(a) The atomic number is the number of protons contained in the nucleus of an atom.

Transcription

1 CHAPTER ATOMIC STRUCTURE AND BONDING. Wht is the mss in grms of () proton, (b) neutron, nd (c) n electron? () mss of proton g (b) mss of neutron g (c) mss of electron g. Wht is the electric chrge in coulombs of () proton, (b) neutron, nd (c) n electron? () proton chrge C (b) neutron chrge 0 C (c) electron chrge C. Define () tomic number, (b) tomic mss unit, (c) Avogdro s number, nd (d) reltive grm tomic mss. () The tomic number is the number of protons contined in the nucleus of n tom. (b) An tomic mss unit, u, is defined s exctly / of the mss of the crbon tom which hs mss of u. (c) Avogdro s number, N A, is toms/mol. (d) The reltive grm tomic mss of n element is the mss in grms of toms of tht element..4 Wht is the mss in grms of one tom of gold? g/mol Au x mss of Au tom ( tom) toms/mol.5 How mny toms re there in g of gold? toms/mol x No. of toms of Au ( g Au) g/mol Au g Au toms Au.6 A gold wire is 0.70 mm in dimeter nd 8.0 cm in length. How mny toms does it contin? The density of gold is 9. g/cm. First determine the mss of gold bsed upon the wire volume nd the density of gold. nd m mss of Au ρv where ρ 9. g/cm Smith Foundtions of Mterils Science nd Engineering Solution Mnul 4

2 V volume of wire (re)(length) πd 4 π 4 cm 0 mm ( L) ( 0.7 mm) ( 8.0 cm) cm Thus, m (9. g/cm Au)(0.008 cm ) g Au The number of toms is then clculted s, No. of toms Au ( g Au) toms Au toms/mol g/mol Au.7 Wht is the mss in grms of one tom of molybdenum? g/mol Mo x mss of Mo tom ( tom) toms/mol.8 How mny toms re there in g of molybdenum? toms/mol x No. of toms of Mo ( g Mo) g/mol Mo g Mo toms Mo.9 A solder contins 5 wt % tin nd 48 wt % led. Wht re the tomic percentges of Sn nd Pb in this solder? Using bsis of 00 g of solder, there re 5 g of Sn nd 48 g of Pb. The number of grm-moles of tin nd led is therefore, No. of grm - moles No.of grm - moles of Sn of Pb 5 g 8.69 g/mol 48 g 07.9 g/mol Totl grm - moles The tomic percentges my then be clculted s, 0.48 mol 0.7 mol mol 0.48 mol Atomic % Sn mol 0.7 mol Atomic % Pb mol ( 00% ) ( 00% ) t% t%.0 A monel lloy consists of 70 wt % Ni nd 0 wt % Cu. Wht re the tomic percentges of Ni nd Cu in this lloy? Smith Foundtions of Mterils Science nd Engineering Solution Mnul 5

3 Using bsis of 00 g of lloy, there re 70 g of Ni nd 0 g of Cu. The number of grm-moles of ech element is thus, No. of grm - moles No. of grm - moles of Cu of Ni 0 g 6.54 g/mol 70 g 58.7 g/mol Totl grm - moles The tomic percentges my then be clculted s, 0.47 mol Atomic % Cu.664 mol.9 mol Atomic % Ni.664 mol ( 00% ) ( 00% ) mol.9 mol.664 mol t% t%. A cupronickel lloy consists of 80 wt % Cu nd 0 wt % Ni. Wht re the tomic percentges of Cu nd Ni in the lloy? Using bsis of 00 g of lloy, there re 80 g of Cu nd 0 g of Ni. The respective number of grm-moles is thus, No. of grm - moles No.of grm - moles of Cu of Ni 80 g 6.54 g/mol 0 g 58.7 g/mol Totl grm - moles The tomic percentges my then be clculted s,.590 mol mol.5997 mol.590 mol Atomic % Cu.5997 mol mol Atomic % Ni.5997 mol ( 00% ) ( 00% ) t% t%. Wht is the chemicl formul of n intermetllic compound tht consists of 49.8 wt % Cu nd 50.8 wt % Au? We need to determine Cu x Au y where x nd y represent the grm-mole frctions of copper nd gold, respectively. Therefore, first clculte the grm-mole frctions of these elements using bsis of 00 g of compound: Smith Foundtions of Mterils Science nd Engineering Solution Mnul 6

4 No. of grm - moles No. of grm - moles x y of Cu Grm - mole frction of Grm - mole frction of 49.8 g 6.54 g/mol 50.8 g of Au 0.58 mol g/mol Totl grm - moles mol.0 mol mol Cu mol 0.58 mol Au mol Thus we hve Cu 0.75 Au 0.5 or, multiplying by 4, Cu Au.. Wht is the chemicl formul of n intermetllic compound tht consists of 5.68 wt % Mg nd 84. wt % Al? The chemicl formul, Mg x Al y, my be determined bsed on the grm-mole frctions of mgnesium nd luminum. Using bsis of 00 g of intermetllic compound, No. of grm - moles No. of grm - moles x y.4 Define photon. of Mg of Al Grm - mole frction of Grm - mole frction of 5.68 g 4. g/mol 84. g 6.98 g/mol Totl grm - moles mol.5 mol.770 mol mol Mg mol.5 mol Al mol Thus we hve Mg 0.7 Al 0.8 or, multiplying by 6, MgAl 5. A photon is discrete mount or quntum of energy in the form of electromgnetic rdition..5 Clculte the energy in joules nd electron volts of the photon whose wvelength is 0.4 nm. Using Plnk s eqution with ν c/λ, Smith Foundtions of Mterils Science nd Engineering Solution Mnul 7

5 E hc (6.6 0 J s)(.00 0 m/s) λ (0.4 nm)(0 m/nm) ( ev J) ev J 9 J.6 Clculte the energy in joules nd electron volts of the photon whose wvelength is 6.4 nm. hc (6.6 0 J s)(.00 0 m/s) E λ (6.4 nm)(0 m/nm) ( ev J) ev J.7 A hydrogen tom exists with its electron in the n 4 stte. The electron undergoes trnsition to the n stte. Clculte () the energy of the photon emitted, (b) its frequency, nd (c) its wvelength in nnometers (nm). () Photon energy emitted is: 9 J E 0.66 ev.06 0 n 4 (b) Photon frequency is found s: 9 J ν E h (c) The wvelength is given s: J J s Hz λ hc E -4 (6.6 0 J s)(.00 0 m/s) -9-9 (.06 0 J)(0 m/nm) nm.8 A hydrogen tom exists with its electron in the n 6 stte. The electron undergoes trnsition to the n stte. Clculte () the energy of the photon emitted, (b) its frequency, nd (c) its wvelength in nnometers. () Photon energy emitted is: E.0 ev n 6 (b) Photon frequency is found s: 9 J Smith Foundtions of Mterils Science nd Engineering Solution Mnul 8

