PAP342-Solid State Physics I Solution 09/10 Seester 2 Wang Shengtao May 10, 2010 Question 1. (a) A scheatic showing the position of the Feri level related to the (b) band edges can be found in [Kittel] p. 162. Metals: Partly filled conduction band (10% 90%) Seiconductors: Slightly full or slightly epty conduction band Insulators: Alost fully filled top energy band ( 100%) ε F = 2 (3π 2 n) 2/3, T F = ε F 2 e k B (1) n = N V = N A a /ρ 2.5 1028 3 (2) ε F 3.15 ev (3) T F 3.65 10 4 K (4) ( ) 2 2π (c) Each state occupies an area of in k-space. Thus, for a circle of radius k, the L nuber of states is N = 2( πk2 2π ) 2 = k2 A 2π, (5) L where the factor of 2 is due to spin degeneracy. Energy in ters of k is ε = 2 k 2 2. We can express the nuber of states in ters of energy: N = A ε. (6) π 2 1
The density of states is thus The density of states is independent of energy. D(ε) = dn dε = A π 2. (7) (d) Assue collisions between the lattice and electrons is independent of collisions between the ipurities and electrons. τ is the relaxation tie and t/τ can be interpreted as the probability an electron will experience collision during a period of tie t. Thus, dt/τ L and dt/τ i can be interpreted as probability of electrons colliding with lattice and ipurity respectively in tie dt. By Newton s second law, we have P (t + dt) = ( 1 dt dt ) ( ) P (t) + f(t)dt. (8) τ L τ i Since we have assued the two scattering echaniss are independent, the probabilities just add directly. Siplify (8), and keep the first order of dt, we have dp ( 1 dt = + 1 ) P + f(t). τ L τ (9) i By coparing to the drift equation for electrons, we can conclude that 1 τ = 1 τ L + 1 τ i. (10) For ore discussions on Mathiessen s rule, please refer to [Ashcroft/Merin] p. 323. (e) The effective ass of electron ( ) is the ass an electron see to assue in a periodic potential in an applied electric or agnetic field ( [Kittel] p. 198). It can be defined as 1 = 1 d 2 ε 2 dk. (11) 2 The effective ass can be deterined by cyclotron resonance by easuring the cyclotron frequency. Refer to [Kittel] p. 200 for ore details. Nanyang Technological University Page 2 / 7
Question 2. (a) The drift equation is ( d dt + 1 ) P (t) = q( E τ + v B), (12) where P (t) = v and B = (0, 0, B). Expand the drift equations in Cartesian coordinates: τ v x = q(e x + v y B) τ v y = q(e y v x B) (13a) (13b) v z = 0 2DEG (13c) Expressing E x and E y in ters of v x and v y, and then substituting v x = J x nq and v y = J y, we can get nq E x = q 2 nτ J x B nq J y E y = q 2 nτ J y + B nq J x. (14a) (14b) Substitute q = e for electrons, and express in vector fors: ( ) ( ) ( B Ex = e 2 ne nτ Jx 1 ωc τ E y B = ρ 0 J ne e 2 y ω c τ 1 nτ where ρ 0 = e 2 nτ and ω c = eb. (b) Fro equation (15), we have ( Ex E y ) == ρ 0 ( 1 ωc τ ω c τ 1 ) ( Jx J y ) ) ( Jx J y ), (15), (16) At steady state, there is no current flowing in the y direction (electrons cannot flow out of the Hall bar in the y direction.) J y = 0. Also, J x = I x. Thus, E x = ρ 0 I x E y = ρ 0 ω c τ I x. (17) Nanyang Technological University Page 3 / 7
Assue E field is constant in the Hall bar, then V x = E x L x = L x ρ 0 I x V y = E y = ρ 0 ω c τi x. (18a) (18b) (c) Fro equation (18), we have V x = L x e 2 nτ I x V y = B en I x. (19) Therefore, n = BI x ev y 3.0 10 15 3. The obility µ is µ = v x E x = J x ne ρ 0 J x = 1 neρ 0 = L x I x en 1.03 10 2 2 /V s, Where the substitution of ρ 0 is fro equation (18). Nanyang Technological University Page 4 / 7
Question 3. (a) For an electron trapped to a donor in seiconductors, the potential is siilar to a hydrogen ato potential with ɛ 0 replaced by ɛ 0 ɛ and 0 replaced by. For ore details, please refer to [Kittel] p. 209 or [Ashcroft/Merin] p. 578. (b) For ground state n = 1, E = 13.6 0.45 1 44.0 ev 11.82 (20) r = 0.529 11.8 1 13.9Å. 0.45 (21) It should be noted that the radius r is large, so that the assuption ade in part (a) is valid (refer to the references for details.) (c) By equation (11), 1 = 1 2 d 2 ε dk 2 = 8 3 0. (22) Thus the effective ass is = 3 8 0. (d) Please refer to [Kittel] p.205-206 for the derivation: ( ) 3/2 e k B T n = 2 e (µ Ec)/kBT. (23) 2π 2 (e) Assue the carrier concentration is intrinsic, and h = e = e. Then, the Feri energy will be in the iddle of the band gap, µ E c = 0.555 ev. By equation (23), we can calculate the carrier concentration n 1.19 10 16 3. Nanyang Technological University Page 5 / 7
Question 4. (a) For diaagnetis, χ = µ 0NZe 2 6 r2. The agnetic susceptibility is independent of teperature. For other types of agnetiss, please refer to [Kittel] p.342 Figure 20. (b) Pauli explained the paraagnetic susceptibility of conduction electrons. Please refer to [Kittel] p. 315 for the derivation. (c) Fro part (b) and question 1(b), we see Substitute the value N = 4.7 10 28 3. χ = 3Nµ2 B 2ɛ F ɛ F = 2 2 e (3π 2 N) 2/3. (24) ɛ F 4.75 ev (25) χ 7.97 (26) (d) N = N A a /ρ 8.49 1028 3 (27) M = Nn B µ B 1.7 10 6 A 2 = 1.7 10 3 erg/g (28) (e) Please refer to [Kittel] p. 330 for the brief description of agnon theory for ferroagnetic aterials. Magnetis will change fro ordered ferroagnetic doain to disordered paraagnetis when the teperature rises close to the Curie teperature. It is a second order phase transition. As teperature rises, theral agitation copetes with the orderly alignent of spins, until the Curie teperature where theral agitation is strong enough to break the orderly alignent of spins. Nanyang Technological University Page 6 / 7
References [Kittel] Charles Kittel, Introduction to Solid State Physics. John Wiley & Sons, Inc, Eighth Edition, 2005. [Ashcroft/Merin] Neil W. Ashcroft and N. David Merin, Solid State Physics. Thoson Learning, Inc, 1976. Nanyang Technological University Page 7 / 7