Solutions to Principles of Electronic Materials Part 1

Chapter 1 Answers to "Why?" in the text Oxygen has an atomic mass of 16 whereas it is 14 for nitrogen. The O2 molecule is therefore heavier than the N2 molecule. Thus, from 12 mv2 =3( 12 kT) , the rms velocity of O2 molecules is smaller than that of N2 molecules. For small extensions, the difference between the engineering and instantaneous strains due to a temperature change are the same. Historically, mechanical and civil engineers measured extension by monitoring the change in length, ΔL; and the instantaneous length L was not measured. It is not trivial to measure both the instantaneous length and the extension simultaneously. However, since we know Lo and measure ΔL, the instantons length L = Lo + ΔL. Is the difference important? Consider a sample of length Lo that extends to a final length L due to a temperature change from To to T. Letε = (L − Lo) / Lo = ΔL/Lo be the engineering strain. The engineering definition of strain and hence the thermal expansion coefficient is Engineering strain = δL Lo = λδ T so that thermal expansion from To to T gives, L dL L Lo o T = λ dT ∴ To ΔL = λ( T − To ) Lo ε =λ( T −To ) (1) ln(1 +ε ) = λ (T −To ) (2) ∴ where ε = ΔL/Lo is the engineering strain as defined above Physics definition of strain and hence the thermal expansion coefficient is Instantaneous train = δL L = λδ T so that thermal expansion from To to T gives, L dL Lo L T To = λ dT ∴ L ln   Lo   = λ( T −To )  ∴ We can expand the ln(1 + ε ) term for small ε , so that Equation (2) essentially becomes Equation (1) 1.1 Virial theorem The Li atom has a nucleus with a +3e positive charge, which is surrounded by a full 1s shell with two electrons, and a single valence electron in the outer 2s subshell. The atomic radius of the Li atom is about 0.17 nm. Using the Virial theorem, and assuming that the valence electron sees the nuclear +3e shielded by the two 1s electrons, that is, a net charge of +e, estimate the ionization energy of Li (the energy required to free the 2s electron). Compare this value with the experimental value of 5.39 eV. Suppose that the actual nuclear charge seen by the valence electron is not +e but a little higher, say +1.25e, due to the imperfect shielding provided by the closed 1s shell. What would be the new ionization energy? What is your conclusion? Solution First we consider the case when the outermost valence electron can see a net charge of +e. From Coulomb’s law we have the potential energy PE = Q1Q2 (+ e)(− e) = 4πε0 r0 4 πε0 r0 −19 =− 2 (1.6 ×10 C) = −1.354 × 10−18 J or −8.46 eV 4π (8.85 ×10 −12 Fm −1 )(0.17 ×10 −9 m) Virial theorem relates the overall energy, the average kinetic energy KE , and average potential energy PE through the relations 1 KE = − PE 2 Thus using Virial theorem, the total energy is E = PE + KE E= and 1 PE = 0.5 × − 8.46eV = − 4.23 eV 2 The ionization energy is therefore 4.23 eV. Consider now the second case where the electron sees +1.25e due to imperfect shielding. Again the Coulombic PE between +e and +1.25e will be PE = Q1Q2 ( +1.25e)(−e) = 4πε 0 r0 4πε 0 r0 =− 1.25⋅ (1.6× 10− 19 C) 2 = −1.692 × 10−18 J or −10.58 eV − − − 4π (85 × 10 12 Fm 1 )(0.17 × 10 9 m) The total energy is, E= 1 PE = − 5.29 eV 2 The ionization energy, considering imperfect shielding, is 5.29 eV. This value is in closer agreement with the experimental value. Hence the second assumption seems to be more realistic. 