6 ν E h (c) The wvelength is given s: J J s Hz λ hc E -4 (6.6 0 J s)(.00 0 m/s) -9-9 ( J)(0 m/nm) 8 40 nm.9 In commercil x-ry genertor, stble metl such s copper (Cu) or tungsten (W) is exposed to n intense bem of high-energy electrons. These electrons cuse ioniztion events in the metl toms. When the metl toms regin their ground stte, they emit x- rys of chrcteristic energy nd wvelength. For exmple, tungsten tom struck by high-energy electron my lose one of its K shell electrons. When this hppens, nother electron, probbly from the tungsten L shell will fll into the vcnt site in the K shell. If such p s trnsition occurs in tungsten, tungsten K α x-ry is emitted. A tungsten K α x-ry hs wvelength λ of 0.08 nm. Wht is its energy? Wht is its frequency? E ν hc λ E h (6.6 0 J s)(.00 0 m/s) 9 (0.08 nm)(0 m/nm) J.40 0 J s 8 9 Hz J.0 Most modern scnning electron microscopes (SEM) re equipped with energy dispersive x-ry detectors for the purpose of chemicl nlysis of the specimens. This x-ry nlysis is nturl extension of the cpbility of the SEM becuse the electrons tht re used to form the imge re lso cpble of creting chrcteristic x-rys in the smple. When the electron bem hits the specimen, x-rys specific to the elements in the specimen re creted. These cn be detected nd used to deduce the composition of the specimen from the well-known wvelengths of the chrcteristic x-rys of the elements. For exmple: Element Wvelength of K α x-rys Cr Mn Fe Co Ni Cu Zn 0.9 nm 0.0 nm 0.97 nm nm nm 0.54 nm 0.46 nm Smith Foundtions of Mterils Science nd Engineering Solution Mnul 9

7 Suppose metllic lloy is exmined in n SEM nd three different x-ry energies re detected. If the three energies re 749, 546, nd 647 ev, wht elements re present in the smple? Wht would you cll such n lloy? (Look hed to Chp. 9 of the textbook.) The elements my be identified by clculting their respective wvelengths. () λ hc E (6.6 0 J s)(.00 0 m/s) -9 (749 ev)(.60 0 J/eV) m nm 8 Ni From the tble provided, wvelength of nm corresponds to nickel, Ni. (b) λ hc E (6.6 0 J s)(.00 0 m/s) -9 (546 ev)(.60 0 J/eV) m 0.9 nm 8 Cr From the tble provided, wvelength of 0.9 nm corresponds to chromium, Cr. (c) λ hc E (6.6 0 J s)(.00 0 m/s) -9 (647 ev)(.60 0 J/eV) m 0.97 nm 8 Fe From the tble provided, wvelength of 0.97 nm corresponds to iron, Fe. The elements present, nickel, chromium, nd iron, re the primry constituents of the ustenitic stinless steels.. Describe the Bohr model of the hydrogen tom. Wht re the mjor shortcomings of the Bohr model? The Bohr model of the hydrogen tom envisions n electron revolving round proton in circulr orbit hving rdius of 0.05 nm. The electron is now believed to hve noncirculr or ellipticl orbits round the hydrogen nucleus; these were not postulted in the originl Bohr theory. The exct position of the hydrogen electron t ny instnt cnnot be precisely determined becuse the electron is such smll prticle. Scientists presently believe tht the hydrogen electron cn ctully devite to some extent in its distnce from the nucleus. However, the highest probbility of finding the electron in spce bout its nucleus is still believed to be t the Bohr rdius of 0.05 nm.. Wht is the ioniztion energy in electron volts for hydrogen tom in the ground stte? The ioniztion energy is given s, E.6 ev n.6 ev.6 ev Smith Foundtions of Mterils Science nd Engineering Solution Mnul 0

8 . Describe the four quntum numbers of n electron nd give their llowed vlues.. The principl quntum number, n, indictes n electron s min energy levels, s well s, its loction in spce; high probbility exists for finding n electron within the n shell. This quntum number my hve positive integer vlues of n,,, 4. The subsidiry quntum number, l, specifies n electron s subenergy levels nd thus the subshell vlues where the electron is likely to be found. The llowed vlues re l 0,,,, n -. The first four of these subenergy vlues hve corresponding letter designtions of l s, p, d, f which represent electron orbitls.. The mgnetic quntum number, m l, describes n tomic orbitl s sptil orienttion nd my hve integrl vlues of l to +l. 4. The electron spin quntum number, m s, provides for two electron spin directions, clockwise nd counterclockwise, which re represented by the permissible vlues of +½ nd -½..4 Write the electron configurtions of the following elements by using spdf nottion: () yttrium, (b) hfnium, (c) smrium, (d) rhenium. () Y (Z 9): [Kr] 4d 5s (c) Sm (Z 6): [Xe] 4f 6 6s (b) Hf (Z 7): [Xe] 4f 4 5d 6s (d) Re (Z 75): [Xe] 4f 4 5d 5 6s.5 Wht is the outer electron configurtion for ll the noble gses except helium? s p 6.6 Of the noble gses Ne, Ar, Kr, nd Xe, which should be the most chemiclly rective? Xenon should be most rective since its outermost electrons (5s 6p 6 ) re further wy from the nucleus thn the other noble gses, nd thus esier to remove..7 Define the term electronegtivity. Electronegtivity is the degree to which n tom ttrcts electrons to itself..8 Which five elements re the most electropositive ccording to the electronegtivity scle? The most electropositive elements re: Fr, Cs, Rb, K, R, nd B. All of these hve the lowest electronegtivity vlue of Which five elements re the most electronegtive ccording to the electronegtivity scle? The most electropositive elements, bsed on the electronegtivity scle of 0 to 4., re: F (4.); O (.5); N (.); Cl (.9); nd Br (.8). Smith Foundtions of Mterils Science nd Engineering Solution Mnul

