半导体物理与器件第四版课后习题答案3

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半导体物理与器件第四版课后习题集答案解析4

Chapter 44.1⎪⎪⎭⎫ ⎝⎛-=kTE N N n gc i exp 2υ ⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=kT E T N N g O cO exp 3003υwhere cO N and O N υ are the values at 300 K.(a) SiliconT (K) kT (eV)i n (cm 3-) 200 400 600 01727.0 03453.0 0518.041068.7⨯ 121038.2⨯ 141074.9⨯(b) Germanium (c) GaAsT (K) i n (cm 3-) i n (cm 3-)200 400 600101016.2⨯ 141060.8⨯ 161082.3⨯38.1 91028.3⨯ 121072.5⨯_______________________________________ 4.2Plot_______________________________________ 4.3(a) ⎪⎪⎭⎫⎝⎛-=kT E N N n g c i exp 2υ ()()()319192113001004.1108.2105⎪⎭⎫⎝⎛⨯⨯=⨯T()()⎥⎦⎤⎢⎣⎡-⨯3000259.012.1exp T()3382330010912.2105.2⎪⎭⎫⎝⎛⨯=⨯T()()()()⎥⎦⎤⎢⎣⎡-⨯T 0259.030012.1expBy trial and error, 5.367≅T K(b)()252122105.2105⨯=⨯=i n()()()()()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=T T 0259.030012.1exp 30010912.2338By trial and error, 5.417≅T K_______________________________________ 4.4At 200=T K, ()⎪⎭⎫⎝⎛=3002000259.0kT017267.0=eVAt 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()()()17222102210025.31040.11070.7200400⨯=⨯⨯=ii nn⎥⎦⎤⎢⎣⎡-⎥⎦⎤⎢⎣⎡-⨯⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛=017267.0exp 034533.0exp 30020030040033g g E E⎥⎦⎤⎢⎣⎡-=034533.0017267.0exp 8g g E E()[]9578.289139.57exp 810025.317-=⨯g Eor()1714.38810025.3ln 9561.2817=⎪⎪⎭⎫⎝⎛⨯=g E or 318.1=g E eVNow ()32103004001070.7⎪⎭⎫⎝⎛=⨯o co N N υ⎪⎭⎫ ⎝⎛-⨯034533.0318.1exp ()()172110658.2370.210929.5-⨯=⨯o co N N υ so 371041.9⨯=o co N N υcm 6-_______________________________________ 4.5()()⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=kT kT kT A n B n i i 20.0exp 90.0exp 10.1exp For 200=T K, 017267.0=kT eV For 300=T K, 0259.0=kT eV For 400=T K, 034533.0=kT eV (a) For 200=T K, ()()610325.9017267.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (b) For 300=T K, ()()41043.40259.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (c) For 400=T K, ()()31005.3034533.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i _______________________________________ 4.6(a) ()⎥⎦⎤⎢⎣⎡---∝kT E E E E f g F c F c exp()⎥⎦⎤⎢⎣⎡---∝kT E E E E c c exp()⎥⎦⎤⎢⎣⎡--⨯kT E E F c exp Let x E E c =-Then ⎪⎭⎫⎝⎛-∝kT x x f g F c expTo find the maximum value: ()⎪⎭⎫⎝⎛-∝-kT x x dx f g d F c exp 212/10exp 12/1=⎪⎭⎫ ⎝⎛-⋅-kT x x kT which yields2212/12/1kTx kT x x =⇒= The maximum value occurs at2kTE E c +=(b)()()⎥⎦⎤⎢⎣⎡---∝-kT E E E E f g F F exp 1υυ()⎥⎦⎤⎢⎣⎡---∝kT E E E E υυexp()⎥⎦⎤⎢⎣⎡--⨯kT E E F υexp Let x E E =-υThen ()⎪⎭⎫ ⎝⎛-∝-kT x x f g F exp 1υTo find the maximum value()[]0exp 1=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-∝-kT x x dx d dx f g d F υ Same as part (a). Maximum occurs at2kTx =or2kTE E -=υ_______________________________________ 4.7()()()()⎥⎦⎤⎢⎣⎡---⎥⎦⎤⎢⎣⎡---=kT E E E E kT E E E E E n E n c c c c 221121exp expwherekT E E c 41+= and 22kTE E c += Then()()()⎥⎦⎤⎢⎣⎡--=kT E E kT kTE n E n 2121exp 24()5.3exp 22214exp 22-=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--=or()()0854.021=E n E n_______________________________________ 4.8Plot_______________________________________ 4.9Plot_______________________________________ 4.10⎪⎪⎭⎫ ⎝⎛=-**ln 43n p midgap Fi m m kT E E Silicon: o p m m 56.0*=, o n m m 08.1*=0128.0-=-midgap Fi E E eV Germanium: o pm m 37.0*=,o n m m 55.0*=0077.0-=-midgap Fi E E eV Gallium Arsenide: o p m m 48.0*=,o n m m 067.0*= 0382.0+=-midgap Fi E E eV _______________________________________ 4.11 ()⎪⎪⎭⎫⎝⎛=-c midgap Fi N N kT E E υln 21()()kT kT 4952.0108.21004.1ln 211919-=⎪⎪⎭⎫ ⎝⎛⨯⨯=T (K) kT (eV)(midgap Fi E E -)(eV) 200 400 600 01727.0 03453.0 0518.0 0086.0- 0171.0- 0257.0-_______________________________________ 4.12(a) ⎪⎪⎭⎫ ⎝⎛=-**ln 43n p midgap Fi m m kT E E ()⎪⎭⎫⎝⎛=21.170.0ln 0259.04363.10-⇒meV(b) ()⎪⎭⎫⎝⎛=-080.075.0ln 0259.043midgap Fi E E47.43+⇒meV_______________________________________ 4.13Let ()==K E g c constant