Idle Mind Solutions

LN–5 LN–5 IDLE MIND SOLUTIONS 1. (a) Schematic X–ray spectrum from a Mo target. Kα and Kβ constitute characteristic target radiation. Intensity MoKα MoKβ Kβ e– Bremsstrahlung Kα 2e– λSWL (b) λKβ λKα Mo λ Characteristic Radiation: After the vacancy is generated in the K shell, it may be filled by an electron transition from either the L shell (Mo Kα line) or from the M shell (Mo Kβ line). Since the Kβ transition is less likely to take place, its intensity on the spectrum is less than that of the Kα line. electron removed from K shell e– Ek 1 mv 21  2 e– Ek  2 mv 22 2 hνBremsstrahlung Mo hν = ∆EK Bremsstrahlung: Bremsstrahlung constitutes conversion of EKin into hν, brought about by the deceleration of an electron upon close approach to the (potential energy barrier of the) nucleus. For an X–ray generator with EKin (electron) = eV, the shortest λ observable corresponds to a deceleration to v=0, or to the total conversion of eV into hν, thus λSWL = hc/eV. (c) 2.
6 4 (m)  1.2 x 10 (Å)  0.3 x 10
10 m
SWL  hc  1.2 x 10 eV V V In a BCC structure the smallest θ observable corresponds to the family of {110} planes. Given a = 10 x 10–10 m, we find d(110) to be: d 110  a  7.07 x 10
10 m 2 continued. (v 2  v 1) LN–5 2. Continued. Looking at a diffraction experiment we realize that θ(110) will increase with increasing λ and can assume a maximum value of 90o for which sinθ = 1. For this maximum diffraction condition, the Bragg relation is reduced to: l  2d (110)  2 x 7.07 x 10 10 m  14.14 x 10 10 m This finding means that any λ > 14.14 x 10–10 m will not be able to lead to (110) diffraction. 3. To determine the lattice constant “a” for the metal requires identification of the Miller indices corresponding to the given diffraction angles – θ – which in turn first requires identification of the particular cubic system involved. To accomplish this identification, let us consider: (a) Selection rules: S.C. All planes diffract B.C.C. Planes with (h+k+l) = even number diffract F.C.C. Planes with only even or only odd (h,k,l) indices diffract (b) The modified Bragg equation: sin 2 q l2   const. (h 2
k 2
l 2) 4a 2 Taking (a) and (b), we find: SC sin 2 q 3 sin 2 q 2 sin 2 q 7 sin 2 q 1     const. 1 3 8 2 BCC sin 2 q 3 sin 2 q 2 sin 2 q 7 sin 2 q 1     const. 4 6 14 2 FCC sin 2 q 3 sin 2 q 1 sin 2 q 2    const. 3 4 8 (continued) LN–5 3. Continued. (a) The problem can now be reduced to finding the constant ratio, which requires finding the sin2θ values. A quick look at the data suggests a BCC system since sin2θ7 is about 7 x sin2θ1. sin2θ (1) (2) (3) (4) (5) (6) (7) 0.118 0.238 0.355 0.474 0.590 0.710 0.828 We find: sin 2 q 7 + 0.059 14 and sin 2 q 1 + 0.059 2 A reminder: For the system to be SC (simple cubic): sin 2 q 7 sin 2 q 1 + 1 8 (0.118 0 0.103) For BCC the smallest θ value corresponds to the {110} family. Therefore: λ = 2d sin θ d (110) + *10 l + 1.541 x 10 + 2.24 x 10 *10 m 2 x 0.344 2 sin q 1 We also have: d (hkl) + a Ǹh 2 ) k 2 ) l 2 and a + d x Ǹ2 a = 3.17 x 10–10 m (b) For a BCC system we find that atoms are in contact along the body diagonal: a Ǹ3 + 4r and r + a Ǹ3 + 1.37 x 10 *10 m 4 LN–5 4. Cr: BCC; r = 1.30 x 10–10 m In BCC systems the relationship between r and a is given by: a 3  4r and