9 .0 Write the electron configurtion of the following ions by using spdf nottion: () Cr +, Cr +, Cr 6+ ; (b) Mo +, Mo 4+, Mo 6+ ; (c) Se 4+, Se 6+, Se () Cr [Ar] d 5 4s (b) Mo [Kr] 4d 5 5s (c) Se [Ar] d 0 4s 4p 4 Cr + [Ar] d 4 Mo + [Kr] 4d Se 4+ [Ar] d 0 4s Cr + [Ar] d Mo 4+ [Kr] 4d Se 6+ [Ar] d 0 Cr 6+ [Ar] Mo 6+ [Kr] 4d Se [Ar] d 0 4s 4p 6. Briefly describe the following types of primry bonding: () ionic, (b) covlent, nd (c) metllic. () Ionic bonding rises from the electrosttic ttrction between oppositely chrged ions. In the process of ion formtion, n electron or number of electrons my be trnsferred from highly electropositive element to highly electronegtive one. The ionic bond in solids is nondirectionl. (b) Covlent bonding is primry type of bonding which rises from the reduction in energy ssocited with the overlpping of hlf-filled orbitls of two toms. In this bond, there is n electron exchnge interction. The covlent bond is directionl type of bond. (c) Metllic bonding is primry type of bonding involving the interction of the vlence electron or electrons of one tom with mny surrounding toms. This interction leds to reduction in energy of the system considered. The vlence bonding electrons of these bonds re sometimes regrded s n electron gs bonding the positive ion cores (toms less their vlence electrons) of toms. The metllic bond is nondirectionl.. Briefly describe the following types of secondry bonding: () fluctuting dipole, nd (b) permnent dipole. () Fluctuting dipole bonding is secondry type of bonding between toms which contin electric dipoles. These electric dipoles, formed due to the symmetricl electron chrge distribution within the toms, chnge in both direction nd mgnitude with time. This type of bond is electrosttic in nture, very wek nd nondirectionl. (b) Permnent dipole bonding is lso secondry type of bonding between molecules possessing permnent electric dipoles. The bonds, formed by the electrosttic ttrction of the dipoles, re directionl in nture.. In generl, why does bonding between toms occur? Bonding between toms generlly occurs becuse the toms energies re lowered through the bonding process..4 Describe the ionic bonding process between pir of N nd Cl toms. Which electrons Smith Foundtions of Mterils Science nd Engineering Solution Mnul

10 re involved in the bonding process? The ionic bonding process between pir of N nd Cl toms involves trnsfer of the outer s electron of the N tom to the p vcncy in the Cl tom. Thus, the N ion formed hs the Ne electron configurtion while the Cl ion hs the Kr electron configurtion..5 After ioniztion, why is the sodium ion smller thn the sodium tom? After ioniztion to the N +, the N tom becomes smller becuse the electron-to-proton rtio of the N tom is decresed when the N + ion forms. Also, the outer third shell no longer exists once the s electron is lost by the N tom..6 After ioniztion, why is the chloride ion lrger thn the chlorine tom? After ioniztion, the Cl ion is lrger becuse the electron-to-proton rtio of the chlorine tom is decresed by the ioniztion process..7 Clculte the ttrctive force ( ) between pir of K + nd Br ions tht just touch ech other. Assume the ionic rdius of the K + ion to be 0. nm nd tht of the Br ion to be 0.96 nm. The ttrctive force between the ion pir is found by pplying Coulomb s lw, Where F ttrctive ZZe 4πε o o Z + for K +, Z - for Br, nd r + + r 0. nm nm 0.9 nm.9 o K Br 0-0 m Substituting, F ttrctive 4π ( ( + )( )( C / N m 9 C) )( m). 0 9 N.8 Clculte the ttrctive force ( ) between pir of B + nd S ions tht just touch ech other. Assume the ionic rdius of the B + ion to be 0.4 nm nd tht of the S ion to be 0.74 nm. The ttrctive force between the ion pir is found by pplying Coulomb s lw, F ttrctive ZZe 4πε o o Where Z + for B +, Z - for S, nd Smith Foundtions of Mterils Science nd Engineering Solution Mnul

11 o Substituting, r B + + r 0.4 nm nm 0.7 nm.7 0 S -0 m F ttrctive 4π ( ( + )( )( C / N m 9 C) )( m) N.9 Clculte the net potentil energy for K + Br pir by using the b constnt clculted from Problem.7. Assume n 9.5. The repulsive energy constnt b is: F b Repulsive n n+ o ( F N m 0 Attrctive n ) n+ o (. 0 9 N)( m) 0.5 Thus the net potentil energy between the ions is, E + K Br + ZZ e 4πε o o + ( + )( )(.60 0 C) 4π ( C /(N m ))(.9 0 ( b + 9 n o J) + ( J) m) ( J 0 0 N m m) Clculte the net potentil energy for B + S ion pir by using the b constnt clculted from Problem.8. Assume n 0.5. The repulsive energy constnt b is: F b Repulsive n n+ o ( F N m 0 Attrctive n ) n+ o ( N)( m).5 Thus the net potentil energy between the ions is, E B S + + ZZ e 4πε o o + ( + )( )(.60 0 C) 4π ( C /(N m ))(.7 0 ( b + n 8 o 9 J) + ( J) m) ( J 9 0 N m m) If the ttrctive force between pir of Cs + nd I ions is N nd the ionic Smith Foundtions of Mterils Science nd Engineering Solution Mnul 4

12 rdius of the Cs + ion is 0.65 nm, clculte the ionic rdius of the I ion in nnometers. From Coulomb s lw, o ZZ e 4πε F o.85 0 Attrctive 0 4π ( m 0.85 nm ( + )( )(.60 0 C /(N m 9 C) ))( N) The ionic rdius of iodine is thus, r I r + o Cs 0.85 nm 0.65 nm 0.0 nm.4 If the ttrctive force between pir of Sr + nd O ions is N nd the ionic rdius of the O ion is 0. nm, clculte the ionic rdius of the Sr + ion in nnometers. From Coulomb s lw, o ZZ e 4πε F o.67 0 Attrctive 0 4π ( m 0.67 nm ( + )( )(.60 0 C /(N m 9 C) ))( N) The ionic rdius of iodine is thus, r Sr + r 0.67 nm 0. nm 0.5 nm o O.4 Describe the two mjor fctors tht must be tken into ccount in the pcking of ions in n ionic crystl.. The structure of ionic solids is prtly determined by the rdii of the ctions nd nions of the ionic solid. Only certin rnges of ction-to-nion rdius rtios re llowed by pcking considertions.. Electricl neutrlity must be mintined. Thus, if the ctions nd nions hve different vlences, the number of nions surrounding prticulr ction will be restricted..44 Describe the covlent bonding process between pir of hydrogen toms. Wht is the driving energy for the formtion of ditomic molecule? The covlent bonding in the hydrogen molecule involves the interction nd overlpping of the s orbitls of the hydrogen toms. The covlent bond forms between the two hydrogen toms becuse their energies re lowered by the bonding process..45 Describe the covlent bonding electron rrngement in the following ditomic molecules: () fluorine, (b) oxygen, (c) nitrogen. () In fluorine molecules, single covlent bond is formed between the p orbitls of two Smith Foundtions of Mterils Science nd Engineering Solution Mnul 5