Then()()dE E fE g n FE co c⎰∞=dE kT E E Kc E F⎰∞⎪⎪⎭⎫⎝⎛-+=exp 11()dE kT E E K cE F ⎰∞⎥⎦⎤⎢⎣⎡--≅exp Let kT E E c-=η so that ηd kT dE ⋅=We can write ()()c F c F E E E E E E -+-=-so that()()()η-⋅⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--e x p e x p e x p kT E E kT E E F c F The integral can then be written as()()ηηd kT E E kT K n F c o ⎰∞-⎥⎦⎤⎢⎣⎡--⋅⋅=0exp exp which becomes()⎥⎦⎤⎢⎣⎡--⋅⋅=kT E E kT K n F c o exp _______________________________________ 4.14Let ()()c c E E C E g -=1 for c E E ≥ Then()()dE E fE g n FE co c⎰∞=()dE kT E E E E C c E Fc ⎰∞⎪⎪⎭⎫⎝⎛-+-=exp 11()()dE kT E E E E C F E C c⎥⎦⎤⎢⎣⎡---≅⎰∞exp 1LetkTE E c-=η so that ηd kT dE ⋅= We can write()()F c c F E E E E E E -+-=-Then()⎥⎦⎤⎢⎣⎡--=kT E E C n F c o exp 1()()dE kT E E E E c E c c⎥⎦⎤⎢⎣⎡---⨯⎰∞exp or()⎥⎦⎤⎢⎣⎡--=kT E E C n F c o exp 1 ()()()[]()ηηηd kT kT -⨯⎰∞exp 0We find that()()()11exp exp 0+=---=-∞∞⎰ηηηηηdSo()()⎥⎦⎤⎢⎣⎡--=kT E E kT C n F c o exp 21 _______________________________________ 4.15We have ⎪⎪⎭⎫⎝⎛=∈*1m m a r o r o For germanium, 16=∈r , o m m 55.0*= Then()()()53.02955.01161=⎪⎭⎫⎝⎛=o a roroA r 4.151=The ionization energy can be written as ()6.132*⎪⎪⎭⎫⎝⎛∈∈⎪⎪⎭⎫ ⎝⎛=s o o m m E eV ()()029.06.131655.02=⇒=E eV_______________________________________ 4.16We have ⎪⎪⎭⎫⎝⎛=∈*1m m a r o r o For gallium arsenide, 1.13=∈r ,o m m 067.0*= Then()()oA r 10453.0067.011.131=⎪⎭⎫⎝⎛=The ionization energy is()()()6.131.13067.06.1322*=⎪⎪⎭⎫ ⎝⎛∈∈⎪⎪⎭⎫ ⎝⎛=s o o m m E or0053.0=E eV_______________________________________ 4.17(a) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1519107108.2ln 0259.02148.0=eV (b) ()F c g F E E E E E --=-υ90518.02148.012.1=-=eV(c) ()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp()⎥⎦⎤⎢⎣⎡-⨯=0259.090518.0exp 1004.119 31090.6⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1107ln 0259.0338.0=eV_______________________________________ 4.18(a) ⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=16191021004.1ln 0259.0162.0=eV(b)()υE E E E E F g F c --=- 958.0162.012.1=-=eV(c) ()⎪⎭⎫⎝⎛-⨯=0259.0958.0exp 108.219o n31041.2⨯=cm 3-(d) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1102ln 0259.0365.0=eV_______________________________________ 4.19(a) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=519102108.2ln 0259.08436.0=eV ()F c g F E E E E E --=-υ 8436.012.1-= 2764.0=-υE E F eV (b)()⎪⎭⎫⎝⎛-⨯=0259.027637.0exp 1004.119o p1410414.2⨯=cm 3-(c) p-type_______________________________________ 4.20(a) ()032375.03003750259.0=⎪⎭⎫⎝⎛=kT eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫ ⎝⎛⨯=032375.028.0exp 300375107.42/317o n 141015.1⨯=cm 3-()28.042.1-=--=-F c g F E E E E E υ 14.1=eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.014.1exp 3003751072/318o p 31099.4⨯=cm 3-(b) ()⎪⎪⎭⎫⎝⎛⨯⨯=-14171015.1107.4ln 0259.0F c E E2154.0=eV()2154.042.1-=--=-F c g F E E E E E υ 2046.1=eV()⎥⎦⎤⎢⎣⎡-⨯=0259.02046.1exp 10718o p21042.4-⨯=cm 3-_______________________________________ 4.21(a) ()032375.03003750259.0=⎪⎭⎫⎝⎛=kT eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.028.0exp 300375108.22/319o n 151086.6⨯= cm 3-()28.012.1-=--=-F c g F E E E E E υ 840.0=eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.0840.0exp 3003751004.12/319o p 71084.7⨯=cm 3-(b) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=151910862.6108.2ln 0259.02153.0=eV9047.02153.012.1=-=-υE E F eV()⎥⎦⎤⎢⎣⎡-⨯=0259.0904668.0exp 1004.119o p 31004.7⨯=cm 3-_______________________________________ 4.22 (a) p-type(b) 28.0412.14===-g F E E E υeV()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp ()⎥⎦⎤⎢⎣⎡-⨯=0259.028.0exp 1004.119141010.2⨯=cm 3- ()υE E E E E F g F c --=- 84.028.012.1=-=eV()⎥⎦⎤⎢⎣⎡--=kT E E N n F c c o exp ()⎥⎦⎤⎢⎣⎡-⨯=0259.084.0exp 108.219 51030.2⨯=cm 3-_______________________________________ 4.23(a) ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi Fi o exp ()⎥⎦⎤⎢⎣⎡⨯=0259.022.0exp 105.110131033.7⨯=cm 