1.2 Virial theorem and the He atom In Example 1.1 we calculated the radius of the H-atom using the Virial theorem. First consider the He+ atom, which as shown in Figure 1.75a, has one electron in the Ksell orbiting the nucleus. Take the PE and the KE as zero when the electrons and the nucleus are infinitely separated. The nucleus has a charge of +2e and there is one electron orbiting the nucleus at a radius r2. Using the Virial theorem show that the energy of the He+ ion is 2e 2 Energy of He + ion [1.48] 4πε o r2 Now consider the He-atom shown in Figure 1.75b. There are two electrons. Each electron interacts with the nucleus (at a distance r1) and the other electron (at a distance 2r1). Using the Virial theorem show that the energy of the He atom is  7 e2  Energy of He atom [1.49] E ( He ) = − (1 / 2)  8πε o r1  The first ionization energy EI1 is defined as the energy required to remove one electron from the He atom. The second ionization energy E I2 is the energy required to remove the second (last) electron from He+ . Both are shown in Figure 1.75 These have been measured and given as EI1 = 2372 kJ mole −1 and EI2 = 5250 kJ mol−1. Find the radii r 1 and r2 for He and He + . Note that the first ionization energy provides sufficient energy to take He to He + , that is, He → He+ + e− absorbs 2372 kJ mol−1 . How does your r 1 value compare with the often quoted He radius of 31 pm? + E (He ) = − (1 / 2) Figure 1.75: (a) A classical view of a He + ion. There is one electron in the K-shell orbiting the nucleus that has a charge +2e. (b) The He atom. There are two electrons in the K-shell. Due to their mutual repulsion, they orbit to void each other. Solution Virial theorem relates the overall energy, the average kinetic energy KE , and average potential energy PE through the relations 1 1 1 E = PE +KE ; KE = − PE ; E = PE ; KE = − E (1) 2 2 2 Now, consider the PE of the electron in Figure 1.75a. The electron interacts with +2e of positive charge, so that (−e )(2e ) 2e 2 PE = =− 4πε o r2 4πε o r2 which means that the total energy (average) is 2 2 e 1 2e E (He+ ) = PE = −(1 / 2) =− (2) 2 4πε o r2 4πε o r2 which is the desired result. Now consider Figure 1.75b. Assume that, at all times, the electrons avoid each other by staying in opposite parts of the orbit they share. They are "diagonally" opposite to each other. The PE of this system of 2 electrons one nucleus with +2e is ∴ PE = PE of electron 1 (left) interacting with the nucleus (+2e), at a distance r1 + PE of electron 2 (right) interacting with the nucleus (+2e), at a distance r1 + PE of electron 1 (left) interacting with electron 2 (right) separated by 2r1 (−e )(2e ) (− e)(2e) (−e )(−e ) PE = + + 4πε o r1 4πε o r1 4 πε o(2 r1) 7e 2 8πε o r2 From the Virial theorem in Equation (1)  7 e2  (3) E (He ) = − (1 / 2)  8πε o r1  We are given, EI1 = Energy required to remove one electron from the He atom = 2372 kJ mole−1 = 25.58 eV EI2 = Energy required to remove the second (last) electron from He+ = 5250 kJ mol−1 = 54.41 eV The eV values were obtained by using 1  E(eV ) = E( J/mole )   eNA  We can now calculate the radii as follows. Starting with Equation 2 for the ionization of He+, e2 (1.602 ×10 −19 C) 2 − + EI 2 = (54.41eV)(1.602 ×10 19 J/eV) = EI 2 = E (He ) = = − 12 −1 4πε o r 2 4π (8.854 ×10 F m ) r2 from which, r2 = 2.65×10−11 or 26.5 pm The calculation of r1 involves realizing that Equation (3) is the energy of the whole He atom, with 2 electrons. If we remove 1 electron we are left with He+ whose energy is Equation (2). Thus the ∴ PE = − EI1 = E(He) − E(He + ) −19  7 e2  + J/eV) = E( He ) − E( He ) = (1 / 2)  − E I2 8πε o r1  ∴ (24 .58 eV)(1.602 ×10 ∴ (54 .41 eV + 24 .58 eV)(1.602 × 10 − 19 −19 2 7 e2  (1.602 ×10 C) J/eV ) = (1 / 2)  = −12 −1 8πε o