13 fluorine toms. (b) In oxygen molecules, double covlent bond is formed between the p orbitls of two oxygen toms. (c) In nitrogen molecules, triple covlent bond is formed between the p orbitls of two nitrogen toms..46 Describe the hybridiztion process for the formtion of four equivlent sp hybrid orbitls in crbon during covlent bonding. Use orbitl digrms. In the hybridiztion process, four equl sp hybrid orbitls re formed in the crbon tom. This is ccomplished through the promotion of one of the s orbitls to p level. s s Two hlf-filled p orbitls Hybridiztion s Four equivlent hlffilled sp orbitls.47 List the number of toms bonded to C tom tht exhibits sp, sp, nd sp hybridiztion. For ech, give the geometricl rrngement of the toms in the molecule. sp hybridiztion: Four toms re bonded to centrl crbon tom in tetrhedrl rrngement. An exmple is methne, CH 4. sp hybridiztion: Three toms re bonded to crbon tom in plnr rrngement. An exmple is ethylene, CH : CH. sp hybridiztion: Two toms re bonded to crbon tom in liner rrngement. An exmple is cetylene, CH:CH. Methne Ethylene Acetylene H H C H H H C C H Tetrhedron.48 Why is dimond such hrd mteril? Plne Liner Dimond is extremely hrd becuse its crbon toms re covlently bonded by single sp hybrid bonds in three dimensionl rrngement..49 Describe the metllic bonding process mong n ggregte of copper toms. Smith Foundtions of Mterils Science nd Engineering Solution Mnul 6

14 Copper hs crystl structure of closely pcked toms; ech tom is surrounded by twelve others. This compct structure is due, in prt, to the fct tht ech tom hs only one vlence electron. Consequently, there is low density electron chrge cloud vilble for bonding..50 How cn the high electricl nd therml conductivities of metls be explined by the electron gs model of metllic bonding? Ductility? The high electricl nd therml conductivities of metls re explined by the mobility of their outer vlence electrons in the presence of n electricl potentil or therml grdient. The ductility of metls is explined by the bonding electron gs which enbles toms to pss over ech other during deformtion, without severing their bonds..5 The melting point of the metl potssium is 6.5ºC, while tht of titnium is 660ºC. Wht explntion cn be given for this gret difference in melting tempertures? The significntly higher melting point of titnium, s compred to potssium, is in prt ttributed to the hybridized covlent bonding between the d nd 4s orbitls of titnium..5 Is there correltion between the electron configurtions of the element potssium (Z 9) through copper (Z 9) nd their melting points? (See Tbles.8 nd.9.) A possible correltion between the melting points nd the electron configurtions of the elements from scndium (Z ) to copper (Z 9) is tht unpired d electrons cuse covlent hybridized bonds, nd hence give higher melting points to these trnsition metls..5 Using the covlent-metllitivity vlues of Tble.9, clculte vlues for the percent metllic nd covlent bonding in the structurl metl titnium. Assuming purely covlent nd metllic bonding in the titnium,.9 % covlent bonding 00% 47.75% 4.00 % metllic bonding 00% % 5.5%.54 Using the covlent-metllitivity vlues of Tble.9, clculte vlues for the percent metllic nd covlent bonding in the metl tungsten. Assuming purely covlent nd metllic bonding in the tungsten,.86 % covlent bonding 00% 96.50% 4.00 % metllic bonding 00% %.50% Smith Foundtions of Mterils Science nd Engineering Solution Mnul 7

15 .55 Define n electric dipole moment. An electric dipole moment is defined s the product of the bsolute chrge of one of the dipoles (positive or negtive) multiplied by the seprtion distnce between the positive nd negtive dipole chrges..56 Describe fluctuting dipole bonding mong the toms of the noble gs neon. Of choice between the noble gses krypton nd xenon, which noble gs would be expected to hve the strongest dipole bonding nd why? A fluctuting electric dipole exists in the toms of noble gses, such s neon, becuse there is, t ny instnt, n symmetricl distribution of electricl chrge mong their electrons. The noble gs xenon would be expected to hve stronger fluctuting dipole moment thn krypton since it hs n dditionl electron shell; the krypton tom hs four electron shells while the xenon tom hs five. The electrons of this fifth shell, being further wy from the xenon nucleus, nd re ble to fluctute more nd thus crete greter symmetry of chrge..57 Describe permnent dipole bonding mong polr covlent molecules. Polr covlent molecules with permnent dipoles bond together becuse of the electrosttic ttrction of their positive nd negtive chrge centers..58 Crbon tetrchloride (CCl 4 ) hs zero dipole moment. Wht does this tell us bout the C Cl bonding rrngement in this molecule? Since the molecule CCl 4 hs zero dipole moment, the C Cl bonding rrngement must be symmetricl bout the crbon nucleus..59 Describe the hydrogen bond. Among wht elements is this bond restricted? The hydrogen bond is permnent dipole bond restricted to hydrogen tom nd highly electronegtive toms such s O, N, F nd Cl. This secondry bond is reltively strong due to the smll size of the hydrogen tom..60 Describe hydrogen bonding mong wter molecules. Hydrogen bonding occurs between wter molecules becuse of the electrosttic ttrction between the negtively chrged oxygen regions of the wter molecules nd the positively chrged hydrogen regions..6 Methne (CH 4 ) hs much lower boiling temperture thn does wter (H O). Explin why this is true in terms of the bonding between molecules in ech of these two substnces. Smith Foundtions of Mterils Science nd Engineering Solution Mnul 8

16 The methne molecules re bonded together by wek C H dipoles. The wter molecules re bonded together by the much stronger O H hydrogen bonded dipoles..6 Sketch tetrhedron nd mke smll model of one from crdbord nd tpe with -inch side. Do the sme for n octhedron. The models should resemble these -D sketches: Tetrhedron Octhedron.6 Wht is Puling s eqution for determining the percentge ionic chrcter in mixed ionic-covlently bonded compound? Puling s eqution relting the percent ionic chrcter of n ionic-covlent bond to the electronegtivites of the bonding toms is: % Ionic Chrcter [- e 0 ). 5(xA xb ][00%] where x A is the electronegtivity of the more electropositive element nd x B is the electronegtivity of the more electronegtive element..64 Compre the percentge ionic chrcter in the semiconducting compound CdTe nd InP. Applying Puling s eqution to CdTe nd InP compounds, 0.5(.5.0) For CdTe ( 6), % Ionic chrcter ( e )(00%) 6.% 0.5(.5.) For InP ( 5), % Ionic chrcter ( e )(00%) 8.6% While 6 compound typiclly hs higher ionic chrcter thn 5, the reltively high electronegtivity of phosphorous cuses InP to be more ionic in nture..65 Compre the percentge ionic chrcter in the semiconducting compound InSb nd ZnTe. Applying Puling s eqution to InSb nd ZnTe compounds, 0.5(.5.8) For InSb ( 6), % Ionic chrcter ( e )(00%).% 0.5(.7.0) For ZnTe ( 5), % Ionic chrcter ( e )(00%).% While 6 compound typiclly hs higher ionic chrcter thn 5, in this cse, the compounds hve identicl ionic chrcter. Smith Foundtions of Mterils Science nd Engineering Solution Mnul 9