3-⎥⎦⎤⎢⎣⎡-=kT E E n p F Fii o exp ()⎥⎦⎤⎢⎣⎡-⨯=0259.022.0exp 105.11061007.3⨯=cm 3-(b) ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi F i o exp()⎥⎦⎤⎢⎣⎡⨯=0259.022.0exp 108.16 91080.8⨯=cm 3-⎥⎦⎤⎢⎣⎡-=kT E E n p F Fii o exp()⎥⎦⎤⎢⎣⎡-⨯=0259.022.0exp 108.16 21068.3⨯=cm 3-_______________________________________ 4.24(a) ⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=151********.1ln 0259.01979.0=eV(b)()υE E E E E F g F c --=- 92212.019788.012.1=-=eV(c) ()⎥⎦⎤⎢⎣⎡-⨯=0259.092212.0exp 108.219o n31066.9⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1105ln 0259.03294.0=eV _______________________________________ 4.25()034533.03004000259.0=⎪⎭⎫⎝⎛=kT eV()2/3193004001004.1⎪⎭⎫⎝⎛⨯=υN1910601.1⨯=cm 3-()2/319300400108.2⎪⎭⎫⎝⎛⨯=c N19103109.4⨯=cm 3-()()1919210601.1103109.4⨯⨯=i n⎥⎦⎤⎢⎣⎡-⨯034533.012.1exp 24106702.5⨯=1210381.2⨯=⇒i n cm 3- (a)⎪⎪⎭⎫ ⎝⎛=-oF pN kT E E υυln ()⎪⎪⎭⎫⎝⎛⨯⨯=151910510601.1ln 034533.02787.0=eV (b) 84127.027873.012.1=-=-F c E E eV(c)()⎥⎦⎤⎢⎣⎡-⨯=034533.084127.0exp 103109.419o n910134.1⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=121510381.2105ln 034533.02642.0=eV _______________________________________ 4.26(a) ()⎥⎦⎤⎢⎣⎡-⨯=0259.025.0exp 10718o p141050.4⨯=cm 3-17.125.042.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.017.1exp 107.417o n21013.1-⨯=cm 3- (b) 034533.0=kT eV ()2/318300400107⎪⎭⎫ ⎝⎛⨯=υN1910078.1⨯=cm 3- ()2/317300400107.4⎪⎭⎫⎝⎛⨯=c N1710236.7⨯=cm 3- ⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=14191050.410078.1ln 034533.03482.0=eV072.13482.042.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=034533.007177.1exp 10236.717o n 41040.2⨯=cm 3-_____________________________________ 4.27(a) ()⎥⎦⎤⎢⎣⎡-⨯=0259.025.0exp 1004.119o p141068.6⨯=cm 3-870.025.012.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.0870.0exp 108.219o n41023.7⨯=o n cm 3- (b)034533.0=kT eV()2/3193004001004.1⎪⎭⎫⎝⎛⨯=υN1910601.1⨯=cm 3- ()2/319300400108.2⎪⎭⎫ ⎝⎛⨯=c N1910311.4⨯=cm 3-⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=14191068.610601.1ln 034533.03482.0=eV7718.03482.012.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=034533.077175.0exp 10311.419o n91049.8⨯=cm 3-_______________________________________ 4.28(a) ()F c o F N n ηπ2/12=For 2kT E E c F +=,5.02==-=kTkT kT E E c F F η Then ()0.12/1≅F F η()()0.1108.2219⨯=πo n191016.3⨯=cm 3-(b) ()F c o F N n ηπ2/12=()()0.1107.4217⨯=π171030.5⨯=cm 3-_______________________________________ 4.29()F o F N p ηπυ'=2/12()()FF ηπ'⨯=⨯2/119191004.12105So ()26.42/1='FF η We find kTE E FF-=≅'υη0.3()()0777.00259.00.3==-F E E υeV_______________________________________ 4.30(a) 44==-=kTkTkT E E c F F ηThen ()0.62/1≅F F η ()F c o F N n ηπ2/12=()()0.6108.2219⨯=π201090.1⨯=cm 3-(b) ()()0.6107.4217⨯=πo n181018.3⨯=cm 3-_______________________________________ 4.31For the electron concentration ()()()E f E g E n F c =The Boltzmann approximation applies, so ()()c nE E hm E n -=32/3*24π()⎥⎦⎤⎢⎣⎡--⨯kT E E F exp or()()()⎥⎦⎤⎢⎣⎡--=kT E E h m E n F c nexp 2432/3*π()⎥⎦⎤⎢⎣⎡---⨯kT E E kT E E kTc c exp DefinekTE E x c-= Then()()()x x K x n E n -=→exp To find maximum ()()x n E n →, set()()x x K dx x dn -⎢⎣⎡==-exp 2102/1 +()()⎥⎦⎤--x x exp 12/1or()⎥⎦⎤⎢⎣⎡--=-x x Kx 21exp 02/1which yieldskT E E kT E E x c c 2121+=⇒-==For the hole concentration ()()()[]E f E g E p F -=1υUsing the Boltzmann approximation ()()E E h m E p p-=υπ32/3*24()⎥⎦⎤⎢⎣⎡--⨯kT E E F exp or()()()⎥⎦⎤⎢⎣⎡--=kT E E h mE pF pυπexp 2432/3*()⎥⎦⎤⎢⎣⎡---⨯kT E E kT E E kTυυexp DefinekTEE x -='υThen()()x x K x p '-''='exp To find maximum value of ()()x p E p '→,set()0=''x d x dp Using the results from above,we find the maximum atkT E E 21-=υ_______________________________________4.32 (a) Silicon: We have()⎥⎦⎤⎢⎣⎡--=kT E E N n F c c o exp We can write()()F d d c F c E E E E E E -+-=- For045.0=-d c E E eV andkT E E F d 3=-eV we can write()⎥⎦⎤⎢⎣⎡--⨯=30259.0045.0exp 108.219o n()()737.4exp 