10 a  4r  4 x 1.30 x 10  3.00 x 10
10 m 3 3 The first diffraction peak for BCC corresponds to {110} planes; thus: λ = 2d sinθ sin q 1  l 2d (d 110  a ) 2 l 2  0.3634  l  2a 2a 2 θ1 = 21.3o Pd: FCC; r = 1.37 x 10–10 m a 2  4r ; a  2 2 r  3.87 x 10
10 m According to the selection rules, the first diffraction peak corresponds to the family of {111} planes: d (111)  a  2.23 x 10
10 m 3 sin q 1  l  0.345 2d θ1 = 20.2o 5. (a) While the diffraction angle for the (112) planes of Au can readily be calculated, this calculation serves no purpose since the selection rules prohibit diffraction at this set of planes! (FCC!) LN–5 5. Continued. (b) Diffraction on {220} planes is encountered. For FCC we have: a 2  4r and a  4r  2 2 r  4.13 x 10
10 m 2 and d (200)  a  1.46 x 10
10 m 8 λ = 2d sin θ sinθ = λ/(2d) = 0.342 θ = 20o 6. (a) The maximum energy (eV) of the electrons corresponds to the generation of λSWL as bremsstrahlung. Thus:
6
SWL  hc  1.24 x 10 (m)  1.82 x 10
10 m eV V Since λKα is smaller than λSWL, it is clear that the energy of the incident electrons is insufficient to remove an electron from the K shell of Cu and thus no Kα radiation can be generated. λSWL = 1.82 x 10–10 m (b) This question was answered in (a): (c) The answer to this question is no! This answer can be explained on the basis of an energy diagram. 0 n=∞ –E ↓ M L CuK β energy required to remove an electron from the K shell! It can be seen that this energy (required to generate Kβ radiation through the removal of an electron from the K shell) is significantly more than that associated with CuKβ radiation. K 7. The system can only be BCC since sin2θ7 is 7 x sin2θ1; in SC systems the value of sin2θ after 0.600 would be 0.800! LN–5 8. The shortest wavelength emanated by a tube is given by: λSWL = hc/eV (Bremsstrahlung) a = 3.26 x 10–10 m BCC structure; first diffraction occurs on {110} planes: d (110) + a + 3.26 + 2.31Å Ǹ2 Ǹh 2 ) k 2 ) l 2 λ = 2d sinθ sinθ is always ≤ 1, so λ ≤ 2d ≤ 4.62 x 10–10 m λSWL = 4.62 x 10–10 = hc/eV V + E + e @ V, so V + 9. 10. (a) SC: θ{100} (b) BCC: θ{110} (c) FCC: θ{111} 6.63 x 10 *34 x 3 x 10 8 1.6 x 10 *19 x 4.62 x 10 *10 4.25 x 10 *16 J + 2690 Volt 1.6 x 10 *19 eVńJ We determine the crystal system and from the constant ratio [(sin2θ)/(h2+k2+l2) = λ2/4a2] we obtain “a” since λ is given. We look at θ1, θ2 (better sin2θ) and, if necessary, at θ7. (1) (2) (7) θ 13.70 15.89 36.62 sinθ 0.2368 0.2738 sin2θ 0.56092 0.74962 LN–5 10. Continued. The sin2θ values do not suggest BCC or SC structure, so that leaves FCC. We need only the first two ratios, but, for the exercise, let us identify all 9 diffraction lines! θB sinθB (1) 13.70 0.2368 0.05609/3 0.0187 (111) (2) 15.89 0.2738 0.07496/4 0.0187 (200) (3) 22.75 0.3867 0.1496/8 0.0187 (220) (4) 26.91 0.4526 0.2048/11 0.0186 (311) (5) 28.25 0.4733 0.2240/12 0.0187 (222) (6) 33.15 0.5468 0.2990/16 0.0187 (400) (7) 36.62 0.5965 0.3558/19 0.0187 (331) (8) 37.60 0.6101 0.3723/20 0.0186 (420) (9) 41.95 0.6685 0.4469/24 0.0186 (422)
