5 Problems Chapter 5: Electrons Subject to a Periodic Potential Band Theory of Solids

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1 E n = :75, so E cont = E E n = :75 = :479. Using E =!, :479 = m e k z =! j e j m e k z! k z = r :479 je j m e = :55 9 (44) (v g ) z = m e k z = m e :55 9 = :95 5 m/s. 4.. A ev electron is to be con ned in a suare uantum dot of side L. What should L be in order for the electron s energy levels to be well-uantized? From (.4), and we need L e. e = h p = h m e v = h = E m e m e h m e j ej m e = :78 nm; (45) 5 Problems Chapter 5: Electrons Subject to a Periodic Potential Band Theory of Solids 5.. To gain an appreciation of the important role of surface e ects at the nanoscale, consider building up a material out of bcc unit cells. (See Section 5.). For one bcc cube, there would be 9 atoms, 8 on the outside and one interior, as depicted on p. 4. If we constrain ourselves to only consider cubes of material, the next largest cube would consist of 8 bcc unit cells, and so on. If one unit cell is :5 nm, how long should the material s side be in order for there to be more interior atoms then surface atoms? # unit cells in the cube Surface Atoms Interior Atoms Ratio (S/I) 8 8 = :9 = :6 4 = : 5 = :8 so that we need 5 unit cells, leading to a material cube having side length 5 (:5) = :5 nm. 5.. Consider the Kronig-Penney model of a material with a = a = 5 Å and V = :5 ev. Determine numerically the starting and ending energies of the rst allowed band. From if < E < V, and cos kt = cos (a) cosh (b) cos kt = cos (a) cos (b) sin (a) sinh (b) ; (46) + sin (a) sin (b) ; (47) if E > V. Then, for < E < V the plot is Band edge energy occurs when cos kt is. It is found numerically that cos kt = when E = :459, cos kt = when E = :, and cos kt = when E = :7. Therefore, the band edges are at E = :7 ev and E = :459 ev. 5.. Use the euation of motion (5.4) to show that the period of Bloch oscillation for a one-dimensional crystal having lattice period a is = h eea : (48)

2 Use dk = ee; (49) dt and assume that is the time reuired for the electron to accelerate across the full Brillouin zone. Then, k Z = a Z = dk dt dt = k () = ee = ee = h eea : eedt (5) 5.4. Determine the probability current density (A/m) from (.87) for the Bloch wavefunction (x) = u (x) e ikx e i!t ; (5) where u is a time-independent periodic function having the period of the lattice, J (r; t) = i m ( (r; t) r (r; t) (r; t) r = i u (x) = u (x + a) : (5) (r; t)) (5) m u (x) e ikx e i!t r u (x) e ikx e i!t u (x) e ikx e i!t r u (x) e ikx e i!t = bx i u (x) e ikx e i!t d m dx u (x) eikx e i!t = bx i m u ik + u u u ik + u u = bx i m u ik = bx k m u = bx p m u : u (x) e ikx e i!t d dx u (x) e ikx e i!t 5.5. If an energy-wavevector relationship for a particle of mass m has the form determine the e ective mass. (Use m = E = m k ; (54) m k 5.6. If the energy-wavenumber relationship for an electron in some material is E = cos ( ; m A = m: (55) determine the e ective mass and the group velocity. (Use (5.9).) Describe the motion (velocity, direction, etc.) of an electron when a d.c. (constant) electric eld is applied to the material, such that the electric eld vector points right to left (e.g., an electron in free space would then accelerate towards the right). In particular, describe the motion as k varies from to. Assume that the electron does not scatter from anything.

