There are 7 crystal systems and 14 Bravais lattices in 3 dimensions.

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1 EXAM IN OURSE TFY40 Solid State Physics Moday 0. May 0 Time: DRAFT OF SOLUTION Problem (0%) Itroductory Questios a) () Primitive uit cell: The miimum volume cell which will fill all space (without holes or overlap) whe traslated with all lattice vectors R. otais oe lattice poit. () rystal system: Bravais lattice poit groups. We make a uit cell with three edges a,b,c ad three agles α, β, γ, which will make up differet crystal systems depedet o the restrictios o legths a,b,c ad agles α, β, γ. Examples of crystal systems are tricliic, moocliic, orthorhombic ) () Bravais lattice: A distict lattice type /special lattice type. A special patter of poits which looks idetical from each poit. There are 7 crystal systems ad 4 Bravais lattices i dimesios. b) Phoo: quatum of eergy of a lattice vibratio; a collective excitatio i a periodic, elastic arragemet of atoms i codesed matter. Optical phoo: masses oscillate with opposite phase; opposite directios. Acoustic phoo: masses oscillate with same phase; same directios. See figure 0, page 98 i Kittel. I total, with s atoms i basis, there are s modes, acoustical ad (s-) optical modes. c) The free electro model ca describe heat capacity, thermal coductivity, electrical coductivity/ resistivity ad electrodyamics of metals. Ad the Hall effect (but ot the sigs i the Hall costat!). d) Metal: Fermi level lies i a partial filled eergy bad. Isulator: A eergy gap occurs (at T=0) betwee a filled lower bad (valece bad) ad a empty higher bad (coductio bad), there is a gap > ev at the Fermi level. Semicoductor: A isulator with a small bad gap (< ev). Thermal excitatios across gap are possible. Electric coductivity of semicoductor ad metal: Semicoductor: A few mobile charge carriers available at room temperature= low coductivity. At higher T will there be more mobile electros/holes ad coductivity icreases.

2 Metal: May mobile electros available. What limits the coductivity is impurities ad phoos. Desity of phoos icreases with T, more phoo-electro collisios ad less coductivity. This also happes for semicoductors, but here the effect of more charge carriers will be more importat tha the icrease phoo-electro scatterig. Problem (5%) Structure ad Diffractio * * * a) Reciprocal lattice vector: Ghkl = ha + kb + lc is a vector betwee lattice poits i the reciprocal lattice, give by the sum of itegers (h, k, l) of primitive vectors i the reciprocal lattice vectors (a*, b*, c*). The vector is ormal to the lattice plae (hkl). We will show that k k ' = Ghkl has to be valid for costructive iterferece. k ad k ' are the icomig ad the scattered k-vector, respectively, for the X-ray scatterig. π k = k ' = k =. We look at scatterig from two lattice poits with a distace R ad look λ at the path differece: k k R k k Differece i path (from figure): k ' k ' λ R R = Rk ( ' k) = Rk ( ' k) k k k π This has to be a iteger umber of λ if costructive iterferece. The equatio over is valid for all R. Rk ( ' k) λ ir( k ' k ) = λ, Rk ( ' k) = π this gives that e =. For this to be valid, π ( k ' k) = Ghkl has ti be valid for all R. This was what should be show. QED. b) The agle betwee k ad k ' is θ. For a cubic crystal we have π π π π a* = b* = c* = which gives ( k ' k) = Ghkl = h+ k+ l a a a a π Square both sides give: ( k) k k ' cos ( θ ) + ( k ') = ( h + k + l ) a π ( θ ) k cos = ( h + k + l ) a π π si θ λ + = a ( h + k + l ) 4 h + k + l si θ = λ a 4a si θ h k l λ = + + λ or si θ = ( h + k + l ) a

3 You will get full pot if you derive si θ from Braggs law, startig from dsiθ = λ, itroducig expressio for d, ad square this (ad obtai the same result!). c) This is what we did i the X-ray lab. We have to fid the (hkl) values correspodig to the 5 lies. This ca be solved i two ways: ) Look at the lowest differece i i the equatio above. λ si ( ) a h k l λ θ a N = + + = i si θ from above ad from this fid the lie si θ si θ i si θi N i N hkl ) We ca assume bcc or fcc ad do a test ad trial: h + k + l (N i ) bcc fcc hkl h+k+l a a hkl h+k+l a a 0 7, ,999 4,859 6, ,9959 6, ,9959 6, , , ,5459 6, ,5906 5,449 40, , ,8599 5, ,69 6,757 I both cases we fid that the reflectios (with itesities) are, 00, 0, ad, correspodig to oe reflectios (extictios) with mixed idices, ad therefore a fcc structure, ad that the lattice parameter is 6.7 Å. d) This is a cubic primitive (P) cell with a basis of A i (000) ad B i ( ½ ½ ½ ); correspods to the sl structure (grey is A ad red is B). Oe AB molecule per Bravais lattice poit. This structure (P) will ot give ay extictios: We have h k l ig ( ) ( ) hkl r i i π hx + ky + lz π + + G = hkl = = = A + B S F fe f e f f e Above T c (statistically disorder o the atom positios) each lattice poit will statistically have ½ A ad ½ B ad the two atoms i the basis will be idetical. We will have a base cetred cubic (bcc) structure ad observe extictios as for the bcc: h k l ig ( ) ( ) for h+k+l=eve hkl r i π + + f A + fb F( hkl) = f e = ( f A + fb) + e = 0 for h+k+l=odd