r1  4π (8.854 ×10 F m ) r1 from r1 = 3.19×10−11 or 31.9 pm very close to the quoted value of 31 pm in various handbooks or internet period tables ____________________________________________________________________________________ 1.3 Atomic mass and molar fractions a. Consider a multicomponent alloy containing N elements. If w1, w2, ..., wN are the weight fractions of components 1,2,..., N in the alloy and M1, M2, ..., MN, are the respective atomic masses of the elements, show that the atomic fraction of the i-th component is given by wi / M i Weight to atomic percentage ni = w1 w2 w + + ... + N M1 M2 MN b. Suppose that a substance (compound or an alloy) is composed of N elements, A, B, C,... and that we know their atomic (or molar) fractions nA, nB nC, .... Show that the weight fractions wA, wB, wC,....are given by n AM A wA = nAM A + nB MB + nC MC + ... n BM B Atomic to weight percentage nAM A + nB MB + nC M C + ... Consider the semiconducting II-VI compound cadmium selenide, CdSe. Given the atomic masses of Cd and Se, find the weight fractions of Cd and Se in the compound and grams of Cd and Se needed to make 100 grams of CdSe. A Se-Te-P glass alloy has the composition 77 wt.% Se, 20 wt.% Te and 3 wt.% P. Given their atomic masses, what are the atomic fractions of these constituents? wB = c. d. Solution a. Suppose that n1, n2, n3,…, ni,…, nN are the atomic fractions of the elements in the alloy, n1 + n2 + n3 +… + nN = 1 Suppose that we have 1 mole of the alloy. Then it has ni moles of an atom with atomic mass Mi (atomic fractions also represent molar fractions in the alloy). Suppose that we have 1 gram of the alloy. Since wi is the weight fraction of the i-th atom, wi is also the mass of i-th element in grams in the alloy. The number of moles in the alloy is then wi/Mi. Thus, Number of moles of element i = wi/Mi Number of moles in the whole alloy = w1/M1 + w2/M2 +…+ wi/Mi +…+wN/MN Molar fraction or the atomic fraction of the i-th elements is therefore, ni = ∴ ni = Numebr of moles of elementi Total numbers of moles in alloy wi / M i w1 w2 w + + ... + N M1 M2 MN b. Suppose that we have the atomic fraction ni of an element with atomic mass Mi. The mass of the element in the alloy will be the product of the atomic mass with the atomic fraction, i.e. niMi. Mass of the alloy is therefore nAMA + nBMB + … + nNMN = Malloy By definition, the weight fraction is, wi = mass of the element i/Mass of alloy. Therefore, n AM A wA = nA M A + nB MB + nC MC + ... wB = n BM B nA M A + nB MB + nC MC + ... c. The atomic mass of Cd and Se are 112.41 g mol−1 and 78.96 g mol−1. Since one atom of each element is in the compound CdSe, the atomic fraction, nCd and nSe are 0.5. The weight fraction of Cd in CdSe is therefore nCd MCd 0.5 ×112.41g mol −1 wCd = = = 0.587 or 58.7% nCd MCd + nSe MSe 0.5 × 112.41g mol −1 + 0.5 × 78.96 g mol −1 Similarly weight fraction of Se is −1 nSe M Se 0.5 × 78.96 g mol wSe = = = 0.4126 or 41.3% −1 nCd MCd + nSe M Se 0.5 × 112.41g mol + 0.5 × 78.96g mol −1 Consider 100 g of CdSe. Then the mass of Cd we need is Mass of Cd = wCdMcompound = 0.587 × 100 g = 58.7 g (Cd) and Mass of Se = wSeMcompound = 0.413 × 100 g = 41.3 g (Se) d. The atomic fractions of the constituents can be calculated using the relations proved above. The atomic masses of the components are MSe = 78.6 g mol−1, MTe = 127.6 g mol−1, and MP = 30.974 g mol−1. Applying the weight to atomic fraction conversion equation derived in part (a) we find, 0.77 −1 wSe / M Se 78.6 g mol = n Se = w Se w w 0.77 0.2 0.03 + Te + P − + − + − M Se M Te M P 78.6 g mol 1 127.6 g mol 1 30.974 g