3 @ m = = m cos ( = m cos k ; (56) v m cos ( = sin k; m For small positive k values the electron moves in the direction of the eld (to the left), and as k increases through positive value from k = to k = =, the electron increases its velocity and mass. At k = =, velocity is maximum and the e ective mass is in nite. As k changes from = to, the velocity decreases, as does the e ective mass. At k =, the velocity is zero. Then, as k increases further, the electron reverses direction, and it s velocity increases again, reaching a maximum at k = =, then decreasing till k = If the energy-wavenumber relationship for an electron in some material is E = E + A cos (ka) ; (58) determine the electron s position as a function of time. Ignore scattering. The solution of the euation of motion (ignoring scattering) is (5.6), Velocity is given as v g (k (t)) (k k (t) = k () + ee t: (59) = Aa sin = Aa sin (k (t) a) (6) (6) e Ea t and position can be determined from the relationship v = dx (t) =dt as Z t Z t Aa x (t) = v g (t) dt = sin e Ea t dt (6) = A cos eea E e t ; (6) and therefore the electron Bloch oscillates in time Consider an electron in a perfectly periodic lattice, wherein the energy-wavenumber relationship in the rst Brillouin zone is E = k 5m e ; where m e is the mass of an electron in free space. Write down the time-independent e ective mass Schrödinger s euation for one electron in the rst Brillouin zone, ignoring all interactions except between the electron and the lattice. De ne all terms in Schrödinger s euation. and so Schrödinger s euation m = d m dx k 5m e A = 5 m e; (64) (x) = E (x) (65) where m is the e ective mass, is the reduced Planck s constant, and E is the energy (V = since there is no potential energy term; potential energy is accounted for by the e ective mass). 4

4 5.9. Assume that a constant electric eld of strength E = kv/m is applied to a material at t =, and that no scattering occurs. (a) Solve the euation of motion (5.4) to determine the wavevector value at t = ; ; 7; and ns. (b) Assuming that the period of the lattice is a = :5 nm, determine in which Brillouin zone the wavevector is in at each time. If the wavevector lies outside the rst Brillouin zone, map it into an euivalent place in the rst zone. (a) The solution of the euation of motion is k (t) = ee t; (66) assuming k () =. Then k ( ns) = :5 9 m, k ( ns) = 4:56 9 m, k (7 ns) = :6 m, and k ( ns) = :5 m. (b) Brillouin zone boundaries occur at k n = n=a, where a is the period and n = ; ; ; :::. Thus, k = 6:8 9 m, k = :6 m, k = :89 m, etc.. Thus, k is in the rst zone for t = and ns, the second zone for t = 7 ns, and the third zone for t = nm. For higher zones, subtracting =a leads to the euivalent point in the rst zone. t (ns) k (t) (m ) zone euiv. point in st zone :5 9 st 4:56 9 st 7 :6 nd : :5 rd : Using the hydrogen model for ionization energy, determine the donor ionization energy for GaAs (m e = :67m e, " r = :). This compares well with measured values. :67m e e 4 E d = j e j 8 (:) = 5: mev. (67) " h 5.. Determine the maximum kinetic energy that can be observed for emitted electrons when photons having = nm are incident on a metal surface with work function 5 ev. E =! = = 5:47 ev c c = 9 = 8:568 9 J (68) So, the maximum kinetic energy is E e = :47 ev. (69) 5.. Photons are incident on silver, which has a work function e = 4:8 ev. The emitted electrons have a maximum velocity of 9 5 m/s. What is the wavelength of the incident light? E KE = m ev = j e j m e 9 5 = : ev (7) E =! = e + E K = 4:8 + : = 7: ev, (7) = 74:67 nm. (7) 7: j e j 5

5 5.. In the band theory of solids, there are an in nite number of bands. If, at T = K, the uppermost band to contain electrons is partially lled, and the gap between that band and the next lowest band is :8 ev, is the material a metal, an insulator, or a semiconductor? Metal 5.4. In the band theory of solids, if, at T = K, the uppermost band to have electrons is completely lled, and the gap between that band and the next lowest band is 8 ev, is the material a metal, an insulator, or a semiconductor? What if the gap is :8 ev. Insulator. What if the gap is :8 ev. Semiconductor 5.5. Describe in what sense an insulator with a nite band gap cannot be a perfect insulator. As long as the band gap is nite, an electron can be elevated to the conduction band, resulting in conduction Draw relatively complete energy band diagrams (in both real-space and momentum space) for a p-type indirect bandgap semiconductor For an intrinsic direct bandgap semiconductor having E g = :7 ev, determine the reuired wavelength of a photon that could elevate an electron from the top of the valance band to the bottom of the conduction band. Draw the resulting transition on both types of energy band diagrams (i.e., energyposition and energy-wavenumber diagrams).! = E g = :7 ev, (7) k = =! c! = c! = c = 7: nm. E g = 5.8. Determine the reuired phonon energy and wavenumber to elevate an electron from the top of the valance band to the bottom of the conduction band in an indirect bandgap semiconductor. Assume that E g = : ev, the photon s energy is E pt = :9 ev, and that the top of the valance band occurs at k =, whereas the bottom of the conduction band occurs at k = k a. E g E pt = E pn = : ev (74) k pn = k a : 5.9. Calculate the wavelength and energy of the following transitions of an electron in a hydrogen atom. Assuming that energy is released as a photon, using Table, on p. 4 classify the emitted light (e.g., X-ray, IR, etc.). (a) n =! n = From and so (b) n = 5! n = 4 E n = :6 n ; (75) E = :6 = : ev, (76) = :6 nm, between visible and UV : j e j E = :6 5 4 = :6 ev, (77) = 454:5 nm, far-infrared :6 j e j 6

