f5ch42 - CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS...

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Unformatted text preview: CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 1129 CHAPTER 42 Answer to Checkpoint Questions 1. 2. 3. 4. 5. (a) larger; (b) same Cleveland: metal; Troy: none; Seattle: semiconductor (a), (b), and (c) (b) (b) Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 4 8 (b) and (c) (c) (a) anywhere in the lattice; (b) in any silicon-silicon bond; (c) in a silicon ion core, at a lattice site (a) and (b) (b) and (d) 4s2 and 4p2 +4e (a) arsenic, antimony; (b) gallium, indium; (c) tin none (a) (a) right to left; (b) back bias zero (a), (b), and (c) 1130 CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 16. (b) Solutions to Exercises & Problems The number of atoms per unit volume is given by n = d=M , where d is the mass density of copper and M is the mass of a single copper atom. Since each atom contributes one conduction electron, n is also the number of conduction electrons per unit volume. Since the molar mass of copper is A = 63:5 g/mol, M = A=NA = (63:5 g/mol)=(6:022 1023 mol 1 ) = 1:054 10 22 g. Thus 3 )(106 3 3 n = (8:96 g/cm 10 cmg =m ) == 8:43 1028 m 3 : 22 1:054 1E 2E Solve p from p = nkT : p = nkT = (8:43 1028 m 3 )(1:38 10 3E 23 J/K)(300 K) = 3:49 108 Pa = 3490 atm : Use your calculator to verify that 3 2=3 = 0:1212153 0:121 : p 16 2 4E This excercise is similar to 1E. The number of atoms per unit volume is given by n = d=M , where d is the mass density of gold and M is the mass of a single gold atom. Since each atom contributes one conduction electron, n is also the number of conduction electrons per unit volume. Since the molar mass of gold is A = 197 g/mol, M = A=NA = (197 g/mol)=(6:022 1023 mol 1 ) = 3:271 10 22 g. Thus 3 )(106 3 3 n = (19:3 g/cm 10 cmg =m ) = 5:90 1028 m 3 : 22 3:271 CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 1131 The temperature coecient of resistivity is de ned as 1 d (T ) = dT (see Eq. 42-8). Thus (a) For copper 5E d = : dT 11 m/K : d = = (2 10 8 m)(4 10 3 K 1 ) = +8 10 Cu dT (b) For silicon d = = (3 103 m)( 70 10 3 K 1 ) = 2:1 102 m/K : Si dT 6E From Eq. 42-6 eV EF = 0:121(chc) n2=3 = 0:121(1240MeVnm) (8:43 1028 m 3 )2=3 = 7:0 eV : 2 m 0:511 2 2 e 2 Use EF = 1 me vF to solve for vF : 2 7E vF = r 2EF = m e r 2EF c = m c2 e r 2(7:0 eV) (3:00 105 km/s) = 1:6 103 km/s : 5:11 105 eV 8E (a) Eq. 42-2 gives for the density of states associated with the conduction electrons of a metal. This can be written n(E ) = CE 1=2 ; 8 2m3=2 E 1=2 n(E ) = h3 e p 1132 CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 8 2m3=2 = 8 2(me c2 )3=2 C = h3 e (hc)3 p 8 2(5:11 105 eV)3=2 = 6:81 1027 m 3 (eV) 3=2 : = (1240 10 9 eV m)3 (b) If E = 5:00 eV then where p p n(E ) = 9E h 6:81 1027 m 3 (eV) 3=2 i (5:00 eV)1=2 = 1:52 1028 (eV) 1 m 3 : (a) At absolute temperature T = 0, the probability is zero that any state with energy above the Fermi energy is occupied. (b) The probability that a state with energy E is occupied at temperature T is given by P (E ) = e(E EF1=kT + 1 ; ) where k is the Boltzmann constant and EF is the Fermi energy. Now E EF = 0:062 eV and (E EF )=kT = (0:062 eV)=(8:62 10 5 eV/K)(320 K) = 2:248, so 1 P (E ) = e2:248 + 1 = 0:0956 : 10E Use the result of 8E: n(E ) = CE 1=2 = h 6:81 1027 m 3 (eV) 2=3 i (8:0 eV)1=2 = 1:9 1028 m 3 eV 1 : This is consistent with Fig.42-5. 