c tes s h e e t f o r X - rillustrated. in the Figure K e e p rh e X - r a y a p p a r a t u s s e c u r e fro m a c c e s s b y u n t ight is given by a u t h o ri z e d p e r s o n s . F ig . 4 Tw o - d i m e n s i o n a l r e p r e s e n t a t i o n o f Bragg r e f l e c t i o n o f X -r a y s a t a s e t o f l a t ti c e p l a n e s in a c u b i c c ry s t a l. T h e l a t t i c e p l a n e s a r e p a r a ll e l t o t h e b i s e c t ri x S b e t w e e n t h e i n c o m i n g a n d t h e d i f f r a c t e d X - r a y. 2.4 on l D o n o t a ll o w t h e a n o d e o f t h e X - r a y t u b e M o t o o v e r h e a t . sin(θ) = √ W h e n s w it c h i n g o n t h e X - r a yhp p + tks , c h e c k t o m a k e a 2 a r a u 2 + l2 s u r e t h a t t h e v e n t il a t o r i n t h e t u b e c h a m b e r i s t u r n i n g . • Using the trigonometric relationships, we determine l tan(θ) = √ h2 + k 2 2 • Finally, comparing the two equations for tan θ, we obtain the following relationship for the ratios of h:k:l ￿ h : k : l = x : y : z where z = x2 + y 2 + L2 − L • Calculate the value of z for all the measured reflections from (a). • Determine the Miller indices (hkl) as the smallest triple of integers that fulfill the above relationship. Number x y z x:y:z h k l 1 12.6 0.2 -13.2 0.0 9.0 -11.0 -9.0 9.0 11.5 4.2 -4.3 -12.2 -11.2 -3.3 3.2 10.5 -0.4 14.5 -0.1 -11.2 9.2 11.0 -9.2 -9.0 3.6 12.4 13.0 4.0 -3.9 -10.0 -10.0 -3.9 4.6 5.9 5.0 3.7 4.8 6.6 4.8 4.7 4.2 4.9 5.3 4.7 4.1 3.4 3.3 3.7 3:0:1 0:3:1 -3:0:1 0:-3:1 2:2:1 -2:2:1 -2:-2:1 2:-2:1 3:1:1 1:3:1 -1:3:1 -3:1:1 -3:-1:1 -1:3:1 1:3:1 3:-1:1 6 0 -6 0 4 -4 -4 4 6 2 -2 -6 -6 -2 2 6 0 6 0 -6 4 4 -4 -4 2 6 6 2 -2 6 6 -2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 Solutions problem set #2, 2011 2.5 • The selection rules for fcc crystal structures require that the Miller indices are either all even or all odd. • As a consequence, the Miller indices of the reflections are multiplied by a factor of 2. d) From the Miller indices (hkl) of the reflection and the diffraction angle θ, calculate the wavelength λ and the spacing of the lattice planes d. • Using the Bragg law and the lattice parameter of NaCl a0 = 5.64Å we can calculate the wavelength λ for each Laue reflection Reflection θ d [Å] λ [Å] 442 620 622 24.1 18.4 17.5 1.15 0.89 0.85 0.94 0.56 0.51 • Note that since Addendumwhite X-rays in the Laue technique, we cannot we are using 1: The larger film 554 896 enables the recording of larger images, independently calculate the lattice parameterofrom the Laue image which provide more resolution, but also require a bit m re exposure time. Solutions problem set #2, 2011 2.6 3) An “excess volume” is associated with high defect concentrations in materials. This is the additional volume compared with a perfect single crystal. a) Suppose that a series of Al samples are prepared by quenching to room temperature from various elevated temperatures up to the melting temperature. The objective is to quench in, at room temperature, the equilibrium vacancy concentrations at the higher temperatures. Make a plot of the density of Al at room temperature as a function of the temperature before quenching, assuming that the experiment works as intended. From the lecture notes: The vacancy formation energy in Al is 0.7 eV. • The mole fraction of vacancies is given by ￿ f￿ ￿ ￿ ￿ ￿ f f Sv Hv Hv xv = exp exp − ≈ exp(1) exp − kB kB T kB T • Assuming an entropy of vacancy formation of 1kB and zero pressure we can calculate the fraction of vacant sites • The resulting change in density is given by ￿ ￿ f EV ∆ρ = ρ0 · xV = ρ0 · exp − kB T • At room temperature the mole fraction is negligible small • We can hence assume that the density of Al ρ0 is 2.70 g/cm3 • The plot below shows how the increased concentration of vacancies near the melting point (red line) decreases the density of Al 2.7000 3 Density [g/cm ] 2.6999 2.6998 2.6997 2.6996 2.6995 200 400 600 Temperature [K] 800 1000 1 1 2] 1] 10 [1 1 1] 12 1] 2.7 1] 1 10] 2 1 1] [111] 1 12 g B 01 01 d [1 1 2 A 1] [1 0 Da 1 b [1 1 2 Solutions problem set #2, 2011 21 11 B g [0 A [1 0 1] 1] C b) Describe