6 (c) n =! n = 9 (d) n = 8! n = (e) n =! n = (f) n =! n = E = :6 9 = :9 ev, (78) :9 j e j = :89 5 m, microwave E = :6 8 = :88 ev, (79) = 89: nm, visible, violet :88 j e j E = :6 = :5 ev, (8) = 9:8 nm, between visible and UV :5 j e j E = :6 = :6 ev, (8) = 9: nm, between visible and UV :6 j e j 5.. Excitons were introduced in Section to account for the fact that sometimes when an electron is elevated from the valance band to the conduction band, the resulting electron and hole can be bound together by their mutual Coulomb attraction. Excitonic energy levels are located just below the band gap, since the usual energy to create a free electron and hole, E g, is lessened by the binding energy of the exciton. Thus, transitions can occur at E = E g m r m e " :6 ev (8) r where E g is in electron volts. (a) For GaAs, determine the reuired photon energy to create an exciton. For m r use the average of the heavy and light hole masses. Using m r = :5m e, " r = :, and E g = :4 ev, we nd that E = E g = :4 m r m e " :6 ev r (8) :5 :6 ev (:) (84) = :46 ev. (85) (b) The application of a d.c. electric eld tends to separate the electron and the hole. Using Coulomb s law, show that the magnitude of the electric eld between the electron and the hole is m jej = r R Y " : (86) r j e j a Really, the uantity :6 should be replaced by :6=n, where n is the energy level of the exciton. Here we consider the lowest level exciton (n = ), which is dominant. m e 7

7 The electric eld due to a charge in a medium characterized by " r is E = br = br jej ; (87) 4" r " r where br is a unit vector that points radially outward from the charge, and r is the radial distance away from the charge. Making the substitutions = e and r = a ex leads to j e j m jej = 4" r " a = r j e j x m e 4" r" a (88) m = r m e j e j m m e 4" h " = r RY ra m e j e j " (89) ra m = r R Y " : (9) r j e j a m e (c) For GaAs, determine jej from (5.79). Determine the magnitude of an electric eld that would break apart the exciton. m jej = r R Y m e " (9) r j e j a = (:5) :6 j e j (:) j e j :5 9 = 5:5 5 V/m. (9) An applied electric eld with a magnitude greater than jej can break apart the exciton. 5.. The E k relationship for graphene is given by (5.6). The Fermi energy for graphene is E F =, and the rst Brillouin zone forms a hexagon (as shown in Fig. 5.5), the six corners of which correspond to E = E F =. The six corners of the rst Brillouin zone at located at and k x = p a ; k x = ; (a) Verify that at these points, E = E F =. v u E (k x ; k y ) = t + 4 cos E ; 4 s = a + 4 cos 4 a = s + 4 cos v u E (k x ; k y ) = t + 4 cos v E p ; u = t a a + 4 cos k y = a ; (9) k y = 4 a : (94) p! kx a a cos + 4 cos a ; + 4 cos 4 a a + 4 cos = a p! kx a a cos + 4 cos a ; p! a p cos a a = r + 4 cos () cos + 4 cos = a + 4 cos a a 8

8 (b) At the six corners of the rst Brillouin zone, jkj = 4=a. Make a two-dimensional plot of the E k relationship for k x ; k y extending a bit past jkj. Verify that the bonding and antibonding bands touch at the six points of the rst Brillouin zone hexagon, showing that graphene is a semi-metal (sometimes called a zero bandgap semiconductor). Also make a one-dimensional plot of E (; k y ) for jkj k y jkj, showing that the bands touch at E = at k y = 4=a. Using (5.6), since a = p (:4 nm) = :46 nm, jkj = 4=a = 7 nm. Thus, E, me in two dimensions (the bands actually touch at the corners, although in the plot a small gap is shown due to using a coarse wavenumber grid). In one-dimension, for we have a E (; k y ) = s + 4 cos s :46 = :5 + 4 cos + 4 cos a + 4 cos :46 E(, ) E(, ) where the vertical scale is in ev and the horizontal scale is in nm. 9