11E According to Eq. 42-6 the Fermi energy is given by 3 2=3 h2 n2=3 ; EF = p me 16 2 where n is the number of conduction electrons per unit volume, me is the mass of an electron, and h is the Planck constant. This can be written EF = An2=3 , where 3 2=3 h2 = 3 2=3 (6:626 10 34 J s)2 = 5:842 10 p A= p me 9:109 10 31 kg 16 2 16 2 38 J2 s2 /kg : CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 1133 Since 1 J = 1 kgm2 /s2 , the units of A can be taken to be m2J. Divide by 1:60210 to obtain A = 3:65 10 19 m2 eV. 12E 19 J/eV The number density of conduction electrons in gold is 3 022 23 n = (19:3 g/cm )(6:g/mol)10 =mol) = 5:90 1022 cm 3 : (197 Thus 121( eV EF = 0:(m chc) n2=3 = 0:121(1240MeVnm) (5:90 1022 cm 3 )2=3 = 5:53 eV : 2) 0:511 2 2 e Let E1 = 63 meV + EF and E2 = 63 meV + EF . Then according to Eq. 42-3 13E P1 = e(E1 EF1)=kT + 1 = ex 1 1 ; + where x = (E1 EF )=kT . Solve for ex : 1 ex = P Thus 1 1 1 = 0:090 1 = 91 : 9 P2 = e(E2 EF1)=kT + 1 = e (E1 E1)=kT + 1 = e x1+ 1 == (91=9)1 1 + 1 = 0:91 ; F where we used E2 EF = 63 meV = EF E1 = (E1 EF ). 14P Solve E from Eq. 42-3: E =EF + kT ln(P 1 1) 5 eV/K)(1000 K) ln 1 = 7:0 eV + (8:62 10 0:90 1 = 6:8 eV : (b) n(E ) = CE 1=2 = (6:81 1027 m 3 eV 3=2 )(6:8 eV)1=2 = 1:77 1028 m 3 (eV) 1 : (c) n0 (E ) = P (E )n(E ) = (0:90)(1:77 1028 m 3 (eV) 1 ) = 1:6 1028 m 3 (eV) 1 : 1134 CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 15P (a) The volume per cubic meter of sodium occupied by the sodium ions is (971 kg)(6:022 1023 = mol)(4=3)(98 10 VNa = (23 g/mol) 12 m)3 = 0:100 m3 ; so the fraction available for conduction electron is 1 (VNa =1:00 m3 ) = 1 0:100 = 0:900: (b) For copper 23 = mol)(4= VCu = (8960 kg)(6:022 10 63:5 g/mol 3)(135 10 12 m)3 = 0:876 m 3 ; so the fraction is 1 (VCu =1:00 m3 ) = 1 0:876 = 0:124: (c) Sodium, because the electrons occupy a greater portion of the space available. According to Eq. 42-3 1 P (EF + E ) = e(EF +E 1 F )=kT + 1 = eE=kT + 1 = ex 1 1 ; E + where x = E=kT . Also 1 P (EF E ) = e(EF E 1 F )=kT + 1 = e E=kT + 1 = e x1+ 1 : E Thus x 1 x+1 P (EF + E ) + P (EF E ) = ex 1 1 + e x1+ 1 = (ee + + + eex + 1) = 1 : x 1)( + 16P A special case of this general result can be found in 13E, where E = 63 meV and P (EF + 63 meV) + P (EF 63 meV) = 0:090 + 0:91 = 1:0. De ne x = E=kT , where E = 1:00 eV. Then the quantum result for P (E ) is given by P (E ) = 1=(ex + 1), while the classical result is P (E ) = e x . The percentage dierence is then 1=(ex + 1) e x = 1=(ex + 1) = e x = e E=kT : 17P Solve for T : T = kE : ln CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 1135 (a) Now = 1:0% so T = kE = (8:62 10 15:00 eV 1:0%) = 2:5 103 K : ln eV/K)(ln (b) Now = 10% so 3 T = kE = (8:62 10 1:500 eV ln eV/K)(ln 10%) = 5:0 10 K : The molar mass of carbon is m = 12:01115 g/mol and the mass of the Earth is Me = 5:98 1024 kg. Thus the number of carbon atoms in a diamond as massive as the Earth is N = (Me =m)NA , where NA is the Avogadro's number. Then from the result of Sample Problem 42-1 the probability in question is given by 18P Me N e Eg =kT P = Ne m A 5:98 1024 kg (6:02 1023 /mol)(3 10 = 12:01115 g/mol Eg =kT = 19P 93 ) = 9 10 43 : (a) Solve n from p = nkT : p atm)(1 105 Pa/atm) n = kT = (1:038 10 :023 J/K)(273 K) = 2:7 1025 m 3 : (1: (b) 3 d 103 n = mCu = (63:8:96 67 kg/m kg) = 8:43 1028 m 3 : 54)(1: 10 27 Cu (c) The ratio is (8:43 1028 m 3 )=(2:7 1025 m 3 ) = 3:1 103 : (d) Use dav = n1=3 . For case (a), dav = (2:7 1025 m 3 )1=3 = 3:3 nm; and for case (b) dav = (8:43 1028 m 3 ) 1=3 = 0:228 nm: Use n0 = n(E )P (E ) = CE 1=2 e(E EF )=kT + 1 1 . At E = 4:00 eV 6:78 1027 m 3 (eV) 3=2 (4:00 eV)1=2 n0 = e(4:00 eV 7:00 eV)=[8:6210 5 eV/K)(1000 K)] + 1 = 1:36 1028 m 3 (eV) 1 ; 20P 1136 CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 27 m 3 (eV) 3=2 )(6:75 eV)1=2 (6:78 n0 = e(6:75 eV710 eV)=[8:6210 5 eV/K)(1000 K)] + 1 :00 = 1:67 1028 m 3 (eV) 1 : Similarly at E = 7:00; 7:25 and 9:00 eV the corresponding values of n0 (E ) are 0:90; 0:10 and 0:00 1028 m 3 (eV) 1 , respectively. and at E = 6:75 eV (a) Use P (E ) = [e(E EF )=kT + 1] 1 . So for E = 4:4 eV. P (E ) = e(4:4 eV 5:5 eV) =[(8:62 1 5 eV/K)(273 K)] + 1 = 1:0 ; 10 for E = 5:4 eV 1 P (E ) = e(5:4 eV 5:5 eV)=[(8:6210 5 eV/K)(273 K)] + 1 = 0:99 ; for E = 5:5 eV 1 P (E ) = e(5:5 eV 5:5 eV)=[(8:6210 5 eV/K)(273 K)] + 1 = 0:50 ; for E = 5:6 eV 1 P (E ) = e(5:6 eV 5:5 eV)=[(8:6210 5 eV/K)(273 K)] + 1 = 0:014 ; and for E = 6:4 eV 1 P (E ) = e(6:4 eV 5:5 eV)=[(8:6210 5 eV/K)(273 K)] + 1 = 2:5 10 17 : Solve the expression for P (E ) for T : E 5 eV T = k ln(P EF 1) = (8:62 105:6 eV 5:ln(0:16 1 1) = 7:0 102 K : 1 5 eV/K) 22P 21P Let N be the number of atoms per unit volume and n be the number of free electrons per unit volume. Then the number of free electrons per atom is n=N . Use the result of 11E to nd n: EF = An2=3 , where A = 3:65 10 19 m2 eV. Thus =2 3 11:6 eV EF 3=2 = = 1:79 1029 m 3 : n= A 3:65 10 19 m2 eV CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 1137 If M is the mass of a single aluminum atom and d is the mass density of aluminum then N = d=M . Now M = (27:0 g/mol)=(6:022 1023 mol 1 ) = 4:48 10 23 g, so N = (2:70 g/cm3 )=(4:48 10 23 g) = 6:03 1022 cm 3 = 6:03 1028 m 3 . Thus the number of free electrons per atom is n = 1:79 1029 m 3 = 2:97 : N 6:03 1028 m 3 The probability Ph that a state is occupied by a hole is the same as the probability the state is unoccupied by an electron. Since the total probability that a state is either occupied or unoccupied is 1, we have Ph + P = 1. Thus 1 e(E EF )=kT Ph = 1 e(E EF1=kT + 1 = 1 + e(E EF )=kT = e (E EF )=kT + 1 : ) 24P 23P (a) (b) 3 :022 22 n = 2Zn = 2(7:133 g/cm :)(6g/mol) 10 =mol) = 1:31 1029 m 3 : m (65 37 Zn 34 J s)2 (1 31 29 m 3 2=3 12 2 EF = 0:m h n2=3 = 0:121(6:63 1031 kg)(1:60 : 10 10 J/eV)) = 9:43 eV : 19 (9:11 10 e (c) s vF = (d) 2EF c2 = m c2 e r 2(9:43 eV)(3:00 108 m/s)2 = 1:82 106 m/s : 0:511 MeV 34 6 = h = mhv = (9:11 10:63 10 :82Js 106 m/s) = 0:40 nm : 31 kg)(1 p e F 25P (a) (b) 3 22 n = mAg = (10:49 g/cm )(6:022 10 =mol) = 5:86 1028 m 3 : 107:87 g/mol Ag 2 34 J s)2 (5 86 28 m 3 2=3 EF = 0:121h n2=3 = 0:121(6:63 1031 kg)(1:60 : 10 10 J/eV)) = 5:52 eV : 19 m (9:11 10 e 1138 CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS (c) s vF = (d) 2EF c2 = m c2 e r 2(5:25 eV)(3:00 108 m/s)2 = 1:39 106 m/s : 0:511 MeV 34 6 = h = mhv = (9:11 