HW04_Solutions - Physics 463 Winter 2012 HW 04 Solutions 1 a The problem assumes that the restoring force is due to the uniform sea of electrons inside
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Physics 463Winter 2012HW # 04 Solutions1. a. The problem assumes that the restoring force is due to the uniform sea of electrons inside the sphere ofradiusr. The electric field from such a sea of electrons is-en(r)4π0r2wheren(r) is the number of electronsinside a sphere of radiusr.n(r) =34πR343πr3=r3R3. Therefore, the restoring force on the ion isF=eE=e-en(r)4π0r2=-e24π0R3rThis is simple harmonic motion and the frequency isω=qCMwhereC=e24π0R3, the “spring”constant.ω=qe24π0MR3b. For sodium,R= 1.91·10-10meters is given by Table 9 on page 71. The atomic weight is about 23g/mol, or about 4·10-26Kg which gives about3·1013s-1.c. The largestKvalue is restricted by the size of the Brillouin zone, soKmax=πawhereais close toR.Therefore,Kmax= 1010m-1. Usingωfrom above, the speed of soundv=ωKmax=3000 m/swhichseems reasonable2. a. SinceC16=C2, two atoms are required to form a basis; therefore the lattice spacing with be 2a, twicethe spacing between atoms.Therefore,a1= 2aˆx, thusb1=2π2aˆx=πaˆxsincea1·b1= 2π.b. Newton’s second law statesFnet=m¨xand Hooke’s law statesF=-Cx.I have opted to use thebook’s notation ofCinstead ofkto avoid confusion withKas the wave vector. Only consider nearest-neighbors for each atom. Work considering different masses was done on pages 95-98, but the methodis similar.We then have the set of equations.m¨us=-C1(us-vs)- -C2(vs-1-us)=-C1us+C1vs+C2vs-1-C2usm¨vs=-C2(vs-us+1)- -C1(us-vs)=-C2vs+C2us+1+C1us-C1vsConsider the Ans¨atzeus=ueisK(2a)e-iωtandvs=veisK(2a)e-iωt.Note, we use 2abecause weconsider waves of thelattice. The equations become.-ω2mu((((((eisK2ae-iωt=(-C1u+C1v+C2ve-iK2a-C2u)((((((eisK2ae-iωt-ω2mv((((((eisK2ae-iωt=(-C2v+C2ueiK2a+C1u-C1v)(((((eisKae-iωtThese equations can be rewritten in a matrix formω2m-(C1+C2)C1+C2e-iK2aC1+C2eiK2aω2m-(C1+C2)uv=00(1)Call this matrixA. If|A| 6= 0 then∃A-1so thenuv=A-100=00, which is a trivialsolution. For nontrivial solutions, we require|A|= 01
The determinate comes out to0=ω4m2-2ω2m(C1+C2) + (C1+C2)2-(C21+C1C2(eiK2a+e-iK2a)+C22)0=ω4m2-2ω2m(C1+C2) + 2C1C2(1-cos(K2a))0=ω4m2-2ω2m(C1+C2) + 4C1C2sin2(Ka)Solve forω2using the quadratic formulaω2±(K)=2m(C1+C2)±q4m2(C1+C2)2-4·4m2C1C2sin2Ka2mm(2)=C1+C2m±1mqC21+C22+ 2C1C2cos(K2a)(3)This gives two dispersion curves, the optical branch (upper,ω+) and acoustical branch (lower,ω-).c. This showsC1/C2 = 2MinusΠ2MinusΠ40Π4Π2012kSlash1aFrequency, arb unitsd. The long wavelength limit isK(2a)1, since it’s equivalent toλ2a.Therefore, cos(K2a)≈1-12K2(2a)2+· · ·To first orderω2±(K)=1m(C1+C2±(C1+C2))ω2+=2m(C1+C2)(4)ω2-vanishes to first order, so include the next order (second order).