108.219-⨯= or171045.2⨯=o n cm 3- We also have()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp Again, we can write()()υυE E E E E E a a F F -+-=- ForkT E E a F 3=- and045.0=-υE E a eV Then()⎥⎦⎤⎢⎣⎡--⨯=0259.0045.03exp 1004.119o p ()()737.4exp 1004.119-⨯= or161012.9⨯=o p cm 3- (b) GaAs: assume 0058.0=-d c E E eV Then()⎥⎦⎤⎢⎣⎡--⨯=30259.00058.0exp 107.417o n ()()224.3exp 107.417-⨯= or161087.1⨯=o n cm 3-Assume 0345.0=-υE E a eV Then()⎥⎦⎤⎢⎣⎡--⨯=30259.00345.0exp 10718o p ()()332.4exp 10718-⨯=or161020.9⨯=o p cm 3-_______________________________________ 4.33Plot_______________________________________ 4.34(a)151510310154⨯=-⨯=o p cm 3- ()415210105.7103105.1⨯=⨯⨯=o n cm 3-(b)16103⨯==d o N n cm 3-()316210105.7103105.1⨯=⨯⨯=o p cm 3-(c)10105.1⨯===i o o n p n cm 3-(d) ()()3191923003751004.1108.2⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯3750259.030012.1exp1110334.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-()8152111034.110410334.7⨯=⨯⨯=on cm 3-(e) ()()3191923004501004.1108.2⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯4500259.030012.1exp1310722.1⨯=⇒i n cm 3-()2132141410722.1210210⨯+⎪⎪⎭⎫ ⎝⎛+=o n1410029.1⨯=cm 3-()12142131088.210029.110722.1⨯=⨯⨯=o p cm 3-_______________________________________ 4.35(a)151510104-⨯=-=d a o N N p15103⨯=cm 3-()3152621008.1103108.1-⨯=⨯⨯==o i o p n n cm 3-(b)16103⨯==d o N n cm 3-()416261008.1103108.1-⨯=⨯⨯=o p cm 3-(c)6108.1⨯===i o o n p n cm 3-(d) ()()318172300375100.7107.4⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯3750259.030042.1exp810580.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-()215281044.110410580.7⨯=⨯⨯=on cm 3-(e) ()()318172300450100.7107.4⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯4500259.030042.1exp1010853.3⨯=⇒i n cm 3- 1410==d o N n cm 3-()7142101048.11010853.3⨯=⨯=op cm 3-_______________________________________ 4.36 (a) Ge: 13104.2⨯=i n cm 3-(i)2222i dd o n N N n +⎪⎪⎭⎫ ⎝⎛+=()21321515104.221022102⨯+⎪⎪⎭⎫ ⎝⎛⨯+⨯=or15102⨯=≅d o N n cm 3-()152132102104.2⨯⨯==o i o n n p111088.2⨯= cm 3- (ii)151610710⨯-=-≅d a o N N p 15103⨯=cm 3- ()152132103104.2⨯⨯==o i o p n n111092.1⨯=cm 3-(b) GaAs: 6108.1⨯=i n cm 3- (i)15102⨯=≅d o N n cm()315261062.1102108.1-⨯=⨯⨯=op cm 3-(ii)15103⨯=-≅d a o N N p cm 3-()315261008.1103108.1-⨯=⨯⨯=on cm 3-(c) The result implies that there is only one minority carrier in a volume of 310cm 3. _______________________________________ 4.37(a) For the donor level⎪⎪⎭⎫ ⎝⎛-+=kT E E N n F d d d exp 2111⎪⎭⎫ ⎝⎛+=0259.020.0exp 2111or41085.8-⨯=dd N n(b) We have()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F F exp 11Now()()F c c F E E E E E E -+-=- or245.0+=-kT E E FThen()⎪⎭⎫ ⎝⎛++=0259.0245.01exp 11E f For()51087.2-⨯=E f F_______________________________________ 4.38 (a) ⇒>d a N N p-type (b) Silicon:1313101105.2⨯-⨯=-=d a o N N p or13105.1⨯=o p cm 3- Then()7132102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- Germanium:2222i da d a o n N N N N p +⎪⎪⎭⎫ ⎝⎛-+-=()21321313104.22105.12105.1⨯+⎪⎪⎭⎫⎝⎛⨯+⎪⎪⎭⎫ ⎝⎛⨯=or131026.3⨯=o p cm 3- Then()131321321076.110264.3104.2⨯=⨯⨯==o i o p n n cm 3-Gallium Arsenide:13105.1⨯=-=d a o N N p cm 3- and()216.0105.1108.113262=⨯⨯==o i o p n n cm 3- _______________________________________ 4.39 (a) ⇒>a d N N n-type(b)1515102.1102⨯-⨯=-≅a d o N N n14108⨯=cm 3-()51421021081.2108105.1⨯=⨯⨯==o i o n n p cm 3-(c)()d a ao N N N p -+'≅ 151515102102.1104⨯-⨯+'=⨯aN 15108.4⨯='⇒aN cm 3-()41521010625.5104105.1⨯=⨯⨯=on cm 3-_______________________________________ 4.40()155210210125.1102105.1⨯=⨯⨯==o i o p n n cm 3- ⇒>o o p n n-type_______________________________________ 4.41()()318192300250100.61004.1⎪⎭⎫⎝⎛⨯⨯=i n()()⎥⎦⎤⎢⎣⎡-⨯3002500259.066.0exp24108936.1⨯=1210376.1⨯=⇒i n cm 3- 2222414i o o i o i o n n n n p n n =⇒==i o n n 21=⇒So 111088.6⨯=o n cm 3-,Then 121075.2⨯=o p cm 3-2222i aa o n N N p +⎪⎪⎭⎫ ⎝⎛+= 212210752.2⎪⎪⎭⎫ ⎝⎛-⨯a N242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=a