Page 1 of 1

Problems (Lecture 8) NFE spectrum Present graphically the band structure of foc met als in the zero order NFE approximat

Posted: Fri Mar 04, 2022 10:17 am
by answerhappygod
Problems Lecture 8 Nfe Spectrum Present Graphically The Band Structure Of Foc Met Als In The Zero Order Nfe Approximat 1
Problems Lecture 8 Nfe Spectrum Present Graphically The Band Structure Of Foc Met Als In The Zero Order Nfe Approximat 1 (30.47 KiB) Viewed 41 times
Problems Lecture 8 Nfe Spectrum Present Graphically The Band Structure Of Foc Met Als In The Zero Order Nfe Approximat 2
Problems Lecture 8 Nfe Spectrum Present Graphically The Band Structure Of Foc Met Als In The Zero Order Nfe Approximat 2 (15.13 KiB) Viewed 41 times
Problems (Lecture 8) NFE spectrum Present graphically the band structure of foc met als in the zero order NFE approximation, i.e. reduce the free electron spectrum e = h?kº/2mo to the 1st Brillouin zone. Answer is presented in Fig. 2. The k-paths in Brillouin zone, along which the band structure is presented, are indicated in Fig. 3. The band structure in the zero order NFE approximation is the free- electron energy spectrum e = fik2/2m, reduced to the first Brillouin zone. Therefore, different energy spectrum branches in Fig. 2 correspond to the dispersion curves 12 (k -9) 2mo

x U W w х w (a) (b) Figure 3: The first Brillouin zones of fec crystals. at different reciprocal lattice vectors g. Let us assume that the first segment of the k-path, IX, is chosen along the (100) direction, 27 k=res where a is the lattice constant, 1 is the quasimomentum modulus in 27/a units (r = [0,1]). Then, the formula e = 1? (k - g)/2mo can be rewritten as a 12 me (19) * [(x + m» – m2 – mo)2 + (m. – ma + ms)2 + (m + m2 - mo)]. + Here m; are indices of the g vector, g = mybı + m2b2 + m363, b; are the primitive reciprocal lattice vectors, and the term in front of brackets is the electron energy at the X-point, ex = ħi(21/a)/2mo. Introducing the dimensionless energy y = €/ex, the s-formula can be presented as the y(x) function y = (x + mi - m2 - m3) + (mı - m2 + m3)² + (mı + m2 - m3), which corresponds to energy bands depicted in Fig. 2 along the TX path. The lowest energy band corresponds to the mi = {0,0,0) in dices, while

upper bands correspond to {0,1,1}, {-1,0,0}, {1,1,1}, {0,1,0}, {0,0,1}. The latter four energy bands are degenerate. (Degeneracy is indicated in Fig. 2 by numbers at energy bands.) Prove that the Fermi energy can be presented in the form 32 2/3 EX 27 EF = (See dashed lines in Fig. 2.)