9 5.. What is the radius of a (9; ) carbon nanotube? Repeat for a (; ) nanotube. Consider a (n; ) zigzag carbon nanotube that has radius :5 nm. What is the value of the index n? The CN s radius is where b = :4 nm. Therefore, r = p bp n + nm + m ; (95) For (n; ), p r (9;) = p r (;) = :4 9 p 9 = :747 nm :4 9 p + () () + = :678 nm. n = p a :5 9 = p b (:4 9 ) = 9: 5.. Since carbon nanotubes are only periodic along their axis, the transverse wavenumber becomes uantized by the nite circumference of the tube. Derive (5.66) and (5.67) by enforcing the condition that an integer number of transverse wavelengths must t around the tube (k? = =? ). For the armchair tube (m = n), tube radius is r = nb=. Thus,? = r = nb = nb k? = k x; = = ; = ; ; :::; n:? nb For zigzag tubes, (n = ), r = p nb=, and p nb? = r = = p nb k? = k y; = =? n p ; = ; ; :::; n: b The limit n on comes from the fact that k x;n = 4=b, and k y;n = 4= p b, and beyond these values one is outside of the rst Brillouin zone of graphene Using (5.68) and (5.69), plot the dispersion curves for the rst eight bonding and antibonding bands in a (5; 5), (9; ), and (; ) carbon nanotube. Let the axial wavenumber vary from k = to k = =a ac for the armchair tube, and from k = to k = =a zz for the zigzag tube. Comment on whether each tube is metallic of semiconducting, and identify the band (i.e., the value) that is most important. If the tube is semiconducting, determine the approximate band gap. For the armchair tube (5; 5), E ac (k y ) = s + 4 cos = s + 4 cos < k y a ac <, = ; ; :::; n, and so cos n cos 5 a a + 4 cos a + 4 cos a

10 .87 E(, E(, E(, E(, E(, E(, E( 4, E( 4, E( 5, E( 5, k π.46 (vertical scale is E= ). The = 5 band (note that n = 5!) is the most important, since these bands cross in the rst Brillouin zone (and hence, there is no band gap). The crossing point is = of the way to the zone boundary, and so k F = =a, such that the Fermi wavelength is F = a = :74 nm. For zigzag tubes, v u E zz (k x ) = t + 4 cos < k y a zz <, = ; ; :::; n. For the (9; ) tube, p! kx a cos n + 4 cos ; n.879 E(, E(, E(, E(, E(, E(, E( 4, E( 4, E( 5, E( 5, E( 6, E( 6, E( 7, E( 7, E( 8, E( 8, k π.46 where the = 6 bands cross at k =, and, hence, this tube is metallic. For the (; ) tube,

11 .9 E(, E(, E(, E(, E(, E(, E( 4, E( 4, E( 5, E( 5, E( 6, E( 6, E( 7, E( 7, E( 8, E( 8, k π.46 no bands cross, hence, the (; ) tube is a semiconductor. The = 7 bands come the closest to each other (at k = ), and so the band gaps is E (k = ) for = 7, which is approximately :88 ev using = :5 ev. It can be shown (See the book by Saito, Dresselhaus, and Dresselhaus, Reference [9] in Chapter 5) that E g = p a ; r which for the (; ) tube (r = :9 nm) leads to :9 ev. 6 Problems Chapter 6: Tunnel Junctions and Applications of Tunneling 6.. Plot the tunneling probability versus electron energy for an electron impinging on a rectangular potential barrier (Fig. 6., p. 85) of height ev and width nm. Assume that the energy of the incident electron ranges from ev to ev. T = 4E (E V ) sin (k a) + 4E (E V ) ; k = m e (E V ) (96) V T = j e j sin m e(ej ej 4E j e j (E j e j j e j) j ej) ( 9 ) + 4E j e j (E j e j j e j) 6.. Plot the tunneling probability verses barrier width for a ev electron impinging on a rectangular potential barrier (Fig. 6., p. 85) of height ev. Assume that the barrier width varies from nm to nm. T = 4E (E V ) sin (k a) + 4E (E V ) ; k = m e (E V ) (97) V T = j e j sin m e(()j ej 4 () j e j (() j e j j e j) j ej) (a 9 ) + 4 () j e j (() j e j j e j)