mol 1 ∴ nSe = 0.794 or 79.4% n Te ∴ 0.2 wTe / M Te 127.6 g mol −1 = = wSe w w 0.77 0.2 0.03 + + + Te + P −1 −1 −1 M Se M Te M P 78.6 g mol 127.6 g mol 30.974 g mol nTe = 0.127 or 12.7 % nP = ∴ wSe M Se 0.03 − wP / M P 30.974 g mol 1 = w w 0.77 0.2 0.03 + + + Te + P −1 −1 −1 M Te M P 78.6 g mol 127.6 g mol 30 .974 g mol nP = 0.0785 or 7.9% 1.4 Mean atomic separation, surface concentration and density There are many instances where we only wish to use reasonable estimates for the mean separation between the host atoms in a crystal and the mean separation between impurities in the crystal. These can be related in a simple way to the atomic concentration of the host atoms and atomic concentration of the impurity atoms respectively. The final result does not depend on the sample geometry or volume. Sometimes we need to know the number of atoms per unit area ns on 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 side L as shown in Figure 1.76. To each atom, we can attribute a portion of the whole volume, which is a cube of side a. Thus, each atom is considered to occupy a volume of a3. Suppose that there are N atoms in the volume L3. Thus, L3 = Na3. a. If nb is the bulk concentration of atoms, show that the mean separation a between the atoms is given by a = 1 / nb3 . b. Show that the surface concentration ns of atoms is given by ns = nb2 / 3 . c. Show that the density of the solid is given by ρ = nb Mat / NA where Mat is 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 1016 cm−3. P atoms substitute for Si atoms and are randomly distributed in the crystal. What is the mean separation between the P atoms? Figure 1.76 Consider a crystal that has volume L3. This volume is proportioned to each atom, which is a cube of side a3. Solution a. Consider a crystal of the material which is a cube of volume L3, so that each side has a length L as shown in Figure 1.76. To each atom, we can attribute a portion of the whole volume. For simplicity, we take the volume proportioned to an atom to be a3, that is, each atom is considered to occupy a volume of a3 . The actual or true volume of the atom does not matter. All we need to know is how much volume an atom has around it given all the atoms are identical and that adding all the atomic volumes must give the whole volume of the crystal. Suppose that there are N atoms in this crystal. Then nb = N/L3 is the atomic concentration in the crystal, the number of atoms per unit volume, the so-called bulk concentration. Since N atoms make up the crystal, we have Na 3 = Crystal volume = L3 The separationbetween any two atoms is a, as shown in Figure 1.76. Thus, 1/ 3 1/ 3 L3  1  1 Mean separation = a =   =   = 1/ 3 nb N  nb  (1) Equation 1 can be derived even more simply because an atom has a volume of a3. In this volume are is only 1 atom. So nba3 must be 1, which leads to Equation 1. The above idea can be extended to finding the separation between impurity atoms in a crystal. Suppose that we wish to determine the separation d between impurities, then we can again follow a similar procedure. In this case, the atoms in Figure 1.76 are impurities which are separated by d. We again assign a portion of the whole volume (for simplicity, a cubic volume) to each impurity. Each impurity atom therefore has a volume d3 (because the separation between impurities is d). Following the above arguments we would find, Mean separation between impurities = d = 1 1/ 3 NI (2) b. We wish to find the number of atoms per unit area ns on the surface of a solid given the atomic concentration nb in the bulk. Consider Figure 1.76. Each atom