2  0.0187 4a 2 sin2θB/(h2+k2+l2) a  4 (1.54)  5.63Å  x 0.0187 A brief check: λ = 2d(111) sin 13.70
 2 a sin 13.70 3 a  ratio 1.54 x 3  5.63Å 2 x 0.2368 2 5.63 x 10
10 m LN–5 11. First we determine λKα(Cu). We also know that Ag crystallizes in an FCC lattice; thus the three smallest θB values are generated by the (111), (200), and (220) planes. Using the Bragg relation, λ = 2d(h,k,l) sinθB, we recognize that we have two unknowns: d(“a”) and θB (to be determined). λ = 2d(111) sinθ1 d (111) + a Ǹ3 a + Ǹ4 x6.0210.3x x1010 3 *6 23 + 4.09Å d(111) = 2.36 x 10–10 m 10.3 x 10–6 = (NA/4)a3 12. q 1 + sin *1 1.54 x 10 *10 + 19.04 o 2 x 2.36 x 10 *10 q 2 + sin *1 1.54 x 10 *10 + 22.1 o 2 x 2.045 x 10 *10 [d (200) + ań Ǹ4 + 2.045Å] q 3 + sin *1 1.54 x 10 *10 + 32.2 o 2 x 1.446 x 10 *10 [d (220) + ań Ǹ8 + 1.446Å] In any X–ray tube, the shortest wavelength of X–rays generated (regardless of the target material) will be: λSWL = hc/eV. Moreover, the “longest” wavelength still generating diffraction will do so with a θB → 90o. Therefore, the limiting wavelength criterion is given by: λlongest = 2d sin 90 = 2d (continued) LN–5 12. Continued. Potassium (K) has BCC structure; thus the smallest θ is associated with {110} diffraction (= the largest interplanar spacing):
SWL + 2d (110) + 2 a + hc Ǹ2 eV 13. 45.46 x 10 Ǹ2 x6.02 x 10 *6 V + hc e x 2a Ǹ2 V + 6.63 x 10 *34 x 3 x 10 8 + 1.65 x 10 3 Volt *10 1.6 x 10 *19 2 x 5.32 x 10 Ǹ2 a + 3 23 + 5.32Å We first determine the lattice constant (a) for Au, and then the wavelength (λp) available for diffraction which, in conjunction with the Bragg relationship, will yield the diffraction angle for (220) planes: Au: FCC; mol. vol. = 10.2 cm3 a + Ǹ4 x6.0210.2x x1010 3 *6 23 + 4.08 x 10 *10 m Electrons accelerated by 45 kV will have a kinetic energy of: EKin = 4.5 x 104 x 1.6 x 10–19 J = (mv2)/2 For convenience, let us determine v, the velocity of the electrons, and insert the value into the deBroglie equation: v + Ǹ9 x 109.1 xx 1.610 x 10 4 *31 *19 + 1.26 x 10 8 mńs (we ignore relativistic effects) 6.63 x 10 *34 h +
p + mv 9.1 x 10 *31 x 1.26 x 10 8 λp = 5.8 x 10–12 m (a short λ, θB will certainly be small) LN–5 13. Continued. l p + 2d (hkl) sin q + 2 a sin q + a sin q Ǹ2 Ǹ8 lp 5.8 x 10 *12 x Ǹ2 q B + sin *1 a + sin *1 4.08 x 10 *10 Ǹ2 θB = 1.15o 14. From the selection rules we recognize that the first diffraction peak in BCC systems is due to (110) planes. For vanadium, νKα = 3/4 R(23–1)2 = 3.98 x 109 m–1 and λKα = 2.51 x 10–10 m λ = 2d sinθ *10 d (110) + 2.51 x 10 + 1.885 x 10 *10 m 2 sin 41.75 d hkl + a Ǹh 2 ) k 2 ) l 2 a + d (110) x Ǹ2 a = 2.66 x 10–10 m 15. First order diffraction refers to “n” in the Bragg equation being one. ν Ag Kα = 3/4 R(46)2 = 1.74 x 1010 m–1 λ Ag Kα = 0.574 x 10–10 m Fe = BCC structure λ = 2d(110) sin θ sin q + l 2d (110) q + sin *1 l + 8.15 o 2d (110) LN–5 16. Ta: BCC structure; the smallest θ is due to (110) reflections. λ = 2d(110) sinθ To determine λ, we must know d(110) – that means “a”, the lattice constant. For Ta: 1.09 x 10–5 (m3/mole) = (NA/2)a3 a  2 x6.021.09x x1010 3
5 23  3.31 x 10
10 m d (110)  a  2.34 x 10