10:63 10 :39Js 106 m/s) = 0:522 nm : 31 kg)(1 p e F Let the energy of the state in question be an amount E above the Fermi energy EF . Then Eq. 42-3 gives the occupancy probability of the state as 1 P = e(EF +E 1 F )=kT + 1 = eE=kT + 1 : E Solve for E to obtain 1 E = kT ln P 26P 1 = (1:38 10 23 J/K)(300 K) ln 1 1 = 9:1 10 0:10 21 J ; which is eqivalent to 5:7 10 2 eV = 57 meV. (a) Substitute the expression for EF , Eq. 42-6, into Eq. 42-2: 27P m3=2 1 m3=2 N (EF ) = 8 2h3 e EF=2 = 8 2h3 e 1=3 2=3 1=3 = (4)(3 )( )me n p " p # " #1=2 3 2=3 h2 n2=3 p me 16 2 h2 (4)(31=3 )(2=3 )(me c2 ) n1=3 = (hc)2 (4)(31=3 )(2=3 )(5:11 105 eV) n1=3 = (1240 eV nm)2 (10 9 m/nm)2 = 4:11 1018 m 2 (eV) 1 n1=3 : (b) For copper n = 8:43 1028 m which agrees well with Fig. 42-5. 3 (see 1E). Thus N (EF ) = [4:11 1018 m 2 (eV) 1 ](8:43 1028 m 3 )1=3 = 1:80 1028 m 3 (eV) 1 ; CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 1139 (a) For simplicity, de ne x = (E EF )=kT . Then P = ex 1 1 + and dP = dP dx = 1 dP : dE dx dE kT dx At E = EF we have x = 0, so dP 1 1 d = kT dP = kT dx ex 1 1 dE E=EF dx x=0 + x=0 x 1 1 = kT (ex e 1)2 = 4kT : + x=0 (b) According to part (a) above the slope of the tangent line to the P (E ) curve is given by s = dP=dE = 1=4kT . Since the line passes through these two points: ( 1 ; EF ) and 2 (0; E ), 1 0 1 s = 4kT = E2 E : 28P Solve for E : E = EF + 2kT . 29P F The average energy of the conduction electrons is given by Z 1 = 1 E En(E )P (E ) dE ; n 0 where n is the number of free electrons per unit volume, n(E ) is the density of states, and P (E ) is the probability function. The density of states is proportional to E 1=2 , so we may write n(E ) = CE 1=2 , where C is a constant of proportionality. The probability function is unity for energies below the Fermi energy and 0 for energies above. Thus C Z EF E 3=2 dE = 2C E 5=2 : E= n 5n F 0 Now n= Z 1 Substitute this expression into the formula for E to obtain ! 0 n(E )P (E ) dE = C Z E F 0 C 3 E 1=2 dE = 23 EF=2 : 3 3 C 5 E = 25 EF=2 3=2 = 5 EF : 2CEF 1140CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 30P 3 Ktotal = N E = (8:43 1028 m 3 )(1:0 10 6 m3 ) 5 (7:0 eV)(1:6 10 = 5:7 104 J = 57 kJ : 31P 19 J/eV) (a) The energy released would be 23 3 E = N E = (3:1 g)(6::02 10 =mol) 5 (7:0 eV)(1:6 10 63 54 g/mol = 2:0 104 J = 20 kJ : 19 J/eV) (a) At absolute zero T = 0, so f / T = 0. (b) At T = 300 K 5 3kT f = 2E = 3(8:62 10 :0eV/K)(300 K) = 5:5 10 3 : 2(7 eV) F 32P (c) At T = 1000 K 5 eV/K)(1000 K) 3kT = 1:9 10 2 : f = 2E = 3(8:62 10 :0 eV) 2(7 F The fraction f of electrons with energies greater than the Fermi energy is given by 3kT f = 2E ; F where T is the temperature on the Kelvin scale, k is the Boltzmann constant, and EF is the Fermi energy. Solve for T : 2(0: 7 eV)) T = 2fEF = 3(8:62013)(4:5 eV/K) = 473 K : 3k 10 34P 33P 5 eV/K)(1000 K) 3kT f = 2E = 3(8:62 10 :5 eV) = 2:9 10 2 : 2(5 F CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 1141 (a) The regular tetrahedron strcuture for the silicon lattice is illustrated in the gure to the right. The center of the tetrahedron is at point O, and the angle to be solved is . Consider point C , the normal projection of point A on the base plane. By symmetry point C is at the center of the equilateral