17 .66 For ech of the following compounds, stte whether the bonding is essentilly metllic, covlent, ionic, vn der Wls, or hydrogen: () Ni, (b) ZrO, (c) grphite, (d) solid Kr, (e) Si, (f) BN, (g) SiC, (h) Fe O, (i) MgO, (j) W, (k) H O within the molecules, (l) H O between the molecules. If ionic nd covlent bonding re involved in the bonding of ny of the compounds listed, clculte the percentge ionic chrcter in the compound. () Ni: Nickel bonding is primrily metllic. (b) ZrO : From Puling s eqution, the Zr O bond is 7.4% ionic nd 6.6% covlent, where x A nd x B re the electronegtivities of zirconium nd oxygen, respectively. (c) Grphite: The bonding is covlent within the lyers nd secondry between the lyers. (d) Solid Kr: The bonding represents vn der Wls due to fluctuting dipoles. (e) Si: Silicon bonding is covlent. (f) BN: The B N bond, from Puling s eqution (-0), is 6.% ionic nd 7.9% covlent. (g) SiC: From Eq. (-0), the Si C bond is % ionic nd 89% covlent. (h) Fe O : From Eq. (-0), the Fe-O bond is 55.5% ionic nd 44.5% covlent. (i) MgO: From Eq. (-0), the Mg-O bond is 70.% ionic nd 9.8% covlent. (j) W: Tungsten bonding primrily consists of metllic bonding with some covlent chrcter. (k) H O within the molecules: The H O bond is 8.7% ionic nd 6.% covlent. (l) H O between the molecules: Hydrogen bonding exists between H O molecules. Smith Foundtions of Mterils Science nd Engineering Solution Mnul 0

18 CHAPTER ATOMIC STRUCTURE AND BONDING. Define crystlline solid. A crystlline solid is one which hs crystl structure in which toms or ions re rrnged in pttern tht repets itself in three dimensions.. Define crystl structure. Give exmples of mterils which hve crystl structures. A crystl structure is identicl to crystlline solid, s defined by the solution of Problem.. Exmples include metls, ionic crystls nd certin cermic mterils.. Define spce lttice. A spce lttice is n infinite three-dimensionl rry of points with ech point hving identicl surrounding points..4 Define unit cell of spce lttice. Wht lttice constnts define unit cell? The unit cell of spce lttice represents repeting unit of tomic sptil positions. The cell is defined by the mgnitudes nd directions of three lttice vectors,, b, nd c: xil lengths, b, nd c; interxil ngles α, β,nd γ..5 Wht re the 4 Brvis unit cells? The fourteen Brvis lttices re: simple cubic, body-centered cubic, fce-centered cubic, simple tetrgonl, body-centered tetrgonl, simple orthorhombic, bse-centered orthorhombic, body-centered orthorhombic, fce-centered orthorhombic, simple rhombohedrl, simple hexgonl, simple monoclinic, bse-centered monoclinic, nd simple triclinic..6 Wht re the three most common metl crystl structures? List five metls which hve ech of these crystl structures. The three most common crystl structures found in metls re: body-centered cubic (BCC), fce-centered cubic (FCC), nd hexgonl close-pcked (HCP). Exmples of metls hving these structures include the following. BCC: α iron, vndium, tungsten, niobium, nd chromium. FCC: copper, luminum, led, nickel, nd silver. HCP: mgnesium, α titnium, zinc, beryllium, nd cdmium..7 How mny toms per unit cell re there in the BCC crystl structure? Smith Foundtions of Mterils Science nd Engineering Solution Mnul

19 A BCC crystl structure hs two toms in ech unit cell..8 Wht is the coordintion number for the toms in the BCC crystl structure? A BCC crystl structure hs coordintion number of eight..9 Wht is the reltionship between the length of the side of the BCC unit cell nd the rdius of its toms? In BCC unit cell, one complete tom nd two tom eighths touch ech other long the cube digonl. This geometry trnsltes into the reltionship 4 R..0 Molybdenum t 0ºC is BCC nd hs n tomic rdius of 0.40 nm. Clculte vlue for its lttice constnt in nnometers. Letting represent the edge length of the BCC unit cell nd R the molybdenum tomic rdius, R or R (0.40 nm) 0. nm. Niobium t 0ºC is BCC nd hs n tomic rdius of 0.4 nm. Clculte vlue for its lttice constnt in nnometers. For BCC unit cell hving n edge length nd contining niobium toms, R or R (0.4 nm) 0.0 nm. Lithium t 0ºC is BCC nd hs n tomic rdius of nm. Clculte vlue for the tomic rdius of lithium tom in nnometers. For the lithium BCC structure, which hs lttice constnt of nm, the tomic rdius is, R (0.509 nm) 0.5 nm 4 4. Sodium t 0ºC is BCC nd hs n tomic rdius of nm. Clculte vlue for the tomic rdius of sodium tom in nnometers. For the sodium BCC structure, with lttice constnt of nm, the tomic rdius is, Smith Foundtions of Mterils Science nd Engineering Solution Mnul

20 R ( nm) 0.86 nm How mny toms per unit cell re there in the FCC crystl structure? Ech unit cell of the FCC crystl structure contins four toms..5 Wht is the coordintion number for the toms in the FCC crystl structure? The FCC crystl structure hs coordintion number of twelve..6 Gold is FCC nd hs lttice constnt of nm. Clculte vlue for the tomic rdius of gold tom in nnometers. For the gold FCC structure, which hs lttice constnt of nm, the tomic rdius is, R ( nm) 0.44 nm Pltinum is FCC nd hs lttice constnt of 0.99 nm. Clculte vlue for the tomic rdius of pltinum tom in nnometers. For the pltinum FCC structure, with lttice constnt of 0.99 nm, the tomic rdius is, R (0.99 nm) 0.9 nm Plldium is FCC nd hs n tomic rdius of 0.7 nm. Clculte vlue for its lttice constnt in nnometers. Letting represent the FCC unit cell edge length nd R the plldium tomic rdius, R or R (0.7 nm) 0.87 nm.9 Strontium is FCC nd hs n tomic rdius of 0.5 nm. Clculte vlue for its lttice constnt in nnometers. For n FCC unit cell hving n edge length n contining strontium toms, R or R (0.5 nm) nm Smith Foundtions of Mterils Science nd Engineering Solution Mnul

