Solution Manual for Principles Of Electronic Materials And Devices, 4th Edition

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 1Solutions Manual toPrinciples of Electronic Materials and DevicesFourth EditionCHAPTER 1Safa KasapUniversity ofSaskatchewanCanadaWater molecules are polar. A water jet can be bent by bringing a charged comb near the jet. The polarmolecules areattracted towards higher fields at the comb'ssurface (Photo by SK)

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 1FourthEditionChapter 1Answers to"Why?" in the textPage 31:Oxygen has an atomic mass of 16 whereas it is 14 for nitrogen. The O2molecule is therefore heavier thanthe N2molecule. Thus, from)(321221kTm=v,the rms velocity of O2molecules is smaller than that of N2molecules.Page 34, footnote 11For small extensions, the difference between the engineering and instantaneous strains due to atemperature change are the same.Historically, mechanical and civil engineers measured extension bymonitoring the change in length,L; and the instantaneous lengthLwas not measured. It is not trivial tomeasure both the instantaneous length and the extension simultaneously. However, since we knowLoandmeasureL, the instantons lengthL=Lo+L. Is the difference important? Considera sample of lengthLothat extends to a final lengthLdue toa temperaturechange fromTotoT.Let= (L−Lo) /Lo=L/Lobe the engineering strain.The engineeringdefinitionof strain and hence the thermal expansion coefficient isEngineering strain =TLLo=so thatthermal expansionfromTotoTgives,=TTLLooodTLdL)(ooTTLL−=)(oTT−=(1)where=L/Lois the engineering strainas defined abovePhysics definition of strain and hence the thermal expansion coefficient isInstantaneous train =TLL=so that thermal expansion fromTotoTgives,=TTLLoodTLdL)(lnooTTLL−=())(1lnoTT−=+(2)We can expand the ln(1 +) term for small,so that Equation (2)essentially becomes Equation (1)1.1 Virial theoremThe Li atom has a nucleus with a +3epositive charge, which is surrounded by a full1sshell withtwoelectrons, and a single valence electron in the outer 2ssubshell. The atomic radius of theLi atom is about 0.17 nm. Using the Virial theorem, and assuming that the valence electron sees the

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 1nuclear +3eshielded by the two 1s electrons, that is,a net charge of +e, estimate the ionization energy ofLi(the energy required to free the 2selectron).Compare this value with the experimental value of 5.39eV. Suppose that the actual nuclear charge seen by the valence electron is not +ebut a little higher, say+1.25e, due to the imperfect shielding provided by the closed 1sshell. What would be the new ionizationenergy? What is your conclusion?SolutionFirst we consider the case when the outermost valence electron can see a net charge of +e. FromCoulomb’s law we have the potential energy0000rπεeerπεQQPE4))((421−+==m)10)(0.17Fm108.85(4C)10(1.69112219−−−−−==−1.35410−18J or−8.46 eVVirial theorem relates the overall energy, the average kineticenergyKE, and average potential energyPEthrough therelationsKEPEE+=andPEKE21−=Thus using Virial theorem, the total energy iseV46.85.021−==PEE=−4.23 eVThe ionization energy is therefore4.23 eV.Considernowthe second case wheretheelectronsees +1.25edue to imperfect shielding. Againthe CoulombicPEbetween +eand +1.25ewill be000021r4πeer4πQQPE))(1.25(−+==m)10)(0.17Fm10(854C)10(1.61.259112219−−−−−=π=−1.69210−18J or−10.58 eVThe total energy is,eV29.521−==PEEThe ionization energy,considering imperfect shielding,is5.29 eV. This value is incloseragreement with the experimental value. Hence the second assumption seems to be more realistic.1.2 Virialtheorem and the He atomIn Example 1.1 we calculated the radius of the H-atom using theVirial theorem. First consider the He+atom, which as shown in Figure 1.75a, has one electron in the K-sell orbiting the nucleus. Take thePEand theKEas zero when the electrons and the nucleus are infinitelyseparated. The nucleus has a charge of +2eand there is one electron orbiting the nucleus at a radiusr2.Using the Virial theorem show that the energy of the He+ion is

