HCP crystal planes are described using the Miller-Bravais indices, (hkil) Draw the hexagonal crystal planes whose Miller-Bravais indices are:

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1 .6 How re crystllogrphic plnes indicted in HCP unit cells? In HCP unit cells, crystllogrphic plnes re indicted using four indices which correspond to four xes: three bsl xes of the unit cell,,, nd, which re seprted by º; nd the verticl c xis..6 Wht nottion is used to describe HCP crystl plnes? HCP crystl plnes re described using the Miller-Brvis indices, (hkil)..6 Drw the hexgonl crystl plnes whose Miller-Brvis indices re: () ( ) (d) ( ) (g) () (j) () (b) ( ) (e) ( ) (h) ( ) (k) ( ) (c) () (f) ( ) (i) ( ) (l) () The reciprocls of the indices provided give the intercepts for the plne (,,, nd c). () () ().,,, c b.,,, c c.,,, c () () ( ) d.,,, c e.,,, c f.,,, c Smith Foundtions of Mterils Science nd Engineering Solution Mnul 9

2 () () () g.,,, c h.,,, c i.,,, c () () () j.,,, c k.,,, c l.,,, c.6 Determine the Miller-Brvis indices of the hexgonl crystl plnes in Fig. P.6. Miller-Brvis Indices for Plnes Shown in Figure P.6() Plne Plne b Plne c Plnr Intercepts - Reciprocls Plnr of Intercepts Intercepts Reciprocls Plnr of Intercepts Intercepts -½ ½ Reciprocls of Intercepts Smith Foundtions of Mterils Science nd Engineering Solution Mnul 4

3 Plnr Intercepts c Reciprocls of Intercepts Plnr Intercepts Reciprocls of Intercepts Plnr Intercepts Reciprocls of Intercepts c c ½ c c The Miller indices of plne re (). The Miller indices of plne b re ( ). The Miller indices of plne c re (). Miller-Brvis Indices for the Plnes Shown in Figure P.6(b) Plne Plne b Plne c Plnr Intercepts - c Reciprocls of Intercepts Plnr Intercepts Reciprocls of Intercepts Plnr Intercepts Reciprocls of Intercepts - - c c c c c The Miller indices of plne re. () The Miller indices of plne b re. ( ) The Miller indices of plne c re ( )..64 Determine the Miller-Brvis direction indices of the, - nd directions. The Miller-Brvis direction indices corresponding to the, - nd directions re respectively, [ ], [ ], nd [ ]..65 Determine the Miller-Brvis direction indices of the vectors originting t the center of the lower bsl plne nd ending t the end points of the upper bsl plne s indicted in Fig..8(d). [ ],[],[ ], [],[],[] [] [ ] [ ] [] [] [] + Smith Foundtions of Mterils Science nd Engineering Solution Mnul 4

4 .66 Determine the Miller-Brvis direction indices of the bsl plne of the vectors originting t the center of the lower bsl plne nd exiting t the midpoints between the principl plnr xes. [4],[4],[4], [4] [ 4] [4] [4] [4] [4] [ 4],[4],[ 4] +.67 Determine the Miller-Brvis direction indices of the directions indicted in Fig. P.67. () (b) For Fig. P.67(), the Miller-Brvis direction indices indicted re [] nd [ ]. Those ssocited with Fig. P.67(b) re [ ] nd [ ]..68 Wht is the difference in the stcking rrngement of close-pcked plnes in () the HCP crystl structure nd (b) the FCC crystl structure? Although the FCC nd HCP re both close-pcked lttices with APF.74, the structures differ in the three dimensionl stcking of their plnes: () the stcking order of HCP plnes is ABAB ; (b) the FCC plnes hve n ABCABC stcking sequence..69 Wht re the densest-pcked plnes in () the FCC structure nd (b) the HCP structure? () The most densely pcked plnes of the FCC lttice re the { } plnes. (b) The most densely pcked plnes of the HCP structure re the { } plnes..7 Wht re the closest-pcked directions in () the FCC structure nd (b) the HCP structure? () The closest-pcked directions in the FCC lttice re the directions. (b) The closest-pcked directions in the HCP lttice re the directions. Smith Foundtions of Mterils Science nd Engineering Solution Mnul 4