21 .0 Clculte the tomic pcking fctor for the FCC structure. By definition, the tomic pcking fctor is given s: volume of toms in FCC unit cell Atomic pcking fctor volume of the FCC unit cell These volumes, ssocited with the four-tom FCC unit cell, re 4 6 Vtoms 4 πr πr nd Vunit cell 4R where represents the lttice constnt. Substituting, 64R Vunit cell The tomic pcking fctor then becomes, πr 6 π APF (FCC unit cell) R How mny toms per unit cell re there in the HCP crystl structure? The hexgonl prism contins six toms.. Wht is the coordintion number for the toms in the HCP crystl structure? The coordintion number ssocited with the HCP crystl structure is twelve.. Wht is the idel c/ rtio for HCP metls? The idel c/ rtio for HCP metls is.6; however, the ctul rtios my devite significntly from this vlue..4 Of the following HCP metls, which hve higher or lower c/ rtios thn the idel rtio: Zr, Ti, Zn, Mg, Co, Cd nd Be? Cdmium nd zinc hve significntly higher c/ rtios while zirconium, titnium, mgnesium, coblt nd beryllium hve slightly lower rtios..5 Clculte the volume in cubic nnometers of the titnium crystl structure unit cell. Titnium is HCP t 0ºC with nm nd c nm. For hexgonl prism, of height c nd side length, the volume is given by: Smith Foundtions of Mterils Science nd Engineering Solution Mnul 4

22 V (Are of Bse)(Height) [(6 Equilterl Tringle Are)(Height)] ( sin 60 )( c) ( nm) (sin 60 )(0.468 nm) 0.06 nm.6 Rhenium t 0ºC is HCP. The height c of its unit cell is nm nd its c/ rtio is.6 nm. Clculte vlue for its lttice constnt in nnometers. The rhenium lttice constnt is clculted s, c nm 0.7 nm c/.6.7 Osmium t 0ºC is HCP. Using vlue of 0.5 nm for the tomic rdius of osmium toms, clculte vlue for its unit-cell volume. Assume pcking fctor of From the definition of the tomic pcking fctor, volume of toms in HCP unit cell HCP unit cell volume APF Since there re six toms in the HCP unit cell, the volume of toms is: V toms 4 6 πr 8 π(0.5) nm The unit cell volume thus becomes, nm HCP unit cell volume nm How re tomic positions locted in cubic unit cells? Atomic positions re locted in cubic unit cells using rectngulr x, y, nd z xes nd unit distnces long the respective xes. The directions of these xes re shown below. +z -x -y +y +x -z Smith Foundtions of Mterils Science nd Engineering Solution Mnul 5

23 .9 List the tom positions for the eight corner nd six fce-centered toms of the FCC unit cell. The tom positions t the corners of n FCC unit cell re: (0, 0, 0), (, 0, 0), (,, 0), (0,, 0), (0, 0, ), (, 0, ), (,, ), (0,, ) On the fces of the FCC unit cell, toms re locted t: (½, ½, 0), (½, 0, ½), (0, ½, ½), (½, ½, ), (, ½, ½), (½,, ½).0 How re the indices for crystllogrphic direction in cubic unit cell determined? For cubic crystls, the crystllogrphic direction indices re the components of the direction vector, resolved long ech of the coordinte xes nd reduced to the smllest integers. These indices re designted s [uvw].. Drw the following directions in BCC unit cell nd list the position coordintes of the toms whose centers re intersected by the direction vector: () [00] (b) [0] (c) [] z (,, ) x y (0, 0, 0) (0, 0, 0) (0, 0, 0) (, 0, 0) (,, 0) () Position Coordintes: (b) Position Coordintes: (c) Position Coordintes: (0, 0, 0), (, 0, 0) (0, 0, 0), (,, 0) (0, 0, 0), (,, ). Drw direction vectors in unit cubes for the following cubic directions: () [ ] (b) [0] (c) [] (d) [] z () (b) y x x + y - z - x + y - z 0 [ ] [ 0] Smith Foundtions of Mterils Science nd Engineering Solution Mnul 6

24 (c) ½ ½ (d) ⅓ [] Dividing by, x -½ y z -½ [] Dividing by, x ⅓ y ⅓ z ⅓. Drw direction vectors in unit cubes for the following cubic directions: () [ ] (d) [0] (g) [0] (j) [0 ] z (b) [ ] (e) [] (h) [ ] (k) [ ] (c) [] (f) [] (i) [] (l) [] y ½ x ½ ⅓ ⅓ () Dividing [] by, x, y, z ⅔ (b) Dividing [] by, x, y, z ½ (c) Dividing [] by, x, y, z ½ (d) Dividing [0] by, x 0, y, z (e) Dividing [ ] by, x, y, z (f) Dividing [] by, x, y, z ⅔ Smith Foundtions of Mterils Science nd Engineering Solution Mnul 7

25 ½ ½ ⅓ ⅔ (g) For [ 0], x, y 0, z (h) Dividing [] by, x, y, z (i) Dividing [] by, x, y, z ⅓ ½ ⅔ ⅔ (j) Dividing [0] by, x, y 0, z (k) Dividing [] by, x, y, z (l) Dividing [] by, x, y, z.4 Wht re the indices of the directions shown in the unit cubes of Fig. P.4? x () ½ ½ z ¾ d ¼ c b ⅓ ¼ y (b) ⅔ ¼ x z ¼ e f g ¾ h ¾ ½ ⅓ y Smith Foundtions of Mterils Science nd Engineering Solution Mnul 8

26 ⅓ / 6 b ¼ c ½ ¼ New 0 ½ New 0. Vector components: x -, y, z 0 Direction indices: [0] b. Moving direction vector down ¼, vector components re: x, y -, z ¼ Direction indices: [44] c. Moving direction vector forwrd ½, vector components re: x - / 6, y, z Direction indices: [66] ¼ ½ ¾ d e New 0 ⅔ f -⅓ ⅓ ¼ New 0 d. Moving direction vector left ¼, vector components re: x, y ½, z Direction indices: [] New 0 e. Vector components re: x -¾, y -, z Direction indices: [44] New 0 f. Moving direction vector up ⅓, vector components re: x -, y, z -⅓ Direction indices: [] ¾ ½ ¾ g -¾ h ¼ ¾ g. Moving direction vector up ½, vector components re: x, y -, z -¼ Direction indices: [44] h. Moving direction vector up ¼, vector components re: x ¾, y -, z -¾ Direction indices: [4] Smith Foundtions of Mterils Science nd Engineering Solution Mnul 9