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 12242)2/1()He(reEo−=+Energy of He+ion[1.48]Now consider the He-atom shown in Figure 1.75b. There are two electrons. Each electron interacts withthe nucleus (at a distancer1) and the other electron (at a distance 2r1). Using the Virial theorem show thatthe energy of the He atom is−=1287)2/1()He(reEoEnergy of He atom[1.49]The first ionization energyEI1is defined as the energy required to remove one electron from the Heatom. The second ionization energyEI2is the energy required to remove the second (last) electron fromHe+. Both are shown in Figure 1.75These have been measured and given asEI1= 2372 kJ mole−1andEI2= 5250 kJ mol−1. Find the radiir1andr2for He and He+. Note that the first ionization energy providessufficient energy to take He to He+, that is, He→He++e−absorbs 2372 kJ mol−1. How does yourr1value compare with the often quoted He radius of 31 pm?Figure 1.75:(a) A classical view of a He+ion. There is one electron in theK-shell orbiting the nucleus that has acharge +2e.(b) The He atom. There are two electrons in theK-shell. Due to their mutual repulsion, they orbit tovoid each other.SolutionVirial theorem relates the overall energy, the average kinetic energyKE, and average potential energyPEthrough the relationsEKEPEEPEKEKEPEE21;21;21;−==−=+=(1)Now, consider thePEof the electron in Figure 1.75a. The electron interacts with +2eof positivecharge, so that222424)2)((rereePEoo−=−=which means that the total energy (average) is2222442)2/1(21)He(rerePEEoo−=−==+(2)whichis the desired result.

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 1Now consider Figure 1.75b. Assume that, at all times, the electrons avoid each other by staying inopposite parts of the orbit they share. They are "diagonally"opposite to each other. The PE of this systemof 2 electrons one nucleus with +2eisPE=PEof electron 1 (left) interacting with the nucleus (+2e), at a distancer1+PEof electron 2 (right) interacting with the nucleus (+2e), at a distancer1+PEof electron 1 (left) interacting with electron 2 (right) separated by 2r1)2(4))((4)2)((4)2)((111reereereePEooo−−+−+−=2287rePEo−=From the Virial theoremin Equation (1)−=1287)2/1()He(reEo(3)We are given,EI1= Energy required to remove one electron from the He atom= 2372 kJ mole−1=25.58eVEI2= Energy required to remove the second (last) electron from He+=5250 kJ mol−1=54.41eVThe eV values were obtained by using=AeNEE1)J/mole()eV(We can now calculate the radii as follows. Starting with Equation 2 for the ionization of He+,2112219222192)mF10854.8(4)C10602.1(4)He()J/eV10602.1)(eV41.54(rreEEEoII−−−+−=====from which,r2= 2.6510−11or26.5 pmThe calculation ofr1involves realizing that Equation (3) is the energy of the whole He atom, with 2electrons. If we remove 1 electron we are left with He+whose energy is Equation (2). Thusthe)He()He(1+−=EEEI2121987)2/1()He()He(J/eV)10602.1(eV)58.24(IoEreEE−=−=+−11122191219)mF10854.8(4)C10602.1(87)2/1()J/eV10602.1(eV)58.24eV41.54(rreo−−−−==+fromr1= 3.1910−11or31.9 pmvery close to the quoted value of 31 pm in various handbooks or internet period tables____________________________________________________________________________________1.3Atomic mass and molar fractions