N()21224210752.2105735.7⎪⎪⎭⎫ ⎝⎛+⨯-⨯aa N N242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=aN so that 1210064.2⨯=a N cm 3-_______________________________________ 4.42Plot_______________________________________ 4.43Plot_______________________________________ 4.44Plot_______________________________________ 4.452222i ad a d o n N N N N n +⎪⎪⎭⎫ ⎝⎛-+-= 2102.1102101.1141414⨯-⨯=⨯2214142102.1102i n +⎪⎪⎭⎫⎝⎛⨯-⨯+()()221321314104104101.1i n +⨯=⨯-⨯22727106.1109.4i n +⨯=⨯ so 131074.5⨯=i n cm 3-1314272103101.1103.3⨯=⨯⨯==o i o n n p cm 3- _______________________________________ 4.46 (a) ⇒>d a N N p-typeMajority carriers are holes1616105.1103⨯-⨯=-=d a o N N p16105.1⨯=cm 3-Minority carriers are electrons()4162102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- (b) Boron atoms must be addedd a ao N N N p -+'=161616105.1103105⨯-⨯+'=⨯aN So 16105.3⨯='aN cm 3-()316210105.4105105.1⨯=⨯⨯=on cm 3-_______________________________________ 4.47 (a) ⇒<<i o n p n-type (b) oi o o i o p n n n n p 22=⇒=on ()16421010125.1102105.1⨯=⨯⨯=cm 3-⇒electrons are majority carriers4102⨯=o p cm 3-⇒holes are minority carriers (c) a d o N N n -= 151610710125.1⨯-=⨯d N so 1610825.1⨯=d N cm 3-_______________________________________ 4.48⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln For Germanium T (K) kT (eV)i n (cm 3-) 200400 60001727.0 03453.0 0518.0101016.2⨯ 141060.8⨯ 161082.3⨯2222i a a o n N N p +⎪⎪⎭⎫⎝⎛+=and 1510=a N cm 3- T (K) op (cm 3-)()F Fi E E -(eV)200400 60015100.1⨯151049.1⨯ 161087.3⨯1855.0 01898.0 000674.0_______________________________________ 4.49(a) ⎪⎪⎭⎫⎝⎛=-d c F c N N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=d N 19108.2ln 0259.0 For 1410cm 3-, 3249.0=-F c E E eV 1510cm 3-, 2652.0=-F c E E eV 1610cm 3-, 2056.0=-F c E E eV 1710cm 3-, 1459.0=-F c E E eV(b) ⎪⎪⎭⎫⎝⎛=-i d Fi F n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=10105.1ln 0259.0d N For 1410cm 3-, 2280.0=-Fi F E E eV 1510cm 3-, 2877.0=-Fi F E E eV 1610cm 3-, 3473.0=-Fi F E E eV 1710cm 3-, 4070.0=-Fi F E E eV _______________________________________ 4.50 (a)2222i d d o n N N n +⎪⎪⎭⎫⎝⎛+= 151005.105.1⨯==d o N n cm 3- ()21515105.01005.1⨯-⨯()2215105.0i n +⨯=so 2821025.5⨯=i nNow()()3191923001004.1108.2⎪⎭⎫ ⎝⎛⨯⨯=T n i()()⎥⎦⎤⎢⎣⎡-⨯3000259.012.1exp T()3382830010912.21025.5⎪⎭⎫ ⎝⎛⨯=⨯T⎥⎦⎤⎢⎣⎡-⨯T 973.12972exp By trial and error, 5.536=T K (b) At 300=T K,⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=-151910108.2ln 0259.0F c E E2652.0=eV At 5.536=T K,()046318.03005.5360259.0=⎪⎭⎫⎝⎛=kT eV()2/3193005.536108.2⎪⎭⎫⎝⎛⨯=c N1910696.6⨯=cm 3-⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=-15191005.110696.6ln 046318.0F c E E5124.0=eV then ()2472.0=-∆F c E E eV (c) Closer to the intrinsic energy level._______________________________________ 4.51⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln At 200=T K, 017267.0=kT eV 400=T K, 034533.0=kT eV 600=T K, 0518.0=kT eVAt 200=T K,()()3191923002001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯017267.012.1exp410638.7⨯=⇒i n cm 3- At 400=T K,()()3191923004001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯034533.012.1exp 1210381.2⨯=⇒i n cm 3- At 600=T K,()()3191923006001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯0518.012.1exp1410740.9⨯=⇒i n cm 3- At 200=T K and 400=T K, 15103⨯==a o N p cm 3- At 600=T K,2222i a a o n N N p +⎪⎪⎭⎫⎝⎛+=()2142151510740.921032103⨯+⎪⎪⎭⎫ ⎝⎛⨯+⨯=1510288.3⨯=cm 3-Then, 200=T K, 4212.0=-F Fi E E eV 400=T K, 2465.0=-F Fi E E eV600=T K, 0630.0=-F Fi E E eV_______________________________________ 4.52(a)()⎪⎪⎭⎫⎝⎛⨯=⎪⎪⎭⎫ ⎝⎛=-6108.1ln 0259.0ln a i a F Fi N n N kT E EFor 1410=a N cm 3-,4619.0=-F Fi E E eV1510=a N cm 3-,5215.0=-F Fi E E eV1610=a N cm 3-,5811.0=-F Fi E E eV1710=a N cm 3-,6408.0=-F Fi E E eV (b)()⎪⎪⎭⎫⎝⎛⨯=⎪⎪⎭⎫ ⎝⎛=-a a F N N N kT E E 18100.7ln 0259.0ln υυ For 1410=a N cm 3-,2889.0=-υE E F eV1510=a