as an area a2, so that within this surface area there is 1 atom. Thus Surface area of 1 atom × Surface concentration of atoms = 1 or a2 ns = 1 ∴ ns = 1 = n2b/ 3 a2 (3) Equation (3) is of course based on the simple arrangement of atoms as shown in Figure 1.75. In reality, 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 for ns given nb. c. We can determine the density ρ from the atomic concentration nb and vice versa. The volume in Figure 1.75 has L3nb atoms. Thus, the density is Mass = ρ= Volume M at ) nM NA = b at L3 NA L3 nb ( (4) For Si, the atomic mas Mat = 28.09 g mol−1, so that with ρ given as 2.33 g cm−3, nb = ρNA M at = (2.33 g cm− 3 )(6.022× 10 23 mol− 1 ) = 5.00×1022 cm−3. −1 (28.09 g mol ) d. The P concentration in the crystal is 1016 cm−3. P atoms substitute for Si atoms and are randomly distributed in the crystal. We can use Equation (2) Mean separation between impurities = d = = 4.64×10−8 m = 46.4 nm 1 1 = 1/ 3 (1× 1016 × 10 6 m − 3 )1/ 3 NI 1.5 The covalent bond Consider the H2 molecule in a simple way as two touching H atoms as depicted in Figure 1.77. Does this arrangement have a lower energy than two separated H atoms? Suppose that electrons totally correlate their motions so that they move to avoid each other as in the snapshot in Figure 1.77. The radius ro of the hydrogen atom is 0.0529 nm. The electrostatic potential energy PE of two charges Q1 and Q2 separated by a distance r is given by Q1Q2/(4πεor). Using the Virial Theorem as in Example 1.1, consider the following: a. Calculate the total electrostatic potential energy (PE) of all the charges when they are arranged as shown in Figure 1.77. In evaluating the PE of the whole collection of charges you must consider all pairs of charges and, at the same time, avoid double counting of interactions between the same pair of charges. The total PE is the sum of the following: electron 1 interacting with the proton at a distance r o on the left, proton at ro on the right, and electron 2 at a distance 2ro + electron 2 interacting with a proton at ro and another proton at 3ro + two protons, separated by 2ro, interacting with each other. Is this configuration energetically favorable? b. Given that in the isolated H-atom the PE is 2 ×(−13.6 eV), calculate the change in PE in going from two isolated H-atoms to the H2 molecule. Using the Virial theorem, find the change in the total energy and hence the covalent bond energy. How does this compare with the experimental value of 4.51 eV? Figure 1.77 A simplified view of the covalent bond in H2. A snapshot at one instant. Solution a. Consider the PE of the whole arrangement of charges shown in the figure. In evaluating the PE of all the charges, we must avoid double counting of interactions between the same pair of charges. The total PE is the sum of the following: Electron 1 interacting with the proton at a distance ro on the left, with the proton at ro on the right and with electron 2 at a distance 2ro + Electron 2 on the far left interacting with a proton at ro and another proton at 3ro + Two protons, separated by 2ro, interacting with each other PE = − e2 e2 e2 − + 4π ε o ro 4π ε oro 4π ε o (2ro ) 2 2 − e e − 4π ε o ro 4π ε o 3ro + e 2 4π ε o 2ro Substituting and calculating, we find PE = 1.0176 × 10−17 J or -63.52 eV The negative PE for this particular arrangement indicates that this arrangement of charges is indeed energetically favorable compared with all the charges infinitely separated (PE is then zero). b. The potential energy of an isolated H-atom is -2× 13.6 eV or -27.2 eV. The difference between the PE of the H2 molecule and two isolated H-atoms is, ΔPE = −(63.52) eV − 2(-27.2) eV = −9.12eV We can write the last expression above as the change in the total energy. 