10 m 2 λ = 2 x 2.34 x 10–10 x sin 29.35 λ = 2.29 x 10–10 m ; ν = 4.36 x 109 m–1 Taking the Moseley relationship, we have: 4.36 x 109 = 3/4 R (Z–1)2 (Z
1)  x 10 4 x 4.36 3R 9  23 Z = 24 The unknown target material is chromium (Cr). 17. (a) First, realize that the two lines nearest the undiffracted beam will be due to the same diffraction cone (use the 3.091 diffraction software if you need to convince yourself of this). This cone will be due to {111} planes, which have the widest spacing in the FCC structure, and thus have the smallest diffraction angle, 2θ. (continued) LN–5 17. Continued. (b) The radius of the camera is 5 cm, so a separation of 7.8 cm between lines is an angle of 7.8/5 radians = 1.56 radians = 89°. This separation represents twice the diffraction angle, so 2θ = 44.5°, and θ111 = 22.25°. lCuK a 1.54Å d 111 + + + 2.034Å 2 sin q 111 2 sin 22.25 a + d 111 Ǹ3 + 3.52Å For the FCC structure: 18. a Ǹ2 + 4r , so r + a Ǹ2 + 1.24Å 4 We first determine λp for the electrons through the deBroglie relationship, λp = h/(mv). This requires knowledge of v: Eel = e x V = (mv2)/2 Thus: lp + h mǸ 2e x V m + v + Ǹ2e mx V h Ǹ2e x V x m λp = 3.9 x 10–12 m λ = 2d(hkl) sinθ; to determine θ we must know d(112), given by ań Ǹ6 . We must first determine the lattice constant (a) for gold (Au). From the P/T we get: molar volume = 10.2 x 10–6 m3; accordingly (Au = FCC): 23 10.2 x 10 *6 + 6.02 x 10 x a 3 4 a + x 10 Ǹ40.8 6.02 x 10 3 *6 23 + 4.08 x 10 *10 m We now find that: *10 d (112) + 4.08 x 10 + 1.66 x 10 *10 m Ǹ6 sin q + 3.9 x 10 *12 l + + 1.17 x 10 *2 2d (hkl) 2 x 1.66 x 10 *10 q + sin *1 1.17 x 10 *2 + 0.67 o After you have determined θ, think for a moment! OK!? Selection rules tell you that diffraction on {112} planes are destructively interfered with! No diffraction on {112}! LN–5 19. We first determine the crystal system, then (knowing λ) we determine a and obtain the asked for molar volume. (a) (b) (c) θ 20.25 29.30 66.30 2θ 40.50o 58.60o 132.60o sinθ 0.3461 0.4894 0.9156 sin2θ 0.1198 0.2395 0.8384 We recognize 0.8384 = 7 x 0.1198; this means (sin2θ7)/14 = (sin2θ1)/2 = const. The system is BCC. l  2d (110) sin q 1  2 a sin q 2 a  2 x 1.541 x 10
10  3.15 x 10
10 m 2 x 0.3461 Molar volume = NA/2(a3 = 9.4 x 10–6 m3/mole = 9.4 cm3/mole 20. From information in the P/T we know the molar volume of Ag at 300K. This allows us to determine θ(111) through the lattice constant. With the given ∆θ we can obtain θ(111) at 1073K (which is 800oC). At 300K: Molar Volume = 10.3 x 10–6 = NA/4(a3) a  4 x6.0210.3x x1010 3
6 23  4.09 x 10
10 m
10 d (111)  4.09 x 10  2.361 x 10
10 3 λ= 2d(111) sinθ q (111)  sin
1 l 2d (111)  sin
1 0.709 x 10
10  8.64 o 2 x 2.361 x 10
10 At 800oC (1073K): θ(111) = 8.64 – 0.11 = 8.53o (continued next page) LN–5 20. Continued: Correspondingly: d (111)  l  2.34 x 10
10 m  a 3 2 sin 8.53 a  3 x 2.39 x 10
10 m  4.14 x 10