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 1a.Consider a multicomponent alloy containingNelements.Ifw1,w2, ...,wNare the weight fractions ofcomponents 1,2,...,Nin the alloy andM1,M2, ...,MN, are the respective atomic masses of theelements, show that the atomic fraction of thei-th component is given byNNiiiMwMwMwMwn+++=.../2211Weight to atomic percentageb.Suppose that a substance (compound or an alloy) is composed ofNelements,A, B, C,... and that weknow their atomic (or molar) fractionsnA,nBnC, ....Show that theweight fractionswA,wB,wC,....aregiven by...+++=CCBBAAAAAMnMnMnMnw...+++=CCBBAABBBMnMnMnMnwAtomic to weight percentagec.Consider the semiconducting II-VI compound cadmium selenide, CdSe. Given the atomic masses ofCd and Se, find the weight fractions of Cd and Se in the compound and grams of Cd and Se neededto make 100 grams of CdSe.d.A Se-Te-P glass alloy has the composition 77 wt.%Se, 20 wt.% Te and 3 wt.% P.Given theiratomicmasses, what are the atomic fractions of these constituents?Solutiona.Suppose thatn1, n2, n3,…, ni,…, nNare the atomic fractions of the elements in the alloy,n1+n2+n3+… +nN= 1Suppose that we have 1 mole of the alloy. Then it hasnimoles of an atom with atomic massMi(atomicfractions also represent molar fractions in the alloy). Suppose that we have 1 gram of the alloy. Sincewiisthe weight fraction of thei-th atom,wiis also the mass ofi-th element in grams in the alloy. The numberof moles in the alloy is thenwi/Mi. Thus,Number of moles of elementi=wi/MiNumber of molesinthe whole alloy =w1/M1+w2/M2+…+wi/Mi+…+wN/MNMolar fraction or the atomic fraction of thei-th elements is therefore,alloyinmolesofnumbersTotalelementofmolesofNumebrini=NNiiiMwMwMwMwn+++=.../2211b.Suppose that we have the atomic fractionniof an element with atomic massMi. The mass of theelement in the alloy will be the product of the atomic mass with the atomic fraction,i.e.niMi. Mass of thealloy is thereforenAMA+ nBMB+…+nNMN=MalloyBy definition, the weight fractionis,wi= mass of the elementi/Mass of alloy.Therefore,...+++=CCBBAAAAAMnMnMnMnw

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 1...+++=CCBBAABBBMnMnMnMnwc.The atomic mass of Cd and Se are 112.41 g mol−1and 78.96 g mol−1. Since one atom of each elementis in the compound CdSe, the atomic fraction,nCdandnSeare 0.5. The weight fraction of Cd in CdSe istherefore111SeSeCdCdCdCdCdmolg96.785.0molg41.1125.0molg41.1125.0−−−+=+=MnMnMnw=0.587 or 58.7%Similarly weight fraction of Se is111SeSeCdCdSeSeSemolg96.785.0molg41.1125.0molg96.785.0−−−+=+=MnMnMnw=0.4126or 41.3%Consider100 g of CdSe.Then the mass of Cd we need isMassof Cd =wCdMcompound= 0.587100 g =58.7 g(Cd)andMass of Se =wSeMcompound= 0.413100 g =41.3g(Se)d.The atomic fractions of the constituents can be calculated using the relations proved above. The atomicmasses of the components areMSe= 78.6 g mol−1,MTe= 127.6 g mol−1,andMP= 30.974 g mol−1.Applying the weight to atomic fraction conversion equation derived in part (a) we find,1111PPTeTeSeSeSeSeSemolg974.3003.0molg6.1272.0molg6.7877.0molg6.7877.0/−−−−++=++=MwMwMwMwnnSe= 0.794or 79.4%1111PPTeTeSeSeTeTeTemolg974.3003.0molg6.1272.0molg6.7877.0molg6.1272.0/−−−−++=++=MwMwMwMwnnTe= 0.127or 12.7 %1111PPTeTeSeSePPPmolg974.3003.0molg6.1272.0molg6.7877.0molg974.3003.0/−−−−++=++=MwMwMwMwnnP= 0.0785or 7.9%1.4 Mean atomic separation, surface concentration and densityThere are many instances where weonly wish to use reasonable estimates for the mean separationbetween the host atoms in a crystal and the