1 1 ΔE = Δ PE = ( −9.12eV) = −4.56eV 2 2 This change in the total energy is negative. The H2 molecule has lower energy than two H-atoms by 4.56 eV which is the bonding energy. This is very close to the experimental value of 4.51 eV. (Note: We used a ro value from quantum mechanics - so the calculation was not totally classical!). 1.6 Ionic bonding and CsCl The potential energy E per Cs+-Cl− pair within the CsCl crystal depends on the interionic separation r in the same fashion as in the NaCl crystal, e 2M B + Energy per ion pair in ionic crystals [1.48] 4πε o r r m 9 −104 where for CsCl, M = 1.763, B = 1.192×10 J m or 7.442×10−5 eV (nm) 9 and m = 9. Find the equilibrium separation (ro) of the ions in the crystal and the ionic bonding energy, that is, the ionic cohesive energy; and compare the latter value to the experimental value of 657 kJ mol−1. Given the ionization energy of Cs is 3.89 eV and the electron affinity of Cl (energy released when an electron is added) is 3.61 eV, calculate the atomic cohesive energy of the CsCl crystal as joules per mole. E (r ) = − Solution Bonding will occur when potential energy E(r) is minimum at r = r0 corresponding to the equilibrium separation between Cs+ and Cl− ions. Thus, differentiating E(r) and setting it equal to zero at r = ro we have dE ( r )  d  e2 M B = − +  =0   dr  r =r o dr  4πε o r r m r = r o ∴  e2 M B  −m m+ 1  =0  2 πε r 4 r  o r =r o ∴ e2 M B − m m+1 =0 ro 4πε o ro2 ∴ 4πε mB m−1 ro =  2 o  e M  1 (1) Thus substituting the appropriate values we have 1 4π (8.8542 ×10 −12 Fm −1) × 9 × (1.192 ×10 −104 J m 9 ) 8  ro =  2 19   1.763 × 1.6 ×10 − C   ( ∴ ro = 3.57 × 10−10 ) m or 0.357 nm. The minimum energy is the energy at r = ro, that is e 2M B E min = − + 4πε or o rom which in terms of eV is E min (eV) = − =− 9 eM B(eV nm ) + 9 4πε o ro ro (nm) (2) (1.6 ×10−19 C)(1.763) 7.442 ×10 −4 eV nm 9 + −12 −1 −10 4π (8.8542 ×10 Fm )(3.57 ×10 m) (0.357 nm) 9 = − 6.32 eV per ion pair, or 3.16 eV per ion. The amount of energy required to break up a CsCl crystal into Cs+ and Cl− ions = 6.32 eV per pair of ions. The corresponding ionic cohesive energy is Ecohesive = (6.32 eV)(1.602 × 10−19 J eV−1)(6.022× 1023 mol−1) = 610 kJ mol─1 of Cs+Cl− ion pairs or 610 kJ mol─1 of Cs+ ions and Cl− ions. (Not far out from the experimental value given the large numbers and the high index, m = 9, involved in the calculations.) The amount of energy required to remove an electron from Cl− ion = 3.61 eV. The amount of energy released when an electron is put into the Cs+ ion = 3.89 eV. Bond Energy per pair of Cs-Cl atoms = 6.32 eV + 3.61 eV – 3.89 eV = 6.04 eV Atomic cohesive energy in kJ/mol is, E cohesive = (6.04 eV)(1.6 × 10−19 J eV−1)(6.022× 1023 mol−1) = 582 kJ mol─1 of Cs or Cl atom (i.e. per mole of Cs-Cl atom pairs) = 291 kJ mol─1 of atoms -----------------------Notes: 1. Various books and articles report different values for B and m for CsCl, which obviously affect the calculated energy; ro is less affected because it requires the (m−1)th root of mB. Richard Christman (Introduction to Solid State Physics, John Wiley and Sons, 1988) in Table 5-1 gives, m = 10.65 and B = 3.44 × 10120, quite different than values here, which are closer to values in Alan Walton, Three Phases of Matter (2nd Edition), Clarendon (Oxford University) Press, 1983 (pp. 258-259). 