10 Molar Volume = NA/4 x a3 = 10.7 x 10–6 m3/mole = 10.7 cm3/mole 21. We recognize that the existence of (100) diffraction indicates a simple cubic system. After determining the lattice constant, we can obtain the central void dimension and answer the question. Required: (100) diffraction = SC SC structure: d(100) = a λ = 2d sinθ = 2a sin θ a = λ/(2sinθ) (The way the question is phrased, it may not appear clear if 14.88o is θ. The diffraction expert would take 14.88o = 2θ, but you may have a different idea. Both assumptions are taken as correct.) Accordingly: (a) 14.88 o  q 1 ; a  l  1.541  3.0Å 2 x 0.257 2 sin 14.88 (b) 14.88 o  2q 1 ; a  l  1.541  5.95Å 2 x 0.129 2 sin 7.44 In the SC system we have a central void; an atom in this location is bounded along the body diagonal by all eight atoms: (continued next page) LN–5 21. Continued. 2r = a a Ǹ3 + 2r ) 2R + a ) 2R 2R + a Ǹ3 * a + a(Ǹ3 * 1) R + a (Ǹ3 * 1) + 0.366 a 2 For (a): R = 0.366 x 3.0 = 1.1Å R = maximum radius for interstitial atoms For (b): R = 0.366 x 5.95 = 2.18Å Since the maximum radius for interstitial atoms is 1.1 and 2.18 Å for (a) and (b) respectively, an atom with r = 1.08Å can be accommodated without lattice distortion in either crystal lattice. 22. In a BCC system, the first diffraction (smallest θ) occurs on {110} planes (see selection rules, LN5–11). Given the lattice constant (a = 5.5Å), we find the d(110) spacing at d (110) + 5.5 + 3.89Å Ǹ2 In order to obtain {110} diffraction (maximum θ = 90o and sinθ = 1), λ ≤ 2d must be satisfied. Since no target material for the tube is indicated, we take the smallest λ available from the X–ray generator, λSWL:
SWL + hc + 10.3Å eV 2d = 3.89 x 2 = 7.78 Å We find 10.3 > 7.78 and therefore no diffraction cone can be generated by this particular powder with the available X–rays. LN–5 23. Grind the material to powder and do a Debye–Scherrer diffraction with CuKα radiation (λ listed or calculated). If the film shows diffraction lines, the material is crystalline [the answer to (a)]. If the sin2θ values for the first and seventh lines yield a constant ratio for: sin 2 q 7 sin 2 q 1

const 14 2 we are dealing with a BCC structure [answer to (b)]. We obtain the lattice constant through the Bragg equation: λ = 2d(110) sinθ1 [ d (110)
a ] 2 l
2a sin q 1 2 a
24. l 2 2 sin q 1 To solve the problem we first determine the longest wavelength still capable of causing diffraction. From the Bragg relation: λ = 2d(hkl) sinθ. To get diffraction, θ may assume a maximum value of 90o (sin 90o = 1). Accordingly: λmax = 2d(hkl). In FCC systems the largest diffraction d is d(111): l max
2d (111)
2(a 3)
6.93Å Using the Moseley relation we know that λCr(Kα) = 2.3Å. Therefore diffraction will take place if Kα radiation is generated by a 1.5 kV accelerating potential, or if λSWL is equal or less than 6.93Å! λSWL = (hc)/eV = (12.4 x 10–6)/V = 8.29 x 10–10 m We recognize that λSWL > λmax. There cannot be any diffraction given the X–rays from a tube operated at V = 1.5 x 103 Volt. LN–5 25. From Moseley’s law we know that λKα ∝ 1/Z2. As a first approximation, by comparing CuKα and MoKα, we therefore have: lMoK a ^ 1.542ǒ29Ǔ 42 2 + 0.735Å Or, more accurately: ǒ 1.542 29 * 1 42 * 1 Ǔ 2 + 0.719Å (The listed value for λKα (Mo) = 0.71Å) Since in replacing a Cu target with a Mo target the resulting λKα is decreased, it must be expected that the first diffraction peak will shift to a smaller θ!! 26. Vanadium is BCC. The largest diffracting d is d(110). From the Bragg relation, λ = 2d(hkl) sinθ, we recognize that θ can go up to 90o (sinθ = 1) and the longest wavelength still giving diffraction is seen to be: l max + 2d (110) + 2(ań Ǹ2) For vanadium: Mol.Vol. = NA/2 x a3 a + Ǹ2 x6.028.78x x1010 3 *6 23 + 3.08 x 10 *10 m l max + 2 a + 4.35 x 10 *10 m Ǹ2 LN–5 27. Mo: Z=42; Kα → (ni=2; nf=1); σ=1 ȱ ȳ n Ka + R(Z * 1) 2ȧ 12 * 12ȧ n 2ȴ Ȳnf ƪ ƫ ƪ ƫ 1 (42 * 1) 2 1 * 1 n Ka + 1.097 @ 10 7 m 12 22 n Ka + 1.38 @ 10 10m *1 l Ka + 28. 