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By Roger James Elliott, Alan Frank Gibson

Elliott and Gibson's vintage creation to sturdy country physics.

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The external field due to the sphere is that of a dipole of moment µ = MV, when V is the volume. In the equatorial plane at the surface of the sphere the field of the sphere is −µ a 3 = −4πM 3 = Ba 2. The total field in this position is 3Ba/2. 10-3 10-4 CHAPTER 11 1. From Eq. (10), χ = −N e2 < r2 > . 2 6mc ∞ −2r a 0 1 2 2 ⋅ π ⋅ = 3a 0 . 02 × 1023 mol–1. 2a. Eu++ has a half-filled f shell. Thus S = 7 × 1/2 = 7/2. The orbitals mL = 3, 2, 1, 0, –1, –2, –3 have one spin orientation filled, so that L = ΣmL = 0.

We have Sρ+δ = Sρ eik ⋅δ . Thus    dSρ ( x ) ⎛ 2JS ⎞ y 6 − ∑ ei k ⋅δ Sρ =⎜ ⎟ δ dt  ⎝ h/ ⎠  ⎛ 2JS ⎞ ⎡ y 6 − 2 ( cos k x a + cos k y a + cos k z a ) ⎤⎦ Sρ ; =⎜ ⎟ ⎣  ⎝ h/ ⎠ dSρ y ⎛ 2JS ⎞ ⎡ x 6 − 2 ( cos k x a + cos k y a + cos k z a ) ⎤⎦ Sρ . = −⎜ ⎟ ⎣ dt  ⎝ h/ ⎠  These equations have a solution with time-dependence ∼ exp(–iωt) if ω = ( 2JS h/ ) ( 6 − 2 cos k x a − 2 cos k y a − 2 cos k z a ) . 2. U = ∑ n k h/ ωk = h/ ∫ dω D( ω) ω < n ( ω) > . If ω = Ak2, then dω dk = 2Ak = 2 A ω1 2 , k  and D( ω) = 4π ω 1 1 ω1 2 .

Thus ωο2 = 4πne 2 3m. 6. md 2r dt 2 = −mω2r = −(e c) ( v × Bzˆ ) − mωο r, where ωo2 = 4πne2/3m, from the 2 solution to A. Thus, with ωc ≡ eB/mc, −ω2 x = iωωc y − ωo x ; 2 −ω2 y = −iωωc x − ωo y . 2 14-1 Form ξ ≡ x + iy; then −ω2ξ − ωωc ξ + ωο ξ = 0, or ω2 + ωωc − ωο = 0, a quadratic 2 2 equation for ω. 7. Eq. (53) becomes c 2 K 2 E = ω2 [ε (∞)E + 4πP], where P is the ionic contribution to the polarization. Then (55) becomes ω2 ε(∞) − c 2 K 2 4πω2 = 0, 2 ω2 Nq 2 M − ωT or 2 2 ω4 ε 2 (∞) − ω2 ⎡⎣c 2 K 2 + ε (ω)ωΤ + 4π Nq 2 M ⎤⎦ + c 2 K 2ωT = 0.

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