5 .7 The lttice constnt for BCC tntlum t ºC is.6 nm nd its density is 6.6 g/cm. Clculte vlue for its tomic mss. The tomic mss cn be ssessed bsed upon the mss of tntlum in unit BCC cell: mss/unit cell ρ (volume/unit cell) ρ υ υ 6 9 (6.6 g/cm )( cm /m )(.6 m) 5.98 g/u.c. Since there re two toms in BCC unit cell, the tomic mss is: (5.98 g/unit cell)(6. toms/mol) Atomic mss toms/unit cell 8.9 g/mol.7 Clculte vlue for the density of FCC pltinum in grms per cubic centimeter from its lttice constnt of.99 nm nd its tomic mss of 95.9 g/mol. First clculte the mss per unit cell bsed on the tomic mss nd the number of toms per unit cell of the FCC structure, (4 toms/unit cell)(95.9 g/mol) mss/unit cell.96 g/unit cell 6. toms/mol The density is then found s, ρ υ mss/unit cell mss/unit cell.96 g/unit cell 9 volume/unit cell [(.99 m) ]/ unit cell m,445, g/m.45 g/cm cm.7 Clculte the plnr tomic density in toms per squre millimeter for the following crystl plnes in BCC chromium, which hs lttice constnt of.8846 nm: () (), (b) (), (c) (). () (b) (c) Smith Foundtions of Mterils Science nd Engineering Solution Mnul 4

6 To clculte the density, the plnr re nd the number of toms contined in tht re must first be determined. () The re intersected by the ( ) plne inside the cubic unit cell is while the number of toms contined is: ( 4 corners) (¼ tom per corner) tom. The density is, ρ p equiv. no. of toms whose centers re intersected by selected re selected re tom 9 m (. toms/m ) ( m) mm. toms/mm (b) For the more densely pcked ( ) plne, there re: tom t center + ( 4 corners) (¼ tom per corner) toms And the re is given s ( )( ). The density is thus, ρ p toms 9 6 (.699 toms/m )( m /mm ) 9 (.8846 m).699 toms/mm (c) The tringulr ( ) plne contins: ( corners) ( / 6 tom per corner) ½ tom. 6 The re is equl to bh ( ). The density is thus, 4 / tom 8 6 ρ p (9.8 toms/m )( m /mm ) 6 ( m) toms/mm.74 Clculte the plnr tomic density in toms per squre millimeter for the following crystl plnes in FCC gold, which hs lttice constnt of.4788 nm: () (), (b) (), (c) (). () (b) (c) Smith Foundtions of Mterils Science nd Engineering Solution Mnul 44

7 () The re intersected by the ( ) plne nd the FCC unit cell is while the number of toms contined is: tom t center + ( 4 corners) (¼ tom per corner) toms The density is therefore, ρ p equiv. no. of toms whose centers re intersected by selected re selected re toms 9 m (. toms/m ) ( m) mm. toms/mm (b) For the more densely pcked ( ) plne, there re: ( fce toms) (½ tom) + ( 4 corners) (¼ tom per corner) toms And the re is given s ( )( ). The density is thus, ρ p toms 8 6 (8.5 toms/m )( m /mm ) 9 (.4788 m) 8.5 toms/mm (c) The tringulr ( ) plne contins: ( fce toms ⅓ tom) + ( corners) ( / 6 tom per corner) toms 6 The re is equl to: bh ( ). The density is therefore, 4 toms 9 6 ρ p (.96 toms/m )( m /mm ) 6 ( m) 4.96 toms/mm.75 Clculte the plnr tomic density in toms per squre millimeter for the () plne in HCP beryllium which hs constnt.856 nm nd c constnt of.58 nm. The re intersected by the ( ) plne nd the HCP unit cell is simply the bsl re, shown in the sketch to the right: Smith Foundtions of Mterils Science nd Engineering Solution Mnul 45

8 Selected Are (6 tringles) (equilterl tringle re) 6 While the number of toms contined is: tom t center + ( 6 corners) (⅓ tom per corner) toms The density is therefore, ρ p equiv. no. of toms whose centers re intersected by selected re selected re toms 9 m (. toms/m ) ( mm m). toms/mm.76 Clculte the liner tomic density in toms per millimeter for the following directions in BCC vndium, which hs lttice constnt of.9 nm: () [], (b) [], (c) []. () (b) (c) [ ] [ ] [ ] In generl, the liner tomic density is derived from: ρ l no. of tomic dim. intersected by selected length of direction line selected length of line () For the [] direction of BCC vndium, no. tom di. tom 6 ρ l.9 mm -9 (.9 nm)( m/nm)( mm/m) (b) For the [] direction of BCC vndium, Smith Foundtions of Mterils Science nd Engineering Solution Mnul 46