1 + 7.25 @ 10 *11m n Ka ȱ ȳ 1 + n + R(Z * 1) 2ȧ 12 * 12ȧ Ka l Ka nf ȴ Ȳni + R(Z * 1) 2 @ 3 4 (Z * 1) + Ǹ3 @ l4 @ R + 22 Z = 23 (Vandium) 29. Ni = FCC; n = 4; AtWt = 58.7; ρ = 8.9 g/cm3 To get the longest wavelength diffracting in Ni – we look for diffracting planes with maximum spacing ({111}) and we operate at the maximum θ possible (90o) l + 2d (111) sin q + 2 a sin 90 o + 2 a Ǹ3 Ǹ3 for nickel: a(Ni) + @4 Ǹ8.9 @58.7 6.02 @ 10 l + 2 @ 3.52 @ 10 *10 Ǹ3 3 23 + 3.52 @ 10 *8cm @ 1m + 3.52 @ 10 *10m 100cm + 4.06 @ 10 *10m LN–5 30. (V) e– beam Mo target (Z = 42) x–rays
6
6 l SWL  hc  1.24  10 [m]  1.24  104 acc.Voltage eV 5  10 = l Ka  2.48 x 10 –11m 4  7.23 x 10–11m 2 3R(z
1) how did I get λKα? n K a  R(Z
1) 2 3 4 1  l  4 n 3R(Z
1) 2 LN–5 31. h ; we need Electron diffraction makes use of the deBroglie relationship: l p + mv d(111) that means we need, a, the lattice constant since: d (111) + a ; the energy Ǹ3 of the electron(s) is eV(acc.)=Mv2/2 . 2a l + 2d (111) sin q + sin q Ǹ3 3Ǹ *6 l p + 2a sin 6 o a + 10.2 @ 10 23@ 4 + 4.08 @ 10 *10m Ǹ3 6.02 @ 10 *10 @ sin 6 o + 4.92 @ 10 *11m l p + 2 @ 4.08 @ 10 Ǹ3 2 E el + eV + mv 2 2 V + mv 2e ; v + h ml p 2 h2 V + m @ h2 2 + + 2e m l p 2eml 2p 32. 623.6 Volt Let’s make use of the data in problem (29): a = 3.52x10–10m; smallest acceleration potential means operating with the shortest λ achievable at a given acceleration potential: *6 l SWL + 1.24 @ 10 ; the longest λ giving diffraction is at θ=90o on planes with Voltage largest interplanar spacing ({111}). l SWL + 2d (111) sin 90 + 2 a + 4.06 @ 10 *10m Ǹ3 *6 4.06 @ 10 *10 + 1.24 @ 10 V *6 V + 1.24 @ 10*10 + 4.06 @ 10 33. 3.06 @ 10 3Volt Operate at the unit cell level with the atomic volume: 3 ; solve for At.wt. Atomic Volume = At.wt + N r na N @ a @ r + 6.02 @ 10 23 @ (0.316 @ 10 *9) 3 @ 19.3 @ 10 6 g n 2 m3 At.wt = 181.6g (the factor of 106 associated with the density term accounts for the change of units from g/cm3 to g/m3) LN–5 34. A diffraction set-up is asked for with specification of conditions for achievement of λKα. You have the choice of material to be used as monochrometer; first determine λKα (for Ni target) then decide on monochrometor material. You could use Ni, or more conveniently, you just determine the atomic weight for tungsten (W; the W stands for Wolfram, the official german name) and you have its lattice constant. ÉÉÉÉ ÉÉÉÉ ÉÉÉÉ ÉÉÉÉ (110) diffraction set up: W monochrometer (110) θ X White Ni radiation collimator lK a (Ni) Fe (to be analyzed) NiK l nK l + 27 2 @ 1.097 @ 10 7 @ 3 (m *1) 4 = 6.0 x 109 (m–1) lK a = 1.67 x 10–10m To isolate Kα from White Ni radiation we establish diffraction for λ = 1.67 x 10–10 m on (110) planes: lK a + 2d (110) sin q + 2 a sin q Ǹ2 Ǹ2 lK a Ǹ2 @ 1.67 @ 10 *10 sin q + + + 0.374 2a 2 @ 3.16 @ 10 *10 q + sin *1 0.374 + 21.9 o LN–5 35. Determine the energy/photon then λ of radiation and use the Mosley relation to find Z the atomic #. E photon + 7.725 @ 10 8Jń6.02 @ 10 23 + 1.28 @ 10 *15Jńphoton hcn + hc l 9 *1 hc l + + 1.55 @ 10 *10m ; n + 6.45 @ 10 m 1.28 @ 10 *15 n + 6.45 @ 10 9 + (Z * 1) 2R 3 4 Z + 36. 