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 1mean separationbetween impurities in the crystal. These can be related in a simple way to the atomicconcentrationof the host atoms and atomic concentrationof the impurity atoms respectively. The finalresult does not depend on the sample geometry or volume. Sometimes we need to know the number ofatoms per unit areanson the surface of a solid given the number of atoms per unit volume in the bulk,nb.Consider a crystal of the material of interest which is a cube of sideLas shown in Figure 1.76. To eachatom, we can attribute a portion of the whole volume, which is a cube of sidea. Thus, each atom isconsidered to occupy a volume ofa3. Suppose that there areNatoms in the volumeL3. Thus,L3=Na3.a.Ifnbis the bulk concentration of atoms, show that the mean separationabetween the atoms isgiven by3/1bna=.b.Show that the surface concentrationnsof atoms is given by3/2bsnn=.c.Show that the density of the solid is given byAbNMn/at=whereMatis the atomic mass.Calculate the atomic concentration in Si from its density (2.33 g cm−3)d.A silicon crystal has been doped with phosphorus. The P concentration in the crystal is 1016cm−3.P atoms substitute for Si atoms and are randomly distributed in the crystal. What is the mean separationbetween the P atoms?Figure 1.76Consider a crystal that has volumeL3. This volume is proportioned to each atom, which is a cube ofsidea3.Solutiona.Consider a crystal of the material which is a cube of volumeL3, so that each side has a lengthLasshown in Figure 1.76. To each atom, we can attributea portion of the whole volume. For simplicity, wetake the volume proportioned to an atom to bea3, that is, each atom is considered to occupy a volume ofa3.The actual or true volume of the atom does not matter. All we need to know is how much volumean atom has around it given all the atoms are identical and that adding all the atomic volumes must givethe whole volume of the crystal.Suppose that there areNatomsin this crystal. Thennb=N/L3is the atomic concentrationin thecrystal, thenumber of atoms per unit volume, the so-called bulk concentration. SinceNatoms make upthe crystal, we haveNa3= Crystal volume =L3

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 1Theseparationbetweenany two atoms isa, as shown in Figure 1.76.Thus,3/13/13/1311separationMeanbbnnNLa====(1)Equation 1 can be derived even more simply because an atom has a volume ofa3. In this volumeare is only 1 atom. Sonba3must be 1, which leads to Equation 1.The above idea can be extended to finding the separation between impurity atoms in a crystal.Suppose thatwe wish to determine the separationdbetween impurities,then we can again follow asimilar procedure. In this case, the atoms inFigure 1.76are impurities which are separated byd. Weagain assign a portion of the whole volume (for simplicity, a cubic volume) to each impurity. Eachimpurity atom therefore has a volumed3(because the separation between impurities isd).Following theabove arguments we would find,3/11impuritiesbetweenseparationMeanINd==(2)b.We wish to findthe number ofatoms per unit areanson the surface of a solid given the atomicconcentrationnbin the bulk.Consider Figure 1.76.Eachatom as an areaa2, so that within this surfacearea there is 1 atom. ThusSurface area of 1 atomSurface concentration of atoms = 1ora2ns= 13/221bsnan==(3)Equation (3) is of course based on the simple arrangement of atoms as shown in Figure1.75. Inreality, the surface concentration of atoms depends on the crystal plane at the surface.Equation (3),however, is a reasonable estimate for the order of magnitude fornsgivennb.c.We candetermine the densityfrom the atomic concentrationnband vice versa.The volume inFigure1.75hasL3nbatoms. Thus, the density isAbAbNMnLNMnLat3at3)(VolumeMass===(4)For Si, the atomic masMat= 28.09 g mol−1, so that withgiven as 2.33 g cm−3,)molg09.28()mol10022.6)(cmg33.2(11233at−−−==MNnAb=5.001022cm−.d.TheP concentration in the crystal is 1016cm−3. P atoms substitute for Si atoms and are randomlydistributed in the crystal.We can use Equation (2)3/136163/1)m10101(11impuritiesbetweenseparationMean−===INd=4.6410−8m = 46.4 nm

