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

HW04_Solutions - Physics 463 Winter 2012 HW 04 Solutions 1...

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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 the sphere of radius r . The electric field from such a sea of electrons is - en ( r ) 4 π 0 r 2 where n ( r ) is the number of electrons inside a sphere of radius r . n ( r ) = 3 4 πR 3 4 3 πr 3 = r 3 R 3 . Therefore, the restoring force on the ion is F = eE = e - en ( r ) 4 π 0 r 2 = - e 2 4 π 0 R 3 r This is simple harmonic motion and the frequency is ω = q C M where C = e 2 4 π 0 R 3 , the “spring” constant. ω = q e 2 4 π 0 MR 3 b. For sodium, R = 1 . 91 · 10 - 10 meters is given by Table 9 on page 71. The atomic weight is about 23 g/mol, or about 4 · 10 - 26 Kg which gives about 3 · 10 13 s - 1 . c. The largest K value is restricted by the size of the Brillouin zone, so K max = π a where a is close to R . Therefore, K max = 10 10 m - 1 . Using ω from above, the speed of sound v = ω K max = 3000 m/s which seems reasonable 2. a. Since C 1 6 = C 2 , two atoms are required to form a basis; therefore the lattice spacing with be 2 a , twice the spacing between atoms. Therefore, a 1 = 2 a ˆ x , thus b 1 = 2 π 2 a ˆ x = π a ˆ x since a 1 · b 1 = 2 π . b. Newton’s second law states F net = m ¨ x and Hooke’s law states F = - C x . I have opted to use the book’s notation of C instead of k to avoid confusion with K as the wave vector. Only consider nearest- neighbors for each atom. Work considering different masses was done on pages 95-98, but the method is similar. We then have the set of equations. m ¨ u s = - C 1 ( u s - v s ) - - C 2 ( v s - 1 - u s ) = - C 1 u s + C 1 v s + C 2 v s - 1 - C 2 u s m ¨ v s = - C 2 ( v s - u s +1 ) - - C 1 ( u s - v s ) = - C 2 v s + C 2 u s +1 + C 1 u s - C 1 v s Consider the Ans¨atze u s = ue isK (2 a ) e - iωt and v s = ve isK (2 a ) e - iωt . Note, we use 2 a because we consider waves of the lattice . The equations become. - ω 2 mu ( ( ( ( (( e isK 2 a e - iωt = ( - C 1 u + C 1 v + C 2 ve - iK 2 a - C 2 u ) ( ( ( ( (( e isK 2 a e - iωt - ω 2 mv ( ( ( ( (( e isK 2 a e - iωt = ( - C 2 v + C 2 ue iK 2 a + C 1 u - C 1 v ) ( ( ( ( ( e isKa e - iωt These equations can be rewritten in a matrix form ω 2 m - ( C 1 + C 2 ) C 1 + C 2 e - iK 2 a C 1 + C 2 e iK 2 a ω 2 m - ( C 1 + C 2 ) u v = 0 0 (1) Call this matrix A . If | A | 6 = 0 then A - 1 so then u v = A - 1 0 0 = 0 0 , which is a trivial solution. For nontrivial solutions, we require | A | = 0 1
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The determinate comes out to 0 = ω 4 m 2 - 2 ω 2 m ( C 1 + C 2 ) + ( C 1 + C 2 ) 2 - ( C 2 1 + C 1 C 2 ( e iK 2 a + e - iK 2 a ) + C 2 2 ) 0 = ω 4 m 2 - 2 ω 2 m ( C 1 + C 2 ) + 2 C 1 C 2 (1 - cos( K 2 a )) 0 = ω 4 m 2 - 2 ω 2 m ( C 1 + C 2 ) + 4 C 1 C 2 sin 2 ( Ka ) Solve for ω 2 using the quadratic formula ω 2 ± ( K ) = 2 m ( C 1 + C 2 ) ± q 4 m 2 ( C 1 + C 2 ) 2 - 4 · 4 m 2 C 1 C 2 sin 2 Ka 2 mm (2) = C 1 + C 2 m ± 1 m q C 2 1 + C 2 2 + 2 C 1 C 2 cos( K 2 a ) (3) This gives two dispersion curves, the optical branch (upper, ω + ) and acoustical branch (lower, ω - ). c. This shows C 1 /C 2 = 2 Minus Π 2 Minus Π 4 0 Π 4 Π 2 0 1 2 k Slash1 a Frequency, arb units d. The long wavelength limit is K (2 a ) 1, since it’s equivalent to λ 2 a . Therefore, cos( K 2 a ) 1 - 1 2 K 2 (2 a ) 2 + · · · To first order ω 2 ± ( K ) = 1 m ( C 1 + C 2 ± ( C 1 + C 2 )) ω 2 + = 2 m ( C 1 + C 2 ) (4) ω 2 - vanishes to first order, so include the next order (second order).
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  • Fall '08
  • Lei Duan
  • Force, Ω, Reciprocal lattice, Brillouin zone, C1 C2, C1 C2 eiK

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