27 .5 A direction vector psses through unit cube from the ( ¾, 0, ¼ ) to the ( ½,, 0 ) positions. Wht re its direction indices? The strting point coordintes, subtrcted from the end point, give the vector components: x y 0 z The frctions cn then be clered through multipliction by 4, giving x, y 4, z. The direction indices re therefore [4]..6 A direction vector psses through unit cube from the (, 0, ¾ ) to the ( ¼,, ¼ ) positions. Wht re its direction indices? Subtrcting coordintes, the vector components re: x y 0 z Clering frctions through multipliction by 4, gives x, y 4, z. The direction indices re therefore [ 4 ]..7 Wht re the crystllogrphic directions of fmily or form? Wht generlized nottion is used to indicte them? A fmily or form hs equivlent crystllogrphic directions; the tom spcing long ech direction is identicl. These directions re indicted by uvw..8 Wht re the directions of the 00 fmily or form for unit cube? [0 0], [00], [00], [00], [00], [00].9 Wht re the directions of the fmily or form for unit cube? [], [], [], [], [], [ ], [], [] [ 0] [0 ].40 Wht 0 -type directions lie on the () plne of cubic unit cell? [0], [0], [0], [0], [0], [0 ] [ 0 ] [ 0] [0] [0] Smith Foundtions of Mterils Science nd Engineering Solution Mnul 0

28 .4 Wht -type directions lie on the (0) plne of cubic unit cell? [], [], [ ], [] [] [] [] [].4 How re the Miller indices for crystllogrphic plne in cubic unit cell determined? Wht generlized nottion is used to indicte them? The Miller indices re determined by first identifying the frctionl intercepts which the plne mkes with the crystllogrphic x, y, nd z xes of the cubic unit cell. Then ll frctions must be clered such tht the smllest set of whole numbers is ttined. The generl nottion used to indicte these indices is (hkl), where h, k, nd l correspond to the x, y nd z xes, respectively..4 Drw in unit cubes the crystl plnes tht hve the following Miller indices: () ( ) (d) () (g) (0) (j) () z (b) (0 ) (e) () (h) () (k) ( ) y (c) ( ) (f) (0 ) (i) () (l) () x - (0, 0, 0) (0, 0, 0) -½ (0, 0, 0) ( ) (0) -½ + ( ) -. For ( ) reciprocls re: x, y -, z - +½ (0, 0, 0) -⅓ () + b. For (0) reciprocls re: x, y, z -½ -½ () + (0, 0, 0) +⅓ c. For ( ) reciprocls re: x, y -½, z - +⅓ (0, 0, 0) (0) -½ d. For ()reciprocls e. For () reciprocls f. For (0) reciprocls re: x ½, y, z -⅓ re: x ⅓, y -½, z re: x ⅓, y, z -½ Smith Foundtions of Mterils Science nd Engineering Solution Mnul

29 +½ (0, 0, 0) - (0) g. For (0)reciprocls re: x ½, y, z - -½ -½ (0, 0, 0) + () h. For ()reciprocls re: x -½, y, z -½ +½ (0, 0, 0) -½ +⅓ () i. For () reciprocls re: x -½, y ⅓, z ½ + (0, 0, 0) -⅓ +⅓ +½ -⅓ (0, 0, 0) +⅓ () j. For () reciprocls re: x, y ⅓, z -⅓ - () (0, 0, 0) +⅓ k. For () reciprocls re: x ⅓, y -, z ½ - () k. For ()reciprocls re: x -⅓, y ⅓, z -.44 Wht re the Miller indices of the cubic crystllogrphic plnes shown in Fig. P.44? () d ½ ⅓ z c ⅔ ⅓ ¾ ¾ ⅓ y (b) ⅔ z ¼ b c d ½ ⅓ y x b x ⅔ Smith Foundtions of Mterils Science nd Engineering Solution Mnul

30 Miller Indices for Figure P.44() Plne bsed on (0,, ) s origin Plne b bsed on (,, 0) s origin Plnr Intercepts x y - z 4 Reciprocls of Intercepts Plnr Intercepts Reciprocls of Intercepts 0 x x - x y 5 y y 5 4 z z 0 z The Miller indices of plne re (04). The Miller indices of plne b re (5 0). Plne c bsed on (,, 0) s origin Plne d bsed on (0, 0, 0) s origin Plnr Intercepts x y - z Reciprocls of Intercepts Plnr Intercepts Reciprocls of Intercepts 0 x x x y y y z z z The Miller indices of plne c re (0 ). The Miller indices of plne d re (). Miller Indices for Figure P.44(b) Plne bsed on (, 0, ) s origin Plne b bsed on (0,, ) s origin Plnr Intercepts x - y z Reciprocls of Intercepts Plnr Intercepts Reciprocls of Intercepts x x x 0 y y z z z The Miller indices of plne re (0). The Miller indices of plne b re (). Smith Foundtions of Mterils Science nd Engineering Solution Mnul

31 Plne c bsed on (0,, 0) s origin Plne d bsed on (0,, 0) s origin Plnr Intercepts x 5 y z Reciprocls of Intercepts Plnr Intercepts Reciprocls of Intercepts x x x y - y 5 y 0 z z z The Miller indices of plne c re ( 5 0). The Miller indices of plne d re ( )..45 Wht is the nottion used to indicte fmily or form of cubic crystllogrphic plnes? A fmily or form of cubic crystllogrphic plne is indicted using the nottion {hkl}..46 Wht re the {00} fmily of plnes of the cubic system? (00), (00), (00), (00), (00), (00).47 Drw the following crystllogrphic plnes in BCC unit cell nd list the position of the toms whose centers re intersected by ech of the plnes: () (00) (b) (0) (c) () z x. (, 0, 0), (, 0, ), (,, 0), (,, ) y x z y b. (, 0, 0), (, 0, ), (0,, 0), (0,, ), (½, ½, ½) x z c. (, 0, 0), (0, 0, ), (0,, 0) y.48 Drw the following crystllogrphic plnes in n FCC unit cell nd list the position coordintes of the toms whose centers re intersected by ech of the plnes: () (00) (b) (0) (c) () Smith Foundtions of Mterils Science nd Engineering Solution Mnul 4