2. The experimental value of 657 kJ mol−1 (6.81 eV per ion pair) for the ionic cohesive energy (the ionic lattice energy) is from T. Moeller et al, Chemistry with Inorganic Qualitative Analysis, Second Edition, 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 to molecular units e.g. Cs+Cl−, which would correspond to the ionic cohesive energy here. Further, most chemists use "energy per mole" to imply energy per mole of chemical units, and hence the atomic cohesive energy per mole would usually refer to energy be per Cs and Cl atom pairs. Some authors refer to the atomic cohesive energy per mole as cohesive energy per mole of atoms, independent of chemical formula. 4. Equation (1) is  e2 M  m−1  B =  πε ro 4 om  which can be substituted into Equation (2) or [1.48] to eliminate B, that is, find the minimum energy of a pair of Cs+-Cl− ions, i.e. e2 M 1  e2 M  m−1  + m E min = −  ro 4πε o r r 4πε om  e2 M  −1  1  E min = − 4πε o or  m which is called the Born-Landé equation. ∴ 1.7 Ionic bonding and LiCl Equation 1.48 can be used to represent the PE of the ion pair inside the LiC crystal. LiCl has the NaCl structure with M = 1.748, m = 7.30, B = 2.34×10−89 J m7.30. Further, the ionization energy of Li (Li → Li+ + e−) is 520.2 kJ mol−1. The electron affinity of Cl (energy associated with 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 of the LiCl crystal. (c) Calculate the density of LiCl. Solution Figure 1Q07-1 shows the crystal structure of NaCl. LiCl has the same crystal structure. (a) The minimum in energy is at  2  dE (r )  = d  − e M + Bm  = 0    dr r =r o dr  4πε o r r r = ro ∴  e2M  −m B  =0  2 r m+1 r = ro 4πε o r ∴ e 2M B − m m+1 =0 ro 4πε o ro2 ∴ 4πε mB m−1 ro =  2 o  e M  1 Thus, substituting the appropriate values we have 4π (8.8542 ×10 Fm ) ×7.30 ×(2.34 ×10 ro =  1.748 × (1.602 ×10 −19 C) 2  −12 ∴ −1 − 89 Jm 7 .30 1 ) 8   ro = 2.612× 10−10 m or 0.261 nm. (b) The minimum energy is the energy at r = ro, that is E min = − e 2M B + 4πε or o rom −19 =− 2 −89 7.8 (1.602 × 10 C) (1.748) 2.34 ×10 J m + − − − − 4π (8.8542 ×10 12 Fm 1 )(2.61×10 10 m) (2.61× 10 10 m)7.8 = −1.33× 10−18 J per of Li+-Cl− ions = −8.32 eV per ion pair, or −4.18 eV per ion. The bond energy is therefore Ebond = − Emin = 8.317 eV per ion pair, or 4.18 eV per ion. (c) In (b) we calculated the amount of energy required to break up Li+-Cl− pair into Li+ and Cl− ions = 8.317 eV per pair of ions. The corresponding ionic cohesive energy is Eionic-cohesive = (8.317 eV)(1.602 × 10−19 J eV−1)(6.022× 10−23 mol−1) = 802 kJ mol─1 of Li+Cl− ion pairs or 802 kJ mol─1 of Li+ ions and Cl− ions. Consider the electron affinity of Cl and the ionization energy of Li, then by definition, we have Energy required to remove an electron from Cl− ion = 348.7 kJ mol−1 = 3.614 eV per ion. Energy released when an electron is put into the Li+ ion = 520.2 kJ mol−1 = 5.392 eV per ion. Bond Energy per pair of Cs-Cl atoms = 8.317 eV + 3.614 eV – 5.392 eV = 6.539 eV or 6.54 eV Atomic cohesive energy in kJ/mol is, E atomic-cohesive = (6.539 eV)(1.602 × 10−19 J eV−1)(6.022×1023 mol−1) = 631 kJ mol─1 of Cs or Cl atom (i.e. per mole of Cs-Cl atom pairs) = 315 kJ mol─1 of atoms (d) We know the interionic separation, which is ro. The lattice parameter a is a = 2ro = 2(2.61×10−10 m) = 5.22×10−10 m We can find the atomic masses of Li and Cl from Appendix C or from any periodic table. The density is given as (Number of atoms of given type in unit cell × Atomic mass of this type ) ρ = Volume of unitcell 4 M Li + 4M Cl = a3 −3 −1 −3 −1 (6.94× 10 kg mol ) (35.45 ×10 kg mol ) +4 4 23 23 −1 −1 (6.022 ×10 mol ) (6.022 ×10 mol ) ∴ ρ = = 1,980 kg m−3 or 1.98 g cm−3 . −10 3 (5.224 ×10 ) Figure 1Q07-1 Left: A schematic illustration