6.45 @ 10 Ǹ34@@1.097 @ 10 9 7 ) 1 + 28 ) 1 + Fe(BCC): sin q + @ 7.1 @ 10 Ǹ2 6.02 @ 10 3 *6 23 + 2.87 @ 10 *10m Ǹ2 @ 1.67 @ 10 *10 Ǹ2 @ l + + 2 @ 2.87 @ 10 *10 2 @ 2.87 @ 10 *10 q + sin *1 0.411 + 0.411 24.3 o The unit cell is defined as a cell bounded by 3 sets of parallel planes which will fill all space when translated by multiples of the plane spacings. Cells that contain a total of one lattice point are called primitive cells, others are called “non–primitive” cells. Each cell is characterized by volume, shape and contents. 38. (a) 39. (copper) This is a simple application of the Bragg relation: We determine “a” for Fe(BCC) and λKα ffor Ni and insert: λKα (Ni) we know from Problem (34); it is 1.67 x 10–10m the lattice constant for a + 37. 29 close packed array of equal sized spheres (FCC) (b) simple cubic structure (SC) (c) body centered cubic lattice (BCC) 12 6 8 If two or more parallel interacting waves are moving through a medium, the resultant wave function at any point is the algebraic sum of the wave functions of the individual waves. As a consequence of this “superposition” principle, the intensity of 2 waves that are coherent and have the same wavelength (coincidence of maxima and minima; ∆x=nλ) is increased (constructive interference); while that of waves, in which the maxima of one coincide with the minima of the other, is reduced (destructive interference). LN–5 39. Continued. Diffraction of x–ray is the result of scattering and interference in ordered structures (crystals). Coherent x–rays, scattered at atomic planes, will exhibit constructive interference in directions for which the scattered rays, at parallel planes, result in path differences of one or more λ. Coherent x–rays, scattered at parallel planes, will experience destructive interference in directions along which the scattered rays experience a path difference of (n–1/2)λ. 40. High quality Al single crystals will always have dislocations and “dislocation generators” which under stress generate more and more dislocations. Under stress, these dislocations move more readily in the slip system from one side of a crystal to the other; thus leading to slip, plastic deformation, limiting the tensile strength of aluminum. In fine grained Al you also have dislocations. In fact, there are many more than in the perfect crystal. When you apply stress, the dislocations move readily. However, they get arrested at the first grain boundary. Here stress may still persist and slip will be initiated, again along a slip plane, which no longer is in the same direction. The arrests of slip and changing slip directions from grain to grain lead to a significant reduction in the magnitude of deformation. The strength is increased.