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By Johann von Neumann

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9-4. The number of atoms per cubic cell of volume a3 in such a lattice: N 1 1 8 = 2⋅[( 8 ⋅ ) + 6 ⋅ ] = 3 3 a 8 2 a . The number of valence electrons per conventional unit cell of diamond lattice = 4 ⋅ 9-5. 9-6. 8 . a3 The primitive translational vectors for; SCC: r r r a = a ex , b = a ey , c = a ez FCC: r a r a r a a = (e x + e y ) , b = (e y + e z ) , a = (e x + e z ) 2 2 2 ; 1 1 1 Diamond lattice = FCC with 2 atoms per basis at ( 0, 0, 0) and  , ,  . It is, therefore, 4 4 4 equivalent to two inter- laced FCC lattice displaced one quarter the distance along the body diagonal of the FCC.

7. Assume a Lorentzian fluorescence linewidth of 10 Ghz. 1x10 x12 hc 2 2 . 4 x10−12 cm 2 . 4 x10−2 cm −1 , if the total population inversion between the 1s and 2p levels of hydrogen in the gaseous medium is 1010 cm-3 . Chapter 9 9-1. 62) : r r  Ze 2  ˆr ˆr ˆ  L ⋅ S ≡ ζ( r ) Lˆ ⋅ Sˆ H s−o =   2m 2 c 2 r 3    The corresponding matrix for l = 1 in the representation in which Lˆ2 , Lˆ z , Sˆ 2, Sˆ z are diagonal is a 6x6 matrix. To diagonalize this matrix within the manifold of degenerate states |n,l = 1, ml , s = 1/2, ms > , the columns and rows corresponding to the pairs of (ml , ms ) values are arranged in a particular order: ( ml , ms ) ( − 1, − 1 2 ) ( − 1, + 1 2 ) ( 0, − 1 2 ) ( 0, + 1 2 ) ( + 1, − 1 2 ) ( + 1, + 1 2 ) (−1,−1/ 2) 1/2 0 0 0 0 0   (−1, +1 / 2 )  0 −1/2 1/ 2 0 0 0 ( 0,−1/ 2)  0 1/ 2 0 0 0 0 2   ⋅ ζ nl h .

The corresponding Schroedinger’s equation is ⎫ ⎧ h2 1 ∂ 2 ∂ [ 2 (r ) ] + V (r )⎬ Rno (r ) = E n Rno (r ) , ⎨− ∂r ⎭ ⎩ 2m r ∂ r for r≤a , ⎧ h2 1 ∂ 2 ∂ ⎫ [ 2 (r ) ]⎬ Rno (r ) = E n Rno (r ) ⎨− ∂r ⎭ ⎩ 2m r ∂ r for r≥a . and , The equation for r ≤ a can be converted to: 2mE d2 U (r ) = − 2 U (r ) 2 h dr , where U(r) = r R(r) . The general solution of this equation is: U (r ) = A cos kr + B sin kr where k = 2mE . To satisfy the boundary condition that Rn0(r) must be finite at r =0, h A must be equal to 0, or 6-7 U (r ) = B sin kr Similarly, , for for r ≥ a , U (r ) = C e α r + D e − α r where α = r ≤ a.

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