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20111ee123B_1_EE123B_HW3_Solutions

# 20111ee123B_1_EE123B_HW3_Solutions - CHAPTER 4 la The...

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Unformatted text preview: CHAPTER 4 la. The kinetic energy is the smn of the individual kinetic energies each of the form 7 M1132. The force 2 between atoms 5 and s+l is 7C(us 7 Us”); the potential energy associated with the stretching of this bond is l — C(us — 11:”)2 . and we sum over all bonds to obtain the total potential energy. 2 1 . l b. The time average of 7 Mus2 15 7 Mmzuz. In the potential energy we have 2 4 n H = n cos[cot — (s +1)Ka] = n{cos(mt — sKa) - cos K3 5 + sin (mt —5Ka)- sin Ka}. Then 11s 71]S+1 : u {cos(mt —sKa)-(1— cos Ka) 7 sin (wt 7 sKa) - sin Ka}. We square and use the mean values over time: . l . <cosz>:<sm2>:5;<cossrn>:0. Thus the square of 11” above is 1 . Eu2[17 Zoos Ka + cosZKa + SlﬂzKﬂ] : u2(l 7 cos Ka). . . 1 2 . . . 1 The potential energy per bond is ECH (l 7 COS Ka), and by the disperswn relation to = (ZCIM) (1 — . . 1 cos Ka) tl'LlS 15 equal to — Mmzuz. Just as for a simple harmonic oscillator, the time average potential 4 energy is equal to the time-average kinetic energy. 3. From Eq. (20) evaluated at K = trial, the zone boundary, we have 7m2Mlu : 72Cu ' 5 402sz : 72Cv . Thus the two lattices are decoupled from one another; each moves independently. At (92 = 2CIM2 the motion is in the lattice described by the displacement v; at (91 = 2CIM1 the u lattice moves. 5. By analogy with Eq. (18), Mdzus/dtz : C1(vs _us) " €20,571 —u§); MdZVs/dtz : C1(“s _ Vs) " C2(us+1 7m2Mu = Cl(v 7 u) 7 C2 (ve'iK‘ 7u); 7(02Mv = Cl(u 7v)7 C2(ue‘K' 7v) , and — vS ), whence (C1+C2)7Mco2 7(C1+C2e’ﬂ“) 70 —(C1 +0261“) (C1+C2)—Mm2 . , ,, _. (205m) ‘ For Ka :0, m2 :0 and 2(C1+C2)/M. For Ka : n, m2 : 2C1/M and ZCZ/M. 6. (a) The Coulomb force on an ion displaced a 4—2 ...
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