9 no. tom di. tom 6 ρ l. mm -6 (.9 nm)( mm/nm) (c) For the [] direction of BCC vndium, no. tom di. toms 6 ρ l.8 mm -6 (.9 nm)( mm/nm).77 Clculte the liner tomic density in toms per millimeter for the following directions in FCC iridium, which hs lttice constnt of.889 nm: () [], (b) [], (c) []. () (b) (c) [ ] [ ] [ ] In generl, the liner tomic density is derived from: ρ l no. of tomic dim. intersected by selected length of direction line selected length of line () For the [] direction of FCC iridium, no. tom di. tom 6 ρ l.6 mm -6 (.889 nm)( mm/nm) (b) For the [] direction of FCC iridium, no. tom di. toms 6 ρ l.68 mm -6 (.889 nm)( mm/nm) (c) For the [] direction of FCC iridium, no. tom di. tom 6 ρ l.5 mm -6 (.889 nm)( mm/nm) Smith Foundtions of Mterils Science nd Engineering Solution Mnul 47

10 .78 Wht is polymorphism with respect to metls? A metl is considered polymorphic if it cn exist in more thn one crystlline form under different conditions of temperture nd pressure..79 Titnium goes through polymorphic chnge from BCC to HCP crystl structure upon cooling through 88ºC. Clculte the percentge chnge in volume when the crystl structure chnges from BCC to HCP. The lttice constnt of the BCC unit cell t 88ºC is. nm nd the HCP unit cell hs.95 nm nd c.468 nm. To determine the volume chnge, the individul volumes per tom for the BCC nd HCP structures must be clculted: nm /unit cell (. nm) V BCC.8 nm /tom toms/unit cell toms o ( c)(sin6 ) nm /unit cell ()(.95 nm) (.468 nm)(sin6 ) V HCP 6 toms/unit cell 6 toms.765 nm /tom Thus the chnge in volume due to titnium s llotropic trnsformtion is, VHCP VBCC % Volume chnge (%) V BCC.765 nm /tom.8 nm /tom (%) -.55%.8 nm /tom.8 Pure iron goes through polymorphic chnge from BCC to FCC upon heting through 9ºC. Clculte the volume chnge ssocited with the chnge in crystl structure from BCC to FCC if t 9ºC the BCC unit cell hs lttice constnt.9 nm nd the FCC unit cell.6. First determine the individul volumes per tom for the iron BCC nd FCC crystl structures: o nm /unit cell (.9 nm) V BCC.58 nm /tom toms/unit cell toms nm /unit cell (.6 nm) V FCC.96 nm /tom 4 toms/unit cell 4 toms Thus the chnge in volume due to iron s llotropic trnsformtion is, Smith Foundtions of Mterils Science nd Engineering Solution Mnul 48

11 VFCC VBCC.96 nm /tom.58 nm /tom % Volume chnge (%) (%) V.58 nm /tom BCC -4.94%.8 Wht re x-rys, nd how re they produced? X-rys re electromgnetic rdition hving wvelengths in the rnge of pproximtely.5 nm to.5 nm. These wves re produced when ccelerted electrons strike trget metl..8 Drw schemtic digrm of n x-ry tube used for x-ry diffrction, nd indicte on it the pth of the electrons nd x-rys. See Figure.5 of textbook..8 Wht is the chrcteristic x-ry rdition? Wht is its origin? Chrcteristic rdition is n intense form of x-ry rdition which occurs t specific wvelengths for prticulr element. The K α rdition, the most intense chrcteristic rdition emitted, is cused by excited electrons dropping from the second tomic shell (n ) to the first shell (n ). The next most intense rdition, K β, is cused by excited electrons dropping from the third tomic shell (n ) to the first shell (n )..84 Distinguish between destructive interference nd constructive interference of reflected x- ry bems through crystls. Destructive interference occurs when the wve ptterns of n x-ry bem, reflected from crystl, re out of phse. Conversely, when the wve ptterns leving crystl plne re in phse, constructive interference occurs nd the bem is reinforced..85 Derive Brgg s lw by using the simple cse of incident x-ry bems being diffrcted by prllel plnes in crystl. Referring to Fig..8 (c), for these rys to be in phse, ry must trvel n dditionl distnce of MP + PN. This extr length must be n integrl number of wvelengths, λ. nλ MP + PN where n,,... Moreover, the MP nd PN distnces must equl dhkl sinθ, where d hkl is the crystl interplnr spcing required for constructive interference. Substituting, MP d sin θ nd PN d sinθ hkl hkl Smith Foundtions of Mterils Science nd Engineering Solution Mnul 49