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 11.5The covalent bondConsider the H2molecule in a simple way as two touching H atoms as depictedin Figure1.77. Does this arrangement have a lower energy than two separated H atoms? Suppose thatelectrons totally correlate their motions so that they move to avoid each other as in the snapshot in Figure1.77.The radiusroof the hydrogen atom is 0.0529 nm. The electrostatic potential energyPEof twochargesQ1andQ2separated by a distanceris given byQ1Q2/(4or).Using the Virial TheoremasinExample 1.1,consider the following:a.Calculate the total electrostatic potential energy (PE) of all the charges when they are arranged asshown in Figure 1.77. In evaluating thePEof the whole collection of charges you must consider allpairs of charges and, at the same time, avoid double counting of interactions between the same pairof charges. The totalPEis the sum of the following: electron 1 interacting with the proton at adistanceroon the left, proton atroon the right, and electron 2 at a distance 2ro+ electron 2interacting with a proton atroand another proton at 3ro+ two protons, separated by 2ro, interactingwith each other. Is this configuration energetically favorable?b.Given that in the isolated H-atom thePEis 2(−13.6 eV), calculate the change inPEin going fromtwo isolated H-atomsto the H2molecule. Using the Virial theorem, find the change in the totalenergy and hence the covalent bond energy. How does this compare with the experimental value of4.51 eV?Figure 1.77A simplified view of the covalent bond inH2. A snapshot at one instant.Solutiona.Consider thePEof the whole arrangement of charges shown in the figure. In evaluating thePEof allthe charges, we must avoid double counting of interactions between the same pair of charges. The totalPEis the sum of the following:Electron 1 interacting with the proton at a distanceroon the left, with the proton atroon the rightand with electron 2 at a distance 2ro+ Electron 2 on the far left interacting with a proton atroand another proton at 3ro+Two protons, separated by 2ro, interacting with each other

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 1PE= −e24oro−e24oro+e24o(2ro)−e24oro−e24o3ro+e24o2roSubstituting and calculating,we findPE=1.017610−17J or-63.52 eVThe negativePEfor thisparticular arrangement indicates that this arrangement of charges is indeedenergetically favorable compared with all the charges infinitely separated (PEis then zero).b.The potential energy of an isolated H-atomis-213.6 eV or-27.2 eV. The difference between thePEof the H2molecule and two isolated H-atoms is,PE=−(63.52) eV−2(-27.2) eV=−9.12eVWe can write the last expression above as thechange in the total energy.eV56.4)eV12.9(2121−=−==PEEThis change in the total energy is negative. The H2molecule has lower energy than two H-atomsby 4.56 eV which is the bonding energy. This is very close to the experimental value of 4.51 eV. (Note:We usedarovalue from quantum mechanics-so the calculation was not totally classical!).1.6Ionic bonding and CsClThe potential energyEper Cs+-Cl−pair within the CsCl crystal depends onthe interionic separationrin the same fashion as in the NaCl crystal,morBrMerE+−=4)(2Energy per ion pair in ionic crystals[1.48]where for CsCl,M= 1.763,B= 1.19210−104J m9or 7.44210−5eV (nm)9andm= 9. Find theequilibrium separation (ro) of the ions in the crystal and the ionic bonding energy, that is, the ioniccohesive energy; and compare the latter value to the experimentalvalue of 657 kJmol−1.Given theionization energyof Cs is 3.89 eV and theelectron affinityof Cl (energy released when an electron isadded) is 3.61 eV, calculate the atomic cohesive energy of the CsCl crystal as joules per mole.SolutionBonding will occur when potential energyE(r) is minimum atr=r0corresponding to the equilibriumseparation between Cs+andCl−ions. Thus,differentiatingE(r) and setting it equal to zero atr = rowehave04)(2=+−===oorrmorrrBrMedrddrrdE04122=−=+orrmorBmrMe