of a cross section from solid NaCl. Solid NaCl is made of Cl − and Na+ 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. The interionic separation r o and the lattice parameter a are shown; clearly a = 2r o . (b) Solid NaCl and the definition of the unit cell with a lattice parameter a. There are 4 Na + and 4 Cl − ions in the unit cell. Clearly, a = 2r o . -----------------------We can compare the results from this calculation with some published experimental and calculated values as in table 1Q07-1. The calculated values above are not too far out from experimental values. Usually m and B are determined by using experimental values for the elastic coefficients as will be apparent in Question 1.9 where elastic coefficients are related to m and B) Lattice energy of an ionic crystal is the energy of formation of the crystal from its ions. Ionic cohesive kJ mol ─1 1Q07 (This question) 802 Atomic cohesive per Li-Cl pair Atomic cohesive energy per atom kJ mol ─1 kJ mol 631 315 ro a ρ (nm) (nm) (g cm−3) 0.261 0.522 ─1 1.98 Wikipedia Kittel, 7Ed, p73, Table 7 0.514 832.6 Moeller 834 Ashcroft and Mermin, Table 20.5, p406 831.1 2.068 0.257 0.257 C. Kittel, Silid State Physics, 7th Edition, John Wiley and Sons, New York, 1996; p73, Table 7. T. Moeller et al, Chemistry with Inorganic Qualitative Analysis, Second Edition, Academic Press, 1984) p. 413, Table 13.5. N.W. Ashcroft and N.D. Mermin, Solid State Physics, Saunders College, 1976; Table 20.5 (p406) data, originally from M.P. Tossi, Solid State Physics, Vol. 16 (Edited by F. Seitz and D. Turnbull), Academic Press, New York, 1964, p54 1.8 Madelung constant If we were to examine the NaCl crystal in three dimensions, we would find that each Na+ ion has 6 Cl− ions as nearest neighbors at a distance r 12 Na+ ions as second nearest neighbors at a distance r 2 8 Cl− ions as third nearest neighbors at a distance r 3 and so on. Show that the electrostatic potential energy of the Na+ atom can be written as e 2  −12 + 8 −  e 2M Madelung constant M for NaCl E( r) = − ...= − 6 4πε or   4πε o r 2 3 where M, called Madelung constant, is given by the summation in the square brackets for this particular ionic crystal structure (NaCl). Calculate M for the first three terms and compare it with M = 1.7476, its value had we included the higher terms. What is your conclusion? Solution From Coulomb’s law of electrostatic attraction we know that the PE between two charges Q1 and Q2 separated by a distance r is given by QQ PE = 1 2 4 πε o r First we consider the interaction between Na+ ion and 6Cl− ions at distance r. Applying Coulomb’s law we have QQ ( −6 e)( +e) − 6e2 PE 1 = 1 2 = = πε 4 or 4πε o r 4πε o r Similarly, we now consider 12 Na+ ions as second nearest neighbors at a distance r 2 QQ ( +12 e)( +e) 12e 2 PE 2 = 1 2 = = 4 πε or 4πε o r 2 4πε o r 2 and 8 Cl─ ions as third nearest neighbors at a distance r 3 PE 3 = − 8e2 Q1Q 2 ( −8e )( +e ) = = 4πε o r 4πε or 3 4πε o r 3 and similarly we can consider the next nearest set of neighbors and so on. Therefore, the overall PE of the Na+ ion is 2 2 2 12e 8e − 6e E( r) = + − + .. . πε 4 or 4πε o r 2 4πε o r 3 or E( r) = − e 2  −12 + 8 −  − Me2 ...= 6 4πε or   4πε or 2 3 where clearly M = 6 − 12 2 + 8 3 + ⋅⋅ ⋅ Considering just the first three terms we have M = 2.133. This is considerably different from the value M = 1.7464, the value obtained when higher order terms are considered. This implies that the next nearest neighbors have substantial effect on the potential energy. Note: See Appendix A for the explanation on how the PE of ions in the crystal is usually written.