triangle that forms the base of the regular tetrahedron. If the side length of the tetrahedron is a, then p jBC j = a= 3. Now consider the right-angled triangle ABC . The angle can be found from 1 sin = jjBC jj = a=a 3 = p ; AB 3 which gives = 35:26 . Thus in the triangle AOB 35P A a O p C a B = 180 2 = 180 2(35:26 ) = 109:5 : (b) In the triangle AOB the bond length b can be obtained as a 388 pm b = jAOj = jBOj = 2 cos = 2 cos 35:26 = 235 pm : (a) 36P Ga As 1142 CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS In the sketch shown in the last page the gallium atoms are represented by the grey spheres while the arsenic atoms are represented by the black ones. (b) Each gallium atom contributes three electrons to the Ga-As bond, while each arsenic atom contributes ve electrons. Therefore the net charge of the gallium ion core is +3e while that of the arsenic ion core is +5e. (c) There are two electrons per bond (see the sketch). To excite an electron across the band gap the energy of the photon of wavelength must be at least equal to Eg , the band gap: hc= Eg . This gives the maximum wavelength 37P hc eV max = E = 1240:5 eVnm = 2:3 102 nm : 5 g (b) The wavelength is shorter than that of any visible light. In fact it is in the ultraviolet (UV) range. 38P (a) At the bottom of the conduction band E = 0:67 eV. Also EF = 0:67 eV=2 = 0:335 eV. So the probability that the bottom of the conduction band is occupied is P (E ) = e(E EF1=kT + 1 = e(0:67 eV 0:335 eV)=[8:621 10 ) = 1:5 10 6 : 5 eV/K)(290 K)] + 1 (b) At the top of the valencen band E = 0, so the probability that the state is unoccupied is given by 1 P (E ) = 1 1 = (E EF )=kT +1 e +1 1 = (0 0:335 eV)=[(8:6210 5 eV/K)(290 K)] e +1 6: = 1:5 10 e(E EF )=kT 1 39P (a) The number of electrons in the valence band is Nev = Nv P (Ev ) = e(Ev EN)v=kT + 1 : F CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 1143 Since there are a total of Nv states in the valence band the number of holes in the valence band is 1 Nhv = Nv Nev = Nv 1 e(Ev EF )=kT + 1 N = (Ev EFv)=kT : e +1 Now, the number of electrons in the conduction band is so from Nev = Nhc we get Nec = Nc P (Ec ) = e(Ec EN)c=kT + 1 ; F (Ev EF )=kT e Nv Nc = (Ec EF )=KT : +1 e +1 (b) In this case e(Ec EF )=kT 1 and e (Ev EF )=kT 1, so from the result of part (a) or e(Ev Ec +2EF )=kT Nv =Nc . Solve for EF : 1 (E + E ) + 1 kT ln Nv : EF 2 c v 2 Nc 40P e(Ec EF )=kT Nc e (Ev EF )=KT Nv ; (a) n-type, since each phosphorous atom has one more valence electron than a silicon atom. (b) The added charge carrier density is nP = 10 7 nSi = 10 7 (5 1028 m 3 ) = 5 1021 m 3 : (c) The ratio is (5 1021 m 3 )=[2(5 1015 m 3 )] = 5 105 : Here the factor of 2 in the denominator re ects the contribution to the charge carrier density from both the electrons in the conduction band and the holes in the valence band. 41P The number of phosphorus atoms needed is 22 3 0 NP = (1:(2g)(10 m ) ) = 4:29 1016 ; :33 g/cm3 so the mass of phosphorus is MP = NP mP = (4:29 1016 )(31)(1:67 10 27 kg) = 0:22 g : 1144 CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS (a) The probability that a state with energy E is occupied is given by P (E ) = e(E EF1=kT + 1 : ) If energies are measured from the top of the valence band, then