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 104122=−+mooorBmrMe1124−=mooMemBr(1)Thus substituting the appropriate values we have()812199104112C106.1763.1)mJ10192.1(9)Fm108542.8(4=−−−−orro= 3.5710−10mor0.357nm.The minimum energy is the energy atr = ro, that ismooorBrMeE+−=42minwhich in terms of eV is99min)nm()nmeV(4)eV(ooorBreME+−=(2)9941011219)nm357.0(nmeV10442.7)m1057.3)(Fm108542.8(4)763.1)(C106.1(−−−−−+−== −6.32eVper ion pair,or 3.16 eV per ion.The amount of energy required to break upa CsCl crystalinto Cs+and Cl−ions=6.32eVper pair of ions.The correspondingionic cohesive energyisEcohesive= (6.32 eV)(1.60210−19J eV−1)(6.0221023mol−1)=610 kJmol─1of Cs+Cl−ionpairsor610kJmol─1ofCs+ions andCl−ions.(Notfar out from the experimental value given the large numbers and the high index,m= 9, involved inthe calculations.)The amount of energy required to removeanelectronfromCl−ion= 3.61 eV.Theamount of energy released when an electron is put into the Cs+ion= 3.89 eV.Bond Energyper pair of Cs-Cl atoms=6.32eV + 3.61 eV–3.89 eV =6.04eVAtomic cohesive energyin kJ/mol is,Ecohesive= (6.04eV)(1.610−19J eV−1)(6.0221023mol−1)=582 kJmol─1of Cs or Cl atom(i.e.per mole of Cs-Cl atom pairs)=291kJmol─1of atoms------------------------Notes:1.Various books and articles report different values forBandmfor CsCl, which obviously affect thecalculated energy;rois less affected because it requires the(m−1)throot ofmB.Richard Christman

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 1(Introduction to Solid State Physics,JohnWileyand Sons, 1988)in Table 5-1 gives,m= 10.65 andB=3.4410120, quite different than values here, which are closertovalues inAlan Walton,Three Phases ofMatter(2ndEdition),Clarendon (Oxford University)Press, 1983 (pp. 258-259).2.The experimental value of657 kJ mol−1(6.81 eV per ion pair)for the ionic cohesive energy (theionic lattice energy)is fromT. Moeller et al,Chemistry with Inorganic Qualitative Analysis,SecondEdition, Academic Press, 1984) p. 413, Table 13.5.3.Some authors use the term molecular cohesive energy to indicate that the crystal is taken apart tomolecular units e.g. Cs+Cl−, which would correspond to the ionic cohesive energy here.Further, mostchemists use "energy per mole" to imply energy permole ofchemical units, and hence the atomiccohesive energy per mole would usually refer to energy be per Cs and Cl atom pairs.Some authors refertothe atomic cohesive energy per moleascohesive energy permole ofatoms, independent of chemicalformula.4.Equation (1) is124−=moormMeBwhich can be substituted into Equation (2) or [1.48] to eliminateB, that is, find the minimum energy of apair of Cs+-Cl−ions,i.e.122min414−+−=moomormMerrMeE −−=mrMeEoo1142minwhich is called theBorn-Landé equation.1.7 Ionic bonding and LiClEquation 1.48can be used to represent thePEof the ion pair inside the LiCcrystal. LiCl has the NaCl structure withM= 1.748,m= 7.30,B= 2.3410−89J m7.30. Further, theionization energy of Li (Li→Li++e−) is 520.2 kJ mol−1. The electron affinity of Cl (energy associatedwith Cl +e−→Cl−) is 348.7 kJ mol−1(a) Calculate the equilibrium separation of ions the LiCl crystal.(b)Calculate the bonding energy per ion pair in the LiCl crystal.(c)Calculate the atomic cohesive energy ofthe LiCl crystal.(c) Calculate the density of LiCl.SolutionFigure 1Q07-1 shows the crystal structure of NaCl. LiCl has the same crystal structure.(a)The minimum in energy is at04)(2=+−===oorrmorrrBrMedrddrrdE04122=−=+orrmorBmrMe