N cm 3-,2293.0=-υE E F eV1610=a N cm 3-,1697.0=-υE E F eV1710=a N cm 3-,1100.0=-υE E F eV_______________________________________ 4.53(a) ⎪⎪⎭⎫ ⎝⎛=-**ln 43n p midgap Fi m m kT E E ()()10ln 0259.043= or0447.0+=-midgap Fi E E eV(b) Impurity atoms to be added so 45.0=-F midgap E E eV (i) p-type, so add acceptor atoms (ii)4947.045.00447.0=+=-F Fi E E eV Then⎪⎪⎭⎫⎝⎛-=kT E E n p F Fi i o exp()⎪⎭⎫⎝⎛=0259.04947.0exp 105or131097.1⨯==a o N p cm 3-_______________________________________4.54()⎥⎦⎤⎢⎣⎡--=-=kT E E N N N n F c c a d o exp so()⎪⎭⎫⎝⎛-⨯+⨯=0259.0215.0exp 108.21051915d N15151095.6105⨯+⨯=or16102.1⨯=d N cm 3-_______________________________________ 4.55 (a) Silicon(i)⎪⎪⎭⎫⎝⎛=-d c F c N N kT E E ln()2188.0106108.2ln 0259.01519=⎪⎪⎭⎫⎝⎛⨯⨯=eV(ii)1929.00259.02188.0=-=-F c E E eV()⎥⎦⎤⎢⎣⎡--=kT E E N N F c c d exp()⎥⎦⎤⎢⎣⎡-⨯=0259.01929.0exp 108.2191610631.1⨯=d N cm 3-15106⨯+'=dN 1610031.1⨯='⇒dN cm 3- Additionaldonor atoms (b) GaAs(i)()⎪⎪⎭⎫⎝⎛⨯=-151710107.4ln 0259.0F c E E15936.0=eV(ii)13346.00259.015936.0=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.013346.0exp 107.417d N1510718.2⨯=cm 3-1510+'=dN 1510718.1⨯='⇒dN cm 3- Additionaldonor atoms _______________________________________4.56 (a) ⎪⎪⎭⎫ ⎝⎛=-a F Fi N N kT E E υln()⎪⎪⎭⎫⎝⎛⨯⨯=16191021004.1ln 0259.01620.0=eV(b) ⎪⎪⎭⎫⎝⎛=-d c Fi F N N kT E E ln ()1876.0102108.2ln 0259.01619=⎪⎪⎭⎫ ⎝⎛⨯⨯=eV (c) For part (a); 16102⨯=o p cm 3-()162102102105.1⨯⨯==o i o p n n410125.1⨯=cm 3-For part (b): 16102⨯=o n cm 3-()162102102105.1⨯⨯==o i o n n p410125.1⨯=cm 3-_______________________________________4.57⎥⎦⎤⎢⎣⎡-=kT E E n n Fi F i o exp ()⎥⎦⎤⎢⎣⎡⨯=0259.055.0exp 108.1615100.3⨯=cm 3-Add additional acceptor impuritiesa d o N N n -= a N -⨯=⨯151510710315104⨯=⇒a N cm 3-_______________________________________ 4.58(a) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln ()3161.0105.1103ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=eV (b) ⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln ()3758.0105.1103ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV(c) Fi F E E =(d) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln ()⎪⎪⎭⎫ ⎝⎛⨯⨯⎪⎭⎫ ⎝⎛=111510334.7104ln 3003750259.0 2786.0=eV(e) ⎪⎪⎭⎫ ⎝⎛=-i o Fi F n n kT E E ln()⎪⎪⎭⎫ ⎝⎛⨯⨯⎪⎭⎫ ⎝⎛=131410722.110029.1ln 3004500259.0 06945.0=eV _______________________________________ 4.59 (a) ⎪⎪⎭⎫ ⎝⎛=-o F p N kT E E υυln()2009.0103100.7ln 0259.01518=⎪⎪⎭⎫ ⎝⎛⨯⨯=eV(b) ()⎪⎪⎭⎫ ⎝⎛⨯⨯=--4181008.1100.7ln 0259.0υE E F 360.1=eV(c) ()⎪⎪⎭⎫⎝⎛⨯⨯=-618108.1100.7ln 0259.0υE E F7508.0=eV(d) ()⎪⎭⎫⎝⎛=-3003750259.0υE E F()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯152/318104300375100.7ln 2526.0=eV(e) ()⎪⎭⎫⎝⎛=-3004500259.0υE E F()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯72/3181048.1300450100.7ln 068.1=eV_______________________________________ 4.60n-type⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln()3504.0105.110125.1ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV ______________________________________ 4.612222i aa o n N N p +⎪⎪⎭⎫ ⎝⎛+= 21051008.51515⨯=⨯22152105i n +⎪⎪⎭⎫ ⎝⎛⨯+()21515105.21008.5⨯-⨯()2215105.2i n +⨯=230301025.6106564.6i n +⨯=⨯29210064.4⨯=⇒i n⎥⎦⎤⎢⎣⎡-=kT E N N n g c i exp 2υ()030217.03003500259.0=⎪⎭⎫⎝⎛=kT eV()1921910633.1300350102.1⨯=⎪⎭⎫ ⎝⎛⨯=c N cm 3-()192191045.2300350108.1⨯=⎪⎭⎫ ⎝⎛⨯=υN cm 3- Now()()1919291045.210633.110064.4⨯⨯=⨯⎥⎦⎤⎢⎣⎡-⨯030217.0exp g ESo()()()⎥⎦⎤⎢⎣⎡⨯⨯⨯=29191910064.41045.210633.1ln 030217.0g E 6257.0=⇒g E eV_______________________________________ 4.62 (a) Replace Ga atoms ⇒Silicon acts as adonor()()1415105.310705.0⨯=⨯=d N cm 3-Replace As atoms ⇒Silicon acts as anacceptor()()15151065.610795.0⨯=⨯=a N cm 3-(b) ⇒>d a N N p-type(c) 1415105.31065.6⨯-⨯=-=d a o N N p 15103.6⨯=cm 3-()4152621014.5103.6108.1-⨯=⨯⨯==o i o p n n cm 3- (d)⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()5692.0108.1103.6ln 0259.0615=⎪⎪⎭⎫⎝⎛⨯⨯=eV_______________________________________。