what consequences a high supersaturation of vacancies can have on a [ b (a) 12 12 11 11 1] [2 [1 1 1] 1 [2 [1 1 1] the metal microstructure. D C [1 1 0 D 1 1 0] • Very high supersaturation of vacancies produces a chemical stress that is suf(b) ficiently large to create new dislocation loopsA Thompson tetrahedron closed (a) that provide( b) the notation in the structure and opened out ( b). In Figure 3.35 [1 1 0 is used in place of the usual notation [1 1 0] to indicate the sense of the vector direction. many new sinks • Figure on right from the lecture notes shows the formation of single-faulted, double-faulted (A) and unfaulted (B) dislocation loops in Al quenched A aluminum from 600°C A • Dislocation loops arise from clustering of vacancies m into disk-shaped cavities B 1 1 2] c) Due to the misfit in the crystal lattice at grain Figure 3.36 Single-faulted, double-faulted (A) and unfaulted (B) dislocation loops in quenched aluminum (after Edington and boundaries there is an excess volume associated with Smallman, 1965; courtesy of Taylor Asgrain boundaries. & Francis). sume that only tilt boundaries are present and that the average grain boundto the (1 1 1) lies and is an edge , β Cγ , Dδ ary angle is 20°. An upper bound of thethe electronplanevolume for represented by Aαdisloca,-αA, etc. Such loops shown excess on which itFigure 3.36 have been producedBin ,aluminum by quenching from about in micrograph of 600 Each loop Burgers vector the vacancies into a disk-shaped cavity, tion is b2/2·l, where b is the magnitude C.of thearises fromtheirclustering ofloops and regular crystallographicwhich then forms l is the a dislocation loop. To reduce energy, the take up forms with their edges parallel to the 1 directions the loop plane. Along 1 1 0 irection it can reduce its length of the dislocation. Make a plot of thedissociating1on 0 of Al in{1 1room forming aa stair-roddat the junction of the two densityan intersectingat 1} plane, temperaenergy by ture as a function of grain size. • Assuming a symmetric tilt boundary with tilt angle θ, we estimate the distance, h, between parallel edge dislocations with Burgers vector b in the boundary to be h = b/θ. • Assuming grains of cubic shape with edge length a, the number of dislocations for each surface of the cube is a/h and the length of each dislocation is l = a. • As a result the total excess volume for a cubic grain is given by ◦ 6 a b2 Vexc = · · ·l 2h2 = 3 · a2 · b · θ where the factor 1/2 accounts for fact, that two grain share each boundary. • The resulting reduction in density is given by Solutions problem set #2, 2011 2.8 ∆ ρ = ρ − ρ0 = m m − V V + Vexc m Vexc · V V + Vexc Vexc 3bθ ≈ ρ0 · = ρ0 · V a • The figure below shows the resulting density as a function of grain size. = 2.8 3 Density [g/cm ] 2.6 2.4 2.2 2.0 0 100 200 300 400 500 Grain size [Å] d) Significant excess volume is generally not associated with dislocations. Based on the geometry of dislocations, explain why this might be true. • Screw dislocations only have shear and hence no dilatation. Hence, little excess volume is associated with screw dislocations. • Edge dislocations have dilatation at the edge of the extra half plane. Above the dislocation line around the extra half plane, the lattice is under compression. Below the dislocation line, the lattice is under expansion. The overall volume change from the compression and tension partially cancels. However, since the volume response to compression and tension is not symmetric, there usually is a net increase in volume due to edge dislocations. Assuming that the region above the dislocation core (compressive region) has the same volume as in bulk, we can make a simple geometric estimate of the excess volume as illustrated in the Figure on the right. The corresponding excess volume for an edge dislocation is about b2/2·l. b b ...
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This note was uploaded on 12/05/2011 for the course MSE 4100 taught by Professor Hennig during the Fall '11 term at Cornell University (Engineering School).

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