12 nλ d hkl sin θ Brgg's Lw.86 A smple of BCC metl ws plced in n x-ry diffrctometer using x-rys with wvelength of λ.54 nm. Diffrction from the {} plnes ws obtined t θ 88.88º. Clculte vlue for the lttice constnt for this BCC elementl metl (Assume first-order diffrction, n.) The interplnr distnce is, d λ.54 nm. nm o sinθ sin(44.49 ) The lttice constnt,, is then, d h + k + l (. nm) + +. nm hkl.87 X-rys of n unknown wvelength re diffrcted by gold smple. The θ ngle ws 64.58º for the {} plnes. Wht is the wvelength of the x-rys used? (The lttice constnt of gold is.4788 nm. Assume first-order diffrction, n.) The interplnr distnce is, d.4788 nm.44 nm h + k + l + + The lttice constnt,, is then, λ d sinθ (.44 nm)sin(.9 ).54 nm.88 An x-ry diffrctometer recorder chrt for n element which hs either the BCC or the FCC crystl structure showed diffrction peks t the following θ ngles: 4.69º, 47.78º, º, nd 84.96º. (The wvelength of the incoming rdition ws.545 nm.) () Determine the crystl structure of the element. (b) Determine the lttice constnt of the element. (c) Identify the element. () Compring the sin θ term for the first two ngles: θ θ sin θ sin θ 4.69 º.55 º º.89 º o Smith Foundtions of Mterils Science nd Engineering Solution Mnul 5

13 sin θ.4 sin θ FCC (b) The lttice constnt lso depends upon the first sin θ term, s well s, the Miller indices of the first set of FCC principl diffrcting plnes, {}. λ h + k + l.545 nm nm sin θ.4 (c) From Appendix I, the FCC metl whose lttice constnt is closest to.84 nm is rhodium (Rh) which hs lttice constnt of.844 nm..89 An x-ry diffrctometer recorder chrt for n element which hs either the BCC or the FCC crystl structure showed diffrction peks t the following θ ngles: 8.6º, 55.7º, 69.7º, 8.55º, 95.º, nd 7.67º. (The wvelength λ of the incoming rdition ws.545 nm.) () Determine the crystl structure of the element. (b) Determine the lttice constnt of the element. (c) Identify the element. () Compring the sin θ term for the first two ngles: θ θ sin θ sin θ 8.6 º 9. º º º sin θ.94 sin θ.8.5 BCC (b) The lttice constnt lso depends upon the first sin θ term, s well s, the Miller indices of the first set of BCC principl diffrcting plnes {}. λ h + k + l.545 nm nm sin θ.94 (c) From Appendix I, the BCC metl whose lttice constnt is closest to.96 nm is niobium (Nb) which hs lttice constnt of.7 nm. Smith Foundtions of Mterils Science nd Engineering Solution Mnul 5

14 .9 An x-ry diffrctometer recorder chrt for n element which hs either the BCC or the FCC crystl structure showed diffrction peks t the following θ ngles: 6.9º, 5.974º, 64.98º, nd 76.66º. (The wvelength λ of the incoming rdition ws.545 nm.) () Determine the crystl structure of the element. (b) Determine the lttice constnt of the element. (c) Identify the element. () Compring the sin θ term for the first two ngles: θ θ sin θ sin θ 6.9 º 8.96 º º º sin θ.9647 sin θ BCC (b) The lttice constnt lso depends upon the first sin θ term, s well s, the Miller indices of the first set of BCC principl diffrcting plnes, {}. λ h + k + l.545 nm nm sin θ.9647 (c) From Appendix I, the BCC metl whose lttice constnt is closest to.57 nm is lithium (Li) which hs lttice constnt of.59 nm..9 An x-ry diffrctometer recorder chrt for n element which hs either the BCC or the FCC crystl structure showed diffrction peks t the following θ ngles: 4.66º, 47.4º, 69.44º, nd 8.448º. (The wvelength λ of the incoming rdition ws.545 nm.) () Determine the crystl structure of the element. (b) Determine the lttice constnt of the element. (c) Identify the element. () Compring the sin θ term for the first two ngles: θ θ sin θ sin θ 4.66 º.5 º º.657 º.46.6 Smith Foundtions of Mterils Science nd Engineering Solution Mnul 5

15 sin θ.7 sin θ.6.75 FCC (b) The lttice constnt lso depends upon the first sin θ term, s well s, the Miller indices of the first set of FCC principl diffrcting plnes, {}. λ h + k + l.545 nm nm sin θ.7 (c) From Appendix I, the FCC metl whose lttice constnt is closest to.897 nm is iridium (Ir) which hs lttice constnt of.889 nm. Smith Foundtions of Mterils Science nd Engineering Solution Mnul 5