the energy associated with a state at the bottom of the conduction band is E = 1:11 eV. Furthermore, kT = (8:62 10 5 eV/K)(300 K) = 0:02586 eV. For pure silicon EF = 0:555 eV and (E EF )=kT = (0:555 eV)=(0:02586 eV) = 21:46. Thus 1 P (E ) = e21:46 + 1 = 4:79 10 10 : For the doped semiconductor (E EF )=kT = (0:11 eV)=(0:02586 eV) = 4:254 and 1 P (E ) = e4:254 + 1 = 1:40 10 2 : (b) The energy of the donor state, relative to the top of the valence band, is 1:11 eV 0:15 eV = 0:96 eV. The Fermi energy is 1:11 eV 0:11 eV = 1:00 eV. Hence (E EF )=kT = (0:96 eV 1:00 eV)=(0:2586 eV) = 1:547 and 1 P (E ) = e 1:547 + 1 = 0:824 : 43P 42P (a) Measured from the top of the valence band, the energy of the donor state is E = 1:11 eV 0:11 eV = 1:0 eV. Solve EF from Eq. 42-3: EF = E kT ln(P 1 1) = 1:0 eV (8:62 10 5 eV/K)(300 K) ln[(5:00 10 5 ) = 0:744 eV : (b) Now E = 1:11 eV so 1 P (E ) = e(E EF1=kT + 1 = e(1:11 eV 0:744 eV)=[(8:6210 ) = 7:13 10 7 : 44P 1 1] 5 eV/K)(300 K)] + 1 To nd the maximum number of electrons that can be excited across the gap, assume that the energy received by each electron is exactly the dierence in energy between the top of CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 1145 Since each electron that jumps the gap leaves a hole behind, this is also the maximum number of electron-hole pairs that can be created. 45P the conduction band and the bottom of the valence band (1:1 eV). Then the number that can be excited is 3 N = 6621110 eV = 6:0 105 : : eV (a) 600 400 I (nA) 200 0 200 0.15 0.1 0.05 0 0.05 0.1 0.15 V (Volts) (b) The ratio is ijv=+0:50 V = i0 [e+0:50 eV=[(8:6210 ijv= 0:50 V i0 [e 0:50 eV=[(8:6210 46P 5 eV/K)(300 K)] 5 eV/K)(300 K)] 1] = 2:5 108 : 1] The valence band is essentially lled and the conduction band is essentially empty. If an electron in the valence band is to absorb a photon, the energy it receives must be sucient to excite it across the band gap. Photons with energies less than the gap width are not absorbed and the semiconductor is transparent to this radiation. Photons with energies greater than the gap width are absorbed and the semiconductor is opaque to this radiation. Thus the width of the band gap is the same as the energy of a photon associated with a wavelength of 295 nm. Thus the width of the gap is Eg = 1240 eV nm = 1240 eV nm = 4:20 eV : 295 nm 1146 CHAPTER 42 CONDUCTION OF ELECTRICITY IN SOLIDS 47P Since The light will be absorbed by the KCI crytal so the crystal is opaque to this light. Denote the maximum dimension (side length) of each transistor as `max , the size of the chip as A, and the number of transistors on the chip as N . Then A = N`2 . Solve for max `max : 2 2 A in.)(2 54 `max = N = (1:0 in. 0:875 3:5 :106 10 m/in.) = 1:3 10 5 m = 13 m : r r 48P Ephoton = hc = 1240 nm eV = 8:86 eV > 7:6 eV ; 140 nm (a) According to Chapter 26 C = 0 A=d. In our case = 4:5, A = (0:50 m)2 , and d = 0:20 m, so 49P 0 A = (4:5)(8:85 10 12 C2=N m2 )(0:50 m)2 = 5:0 10 C= d 0:20 m 17 F : (b) Let the number of elementary charges in question be N . Then the total amount of charges that appear in the gate is q = Ne. Thus q = Ne = CV , which gives : N = CV = (5:01 10 10F)(1C0 V) = 3:1 102 : 19 e :60 17 ...
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