32 z y x. (, 0, 0), (, 0, ), (,, 0), (,, ) (, ½, ½) x z y y b. (, 0, 0), (, 0, ), (0,, 0), (0,, ), (½, ½, 0), ( ½, ½, ) x z c. (, 0, 0), (0, 0, ), (0,, 0), (½, 0, ½) (½, ½, 0), (0, ½, ½).49 A cubic plne hs the following xil intercepts: ⅓, b -⅔, c ½. Wht re the Miller indices of this plne? Given the xil intercepts of (⅓, -⅔, ½), the reciprocl intercepts re:,,. x y z Multiplying by to cler the frction, the Miller indices re (64)..50 A cubic plne hs the following xil intercepts: -½, b -½, c ⅔. Wht re the Miller indices of this plne? Given the xil intercepts of (-½, -½, ⅔), the reciprocl intercepts re:,,. x y z Multiplying by, the Miller indices re (44)..5 A cubic plne hs the following xil intercepts:, b ⅔, c -½. Wht re the Miller indices of this plne? Given the xil intercepts of (, ⅔, -½), the reciprocl intercepts re:,,. x y z Multiplying by, the Miller indices re (4)..5 Determine the Miller indices of the cubic crystl plne tht intersects the following position coordintes: (, 0, 0 ); (, ½, ¼ ); ( ½, ½, 0 ). First locte the three position coordintes s shown. Next, connect points nd b, extending the line to point d nd connect to c nd extend to e. Complete the plne by Smith Foundtions of Mterils Science nd Engineering Solution Mnul 5

33 connecting point d to e. Using (,, 0) s the plne origin, x -, y - nd z ½. The intercept reciprocls re thus,,. The Miller indices re ( ). x y z (, 0, 0) (0, 0, 0) (, ½, ¼) (½, ½, 0) b d c e (,, 0).5 Determine the Miller indices of the cubic crystl plne tht intersects the following position coordintes: ( ½, 0, ½ ); ( 0, 0, ); (,, ). First locte the three position coordintes s shown. Next, connect points nd b nd extend the line to point d. Complete the plne by connecting point d to c nd point c to b. Using (, 0, ) s the plne origin, x -, y nd z. The intercept reciprocls re thus,,. The Miller indices re (). x y z (0, 0, ) (,, ) (0, 0, 0) (, 0, ) b c (½, 0, ½ ) d.54 Determine the Miller indices of the cubic crystl plne tht intersects the following position coordintes: (, ½, ); (½, 0, ¾ ); (, 0, ½ ). After locting the three position coordintes, connect points b nd c nd extend the line to point d. Complete the plne by connecting point d to nd to c. Using (, 0, ) s the plne origin, x -, y ½ nd z -½. The intercept reciprocls then become,,. The Miller indices re (). x y z Smith Foundtions of Mterils Science nd Engineering Solution Mnul 6

34 d (½, 0, ¾) b (, 0, ) (0, 0, 0) (, 0, ½,) c (, ½, ).55 Determine the Miller indices of the cubic crystl plne tht intersects the following position coordintes: ( 0, 0, ½ ); (, 0, 0 ); ( ½, ¼, 0 ). After locting the three position coordintes, connect points b nd c nd extend the line to point d. Complete the plne by connecting point d to nd to b. Using (0, 0, 0) s the plne origin, x, y ½ nd z ½. The intercept reciprocls re thus,,. The Miller indices re therefore ( ). x y z (0, 0, 0) d (0, 0, ½,) c (, 0, 0) (½, ¼, 0) b (0, ½, 0).56 Rodium is FCC nd hs lttice constnt of nm. Clculte the following interplnr spcings: () d (b) d 00 (c) d nm nm d 0.0 nm + + () nm nm d 0.90 nm (b) nm nm d 0.5 nm (c) 0 Smith Foundtions of Mterils Science nd Engineering Solution Mnul 7

35 .57 Tungsten is BCC nd hs lttice constnt of nm. Clculte the following interplnr spcings: () d 0 (b) d 0 (c) d nm nm d 0.4 nm () nm nm d 0. nm (b) nm nm d 0.00 nm (c) 0.58 The d 0 interplnr spcing in BCC element is nm. () Wht is its lttice constnt? (b) Wht is the tomic rdius of the element? (c) Wht could this element be? () (b) d0 h + k + l (0.587 nm) nm (0.50 nm) R 0.7 nm 4 4 (c) The element is brium (B)..59 The d 4 interplnr spcing in FCC metl is nm. () Wht is its lttice constnt? (b) Wht is the tomic rdius of the metl? (c) Wht could this metl be? () (b) d4 h + k + l ( nm) nm (0.408 nm) R 0.44 nm 4 4 (c) The element is gold (Au). Smith Foundtions of Mterils Science nd Engineering Solution Mnul 8

36 .60 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 0º; 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: () ( 0 ) (d) ( ) (g) () (j) (00) (b) (0 ) (e) ( ) (h) ( 00) (k) ( ) (c) (0) (f) ( 0 ) (i) (0 ) (l) (0) The reciprocls of the indices provided give the intercepts for the plne (,,, nd c). (0) (0) (0).,,, c b.,,, c c.,,, c () () ( 0) d.,,, c e.,,, c f.,,, c Smith Foundtions of Mterils Science nd Engineering Solution Mnul 9

37 () (00) (0) g.,,, c h.,,, c i.,,, c (00) () (0) 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 0 Reciprocls Plnr of Intercepts Intercepts -½ 0 ½ Reciprocls of Intercepts 0 Smith Foundtions of Mterils Science nd Engineering Solution Mnul 40

38 Plnr Intercepts c Reciprocls of Intercepts Plnr Intercepts Reciprocls of Intercepts Plnr Intercepts Reciprocls of Intercepts 0 c c ½ c 0 c The Miller indices of plne re (00). The Miller indices of plne b re (0 ). The Miller indices of plne c re (00). 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. (00) The Miller indices of plne b re. ( 0) The Miller indices of plne c re ( 0)..64 Determine the Miller-Brvis direction indices of the, - nd directions. The Miller-Brvis direction indices corresponding to the, - nd directions re respectively, [ 0 0 0], [0 0 0], nd [0 0 0 ]..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

39 .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. [04],[04],[04], [04] [ 04] [04] [04] [04] [04] [0 4],[04],[0 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 [ 0 ] nd [ 0 ]..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 0.74, 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 {0 0 0 } plnes..70 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 0 directions. (b) The closest-pcked directions in the HCP lttice re the 0 directions. Smith Foundtions of Mterils Science nd Engineering Solution Mnul 4

40 .7 The lttice constnt for BCC tntlum t 0ºC is 0.06 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 )(0 cm /m )( m) g/u.c. Since there re two toms in BCC unit cell, the tomic mss is: ( g/unit cell)(6.0 0 toms/mol) Atomic mss toms/unit cell g/mol.7 Clculte vlue for the density of FCC pltinum in grms per cubic centimeter from its lttice constnt of 0.99 nm nd its tomic mss of 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)(95.09 g/mol) mss/unit cell.96 0 g/unit cell toms/mol The density is then found s, ρ υ mss/unit cell mss/unit cell.96 0 g/unit cell 9 volume/unit cell [( m) ]/ unit cell m,445, g/m.45 g/cm 00 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 nm: () (00), (b) (0), (c) (). () (b) (c) Smith Foundtions of Mterils Science nd Engineering Solution Mnul 4