半导体物理与器件(尼曼第四版)答案

半导体物理与器件(尼曼第四版)答案第一章：半导体材料与晶体1.1 半导体材料的基本特性半导体材料是一种介于导体和绝缘体之间的材料。

它的基本特性包括：1.带隙：半导体材料的价带与导带之间存在一个禁带或带隙，是电子在能量上所能占据的禁止区域。

2.拉伸系统：半导体材料的结构是由原子或分子构成的晶格结构，其中的原子或分子以确定的方式排列。

3.载流子：在半导体中，存在两种载流子，即自由电子和空穴。

自由电子是在导带上的，在外加电场存在的情况下能够自由移动的电子。

空穴是在价带上的，当一个价带上的电子从该位置离开时，会留下一个类似电子的空位，空穴可以看作电子离开后的痕迹。

4.掺杂：为了改变半导体材料的导电性能，通常会对其进行掺杂。

掺杂是将少量元素添加到半导体材料中，以改变载流子浓度和导电性质。

1.2 半导体材料的结构与晶体缺陷半导体材料的结构包括晶体结构和非晶态结构。

晶体结构是指材料具有有序的周期性排列的结构，而非晶态结构是指无序排列的结构。

晶体结构的特点包括：1.晶体结构的基本单位是晶胞，晶胞在三维空间中重复排列。

2.晶格常数是晶胞边长的倍数，用于描述晶格的大小。

3.晶体结构可分为离子晶体、共价晶体和金属晶体等不同类型。

晶体结构中可能存在各种晶体缺陷，包括：1.点缺陷：晶体中原子位置的缺陷，主要包括实际缺陷和自间隙缺陷两种类型。

2.线缺陷：晶体中存在的晶面上或晶内的线状缺陷，主要包括位错和脆性断裂两种类型。

3.面缺陷：晶体中存在的晶面上的缺陷，主要包括晶面位错和穿孔两种类型。

1.3 半导体制备与加工半导体制备与加工是指将半导体材料制备成具有特定电性能的器件的过程。

它包括晶体生长、掺杂、薄膜制备和微电子加工等步骤。

晶体生长是将半导体材料从溶液或气相中生长出来的过程。

常用的晶体生长方法包括液相外延法、分子束外延法和气相外延法等。

掺杂是为了改变半导体材料的导电性能，通常会对其进行掺杂。

常用的掺杂方法包括扩散法、离子注入和分子束外延法等。

半导体物理与器件第四版课后习题答案4复习进程

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半导体物理与器件第四版课后习题标准答案

半导体物理与器件第四版课后习题答案————————————————————————————————作者：————————————————————————————————日期：2______________________________________________________________________________________3Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would beginto behave less like a semiconductor and morelike a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V x t x m ,,2222ψ⋅+∂ψ∂-η()tt x j ∂ψ∂=,ηAssume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x ηexp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m ηηexp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j ηηηexp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m ηηexp 222()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jk ηexp 2()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp 22()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu ηexp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u k ηSetting ()()x u x u 1= for region I, the equation becomes: ()()()()021221212=--+x u k dx x du jk dxx u d α where222ηmE=αQ.E.D.In Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x ηexp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m ηηexp 222()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jk ηexp 2()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp 22()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O ηexp()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu ηexp This equation can be written as:______________________________________________________________________________________4()()()2222xx u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV O ηη Setting ()()x u x u 2= for region II, this equation becomes()()dx x du jkdx x u d 22222+()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O ηα where again222ηmE=αQ.E.D._______________________________________ 3.3We have ()()()()021221212=--+x u k dx x du jk dxx u d α Assume the solution is of the form: ()()[]x k j A x u -=αexp 1 ()[]x k j B +-+αexpThe first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexpand the second derivative becomes()()[]()[]x k j A k j dxx u d --=ααexp 2212()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differential equation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k()[]0exp =+-⨯x k j B α We find that00= Q.E.D.For the differential equation in ()x u 2 and theproposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2()[]x k j D +-+βexp for 0<<-x b .The first boundary condition is ()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dx dudx du which yields()()()C k B k A k --+--βαα ()0=++D k βThe third boundary condition is ()()b u a u -=21 which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp ()()[]b k j D -+-+βexp______________________________________________________________________________________5and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp ()[]0exp =+-b k j D βThe fourth boundary condition isbx a x dx dudx du -===21 which yields()()[]a k j A k j --ααexp()()[]a k j B k j +-+-ααexp()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexp and can be written as()()[]a k j A k --ααexp()()[]a k j B k +-+-ααexp ()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________ 3.5(b) (i) First point: πα=aSecond point: By trial and error, πα729.1=a (ii) First point: πα2=aSecond point: By trial and error, πα617.2=a _______________________________________ 3.6(b) (i) First point: πα=aSecond point: By trial and error, πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a _______________________________________ 3.7ka a aaP cos cos sin =+'ααα Let y ka =, x a =α Theny x x xP cos cos sin =+'Consider dydof this function.()[]{}y x x x P dyd sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydx x sin sin -=-Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-' For πn ka y ==,...,2,1,0=n 0sin =⇒y So that, in general,()()dkd ka d a d dy dxαα===0 And22ηmE=α SodkdEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221ηηα This implies thatdk dE dk d ==0α for an k π= _______________________________________ 3.8(a) πα=a 1π=⋅a E m o 212η______________________________________________________________________________________6()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o η19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o η()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J34E E E -=∆1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________ 3.9(a) At π=ka , πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πo E19105172.2-⨯=J o E E E -=∆11919105172.2104114.3--⨯-⨯= 2010942.8-⨯=Jor 559.0106.110942.81920=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J At π=ka . From Problem 3.5,πα729.12=aπ729.