4 That meas; extictios if sum of hkl is a eve iteger. Problem (5%) Phoos a) This derivatio is doe i Kittel pages ad also i the lecture. I therefore show it very brief here. We put up the force o the atom, assumig oly earest eighbour iteractio, ad the Newto s secod law for each of the two masses i the basis. Fm = Kv ( u) + Kv ( u) ad FM = Ku ( v) + Ku ( + v). With Newtos. law, we get the two coupled differetial equatios v u m = Ku ( + + u v) ad M = Kv ( + v u) t t i( ka ωt ) i( ka ωt ) The we use the solutios u = u e og v = v e ad get ika K mω K( + e ) u 0 ika = K( + e ) K Mω v The determiat has to be zero, which gives the equatio ( )( ) ( ika ika K mω K Mω K + e )( + e ) = ( ika ika K Kmω KMω + mmω K + e )( + e ) = 0 This is a secod order equatio i ω, a = mm b ± b 4ac x =, b= Km ( + M) a c = K (+ cos ka) K( m+ M) ± 4 K ( m+ M) 4mM K (+ cos ka) ω = mm. ad gives the dispersio relatioship K( m + M ) 4mM ka ω = ± si mm ( m+ M) as we were supposed to show. (*) NB! I the exercise the root stopped before the si I hope all got the message about this correctio! b) Whe k 0 we get K( + ) optical brach m M ω = ka K M + m ( ) acoustical brach 4

5 O the BZ we have k.. π = ad this gives, whe iserted ito (*) ; a K m ω = K M Which is correspodig to acoustical ad optical depeds o which mass is heaviest. Group velocity for log wavelegths: δω v g = = 0 for optical brach δ k v g δω K = = a δ k ( m+ M) for k 0 acoustic brach.. If the material is exposed to waves with ω, the wave will be damped very quickly; o propagatio. c) Whe M = m the forbidde gap disappears. We are back to the result for the mooatomic lattice; the two braches will be oe folded ito the Brillui zoe (BZ). The acoustic a K velocity is, the same as for the mooatomic lattice with lattice costat a/, as m expected. The dispersio relatio becomes as for the mooatomic chai: 4K si ka ω = m 4 figure! d) We ca measure the soud velocity i ioic materials by sedig i IR light ad look at the resoace curve. We did i the lattice vibratio lab for Si.. figure! Problem 4 (0%) Free-electro model, eergy bads ad semicoductors a) Free electro model: Assume coductio electros are a free electro gas. No iteractio with ios or lattice, or iteractio betwee coductio electros. harged particles; follow the Pauli priciple. Ek ( ) = k. The Fermi eergy is the highest occupied eergy, m correspodig to the Fermi eergy wave vector k F, correspodig to a sphere i D. The umber of states iside the sphere with radius k F is N. The volume i k-space which is occupied by oe state is give so that the size of the material is L*L*L=L =V i real space; π which is L. There is oe allowed wave vector for each π this will the give: L 5

6 4 π k N = = π π L F kv F which gives k F Nπ = V. Nπ This gives the Fermi eergy EF =. m V From this expressio we ca ow fid the umber of states less tha E, give by N(E): E F N( E) π = m V gives N( E) V Em π = The we ca fid This gives V m π = dn V m V m DE ( ) = = E = E de π π b) Empty lattice approximatio; the parabola for free electros are made above each reciprocal lattice poit, as show i figure (b) uder, the bads ca be umbered from where they cross the BZ boudaries.. The figure shows the three differet schemes used: (a) exteded zoe scheme (b) periodical zoe scheme (c) reduced zoe scheme All three descriptios are fully equivalet. c) I words: We start with the Schrödiger equatio; use a potetial i this which is periodic with the lattice ad also a sum of Bloch fuctios as the wave fuctio. Bloch fuctios are plae waves multiplied with a fuctio with period as the lattice. Whe the periodic potetial ad the Bloch fuctios are put ito the S-equatio, (both as Fourier sums) we get the etral equatio -coupled set of equatios. Everythig happes i the BZ, eergy bad, N states i each bad, we get a electroic bad structure; forbidde eergies, Bragg scatterig is the 6

7 reaso Stadig waves o BZ boarder; the same k gives differet eergies. The eergy gap is proportioal to the first Fourier compoet of the potetial. Mathematically (from lectures) (I do t expect all this!!!) d) Itrisic: udoped, o states i bad. Extrisic: doped, eergy level of impurity states i bad gap p-dopig: empty states ear the top of the valece bad -dopig: filled electro states ear the bottom of the coductio bad Explaied i Kittel page 09-. Effective mass of electro or hole reflects that the properties of semicoductors also deped o bad structure ear the gap, ot oly o the gap. The electro i a periodic potetial will be accelerated relative to the lattice as if it has aother mass. It is iversely proportioal to the curvature of the eergy bad. I geeral a tesor... Explaied i lecture ad Kittel page e) Expressio for the desity of states D(E) with the lowest eergy coductio bad equal to E c : V m DE ( ) = ( E E ) (this ca be take from problem 4 a). π ( E µ )/ kt B We ca do this f( E) = e because the bad gap i semicoductors (~ ( E µ )/ kt B e + ev) is usually much larger tha k B T ad therefore ( E µ ) >> kt B. 7

8 ocetratio of electros i the coductio bad: N ( E µ )/ kt V m B = f ( E) D( E) de e ( E E ) de V = V π E E µ m kt B EkT / B = e e ( E E ) de π E This was doe i the lectures ad i Kittel o page 06. We have to use the itegral u u e du = 0 π (This is the Gammafuctio of /) ( ad ca be foud i Rottma. The aswer becomes: ( µ E ) mkbt kt B = e π 8