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 104122=−+mooorBmrMe1124−=mooMemBrThus,substituting the appropriate values we have8121930.789112C)10602.1(748.1)mJ1034.2(30.7)Fm108542.8(4=−−−−orro=2.61210−10mor0.261nm.(b)The minimum energy is the energy atr = ro, that ismooorBrMeE+−=42min8.7108.78910112219)m1061.2(mJ1034.2)m1061.2)(Fm108542.8(4)748.1()C10602.1(−−−−−−+−==−1.3310−18J per of Li+-Cl−ions=−8.32eV per ion pair, or−4.18eV per ion.The bond energy isthereforeEbond=−Emin=8.317eV per ion pair, or4.18 eV per ion.(c)In (b) we calculatedthe amount of energy required to break upLi+-Cl−pair intoLi+and Cl−ions=8.317eV per pair of ions.The correspondingionic cohesive energyisEionic-cohesive= (8.317eV)(1.60210−19J eV−1)(6.02210−23mol−1)=802kJmol─1ofLi+Cl−ion pairsor802kJmol─1ofLi+ions andCl−ions.Consider the electron affinity of Cl and the ionization energy of Li, then by definition, we haveEnergy required to removeanelectron fromCl−ion=348.7 kJ mol−1=3.614eVper ion.Energy released when an electron is put into theLi+ion=520.2 kJmol−1= 5.392eVper ion.Bond Energy per pair of Cs-Cl atoms =8.317eV + 3.614eV–5.392eV =6.539eVor 6.54 eVAtomic cohesive energyin kJ/mol is,Eatomic-cohesive= (6.539eV)(1.60210−19J eV−1)(6.0221023mol−1)=631kJmol─1of Cs or Cl atom (i.e.per mole of Cs-Cl atom pairs)=315kJmol─1of atoms(d)We know the interionic separation, which isro. The lattice parameteraisa= 2ro=2(2.6110−10m) =5.2210−10m

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Solutions toPrinciples of Electronic Materials and Devices:4thEdition(25April 2017)Chapter 1We can find the atomic masses of Li and Cl from Appendix C or from any periodic table.The density isgiven as()3ClLi44unitcellofVolumetypethisofmassAtomiccellunitingiven typeofatomsofNumberaMM+==3101231312313)10224.5()mol10022.6()molkg1045.35(4)mol10022.6()molkg1094.6(4−−−−−−−+== 1,980 kg m−3or1.98 g cm−3.Figure 1Q07-1Left:A schematic illustration of a cross section from solid NaCl. Solid NaCl is made of Cl−andNa+ions arranged alternatingly, so the oppositely charged ions are closest to each other and attract each other.There are also repulsive forces between the like-ions. In equilibrium, the net force acting on any ion is zero.Theinterionic separationroand the lattice parameteraare shown; clearlya= 2ro.(b) Solid NaCland the definition ofthe unit cell with a lattice parametera.There are 4 Na+and 4 Cl−ions in the unit cell.Clearly,a= 2ro.------------------------Comments:The values form, andBare from Richard Christman,Fundamental of Solid State Physics, WileyandSons (New York), 1987, Ch 5, Table 5-1, p130. Cl electron affinity fromhttps://en.wikipedia.org/wiki/Electron_affinity_(data_page)and Li ionization energy fromhttps://en.wikipedia.org/wiki/Molar_ionization_energies_of_the_elements(22 October 2016)We can compare the results from this calculation with some publishedexperimental and calculated values as intable 1Q07-1.The calculated values above are not too far out from experimental values.UsuallymandBaredetermined by using experimental values for theelastic coefficientsas will be apparent inQuestion 1.9whereelastic coefficients arerelatedtomandB)Lattice energyof an ionic crystal is the energy of formation of the crystal from its ions.Ionic cohesivekJ mol─1Atomic cohesiveper Li-Cl pairkJ mol─1Atomic cohesiveenergy per atomkJ mol─1ro(nm)a(nm)(g cm−)1Q07(This question)8026313150.2610.5221.98

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