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV_____________________________________________________________________________________________________________________________73.10(a) πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 12E E E -=∆1919104114.310830.7--⨯-⨯= 19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o η()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J 34E E E -=∆1818103646.1109242.1--⨯-⨯=1910597.5-⨯=Jor 50.3106.110597.51919=⨯⨯=∆--E eV _____________________________________3.11(a) At π=ka , πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=J At 0=ka , By trial and error, πα727.0=a o π727.022=⋅a E m o o η()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=J o E E E -=∆11919108030.1104114.3--⨯-⨯= 19106084.1-⨯=Jor 005.1106.1106084.11919=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6, πα515.12=aπ515.1222=⋅a E m o η()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 23E E E -=∆191810830.7103646.1--⨯-⨯=______________________________________________________________________________________81910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV_______________________________________ 3.12For 100=T K,()()⇒+⨯-=-1006361001073.4170.124g E164.1=g E eV200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV 600=T K, 032.1=g E eV _______________________________________ 3.13The effective mass is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m ηWe have()()B curve dkEd A curve dk E d 2222>so that ()()B curve m A curve m **<_______________________________________ 3.14The effective mass for a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p η We have that()()B curve dkEd A curve dk E d 2222> so that ()()B curve m A curve m p p **<_______________________________________ 3.15Points A,B: ⇒<0dkdEvelocity in -x direction Points C,D: ⇒>0dkdEvelocity in +x directionPoints A,D: ⇒<022dk Ednegative effective massPoints B,C: ⇒>022dkEd positive effective mass_______________________________________ 3.16For A: 2k C E i =At 101008.0+⨯=k m 1-, 05.0=E eV Or()()2119108106.105.0--⨯=⨯=E JSo ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m η 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4 o m m 488.0=*For B: 2k C E i =At 101008.0+⨯=k m 1-, 5.0=E eV Or ()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C 3711025.1-⨯=⇒CNow ()()37234121025.1210054.12--*⨯⨯==C m η321044.4-⨯=kgor o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_____________________________________________________________________________________________________________________________93.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C3921025.6-⨯=⇒C ()()39234221025.6210054.12--*⨯⨯-=-=C m η31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m η3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________ 3.18(a) (i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν1410429.3⨯=Hz(ii) 141010429.3103⨯⨯===νλc E hc 51075.8-⨯=cm 875=nm(b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν 1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm _______________________________________ 3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0=k , and is negativearound 2π±=k ._______________________________________ 3.20()[]O O k k E E E --=αcos 1 Then()()()[]O k k E dkdE---=ααsin 1()[]O k k E -+=ααsin 1 and()[]O k k E dkEd -=ααcos 2122Then221222*11ηηαE dk Ed m o k k =⋅== or212*αE m η=_______________________________________ 3.21(a) ()[]3/123/24l t dnm m m =*()()[]3/123/264.1082.04o o m m =o dnm m 56.0=*(b)oo l t cn m m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cnm m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lh hh dp m m m +=*______________________________________________________________________________________10()()[]3/22/32/3082.045.0o o m m +=[]o m ⋅+=3/202348.030187.0o dpm m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=* ()()()()om ⋅++=2/12/12/32/3082.045.0082.045.0 o cpm m 34.0=*_______________________________________3.23For the 3-dimensional infinite potential well, ()0=x V when a x <<0, a y <<0, and a z <<0. In this region, the wave equation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψηUse separation of variables technique, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=⋅+XYZ mEηDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ηmE z Z Z y Y Y x X X Let01222222=+∂∂⇒-=∂∂⋅X k x X k x X X xx The solution is of the form: ()x k B x k A x X x x cos sin += Since ()0,,=z y x ψ at 0=x , then ()00=Xso that 0=B .Also, ()0,,=z y x ψ at a x =, so that()0=a X . Then πx x n a k = where...,3,2,1=x n Similarly, we have2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅From the boundary conditions, we find πy y n a k = and πz z n a k = where...,3,2,1=y n and ...,3,2,1=z n From the wave equation, we can write022222=+---ηmE k k k z y xThe energy can be written as()222222⎪⎭⎫ ⎝⎛++==a n n n m E E z y x n n n z y x πη _______________________________________ 3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by()332a dk k dk k g T ⋅=ππ where222ηmEk =We can then writeηmEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121ηηSubstituting these expressions into the densityof states function, we have()dE EmmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233ηηππ Noting thatπ2h=ηthis density of states function can be simplified and written as______________________________________________________________________________________()()dE E m h a dE E g T ⋅⋅=2/33324πDividing by 3a will yield the density of states so that()()E hm E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k an E m n ==*πη Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n*⋅=21ηdE Em dk n⋅⋅⋅=*2211η Then()dE Em a dE E g n T ⋅⋅⋅=*2212ηπDivide by the "volume" a , so()Em E g n *⋅=21πηSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o nm m 08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E h m 22/332/33224+*-⋅⋅=π()()2/332/323224kT h m n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV ()()19106.10259.0-⨯=2110144.4-⨯=J Then()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3- or 19100.6⨯=c g cm 3-(ii) At 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯=21105253.5-⨯=J Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3-or 191024.9⨯=c g cm 3-(b) GaAs, o nm m 067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π ()()2/3542102288.1kT ⨯=______________________________________________________________________________________(i) At 300=T K, 2110144.4-⨯=kT J()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m 3- or 171027.9⨯=c g cm 3-(ii) At 400=T K, 21105253.5-⨯=kT J()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3- 181043.1⨯=c g cm 3-_______________________________________ 3.27(a) Silicon, o p m m 56.0=* ()()E E h mE g p-=*υυπ32/324()dE E E h m g E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E h m 32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT h mp-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.4310969.2-⨯⨯=υg2510116.4⨯=m 3- or 191012.4⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J ()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m 3-or 191034.6⨯=υg cm 3- (b) GaAs, o p m m 48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3- or 191027.3⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υg cm 3-_______________________________________ 3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1 For c E E =; 0=c g1.0+=c E E eV; 4610509.1⨯=c g m 3-J 1-2.0+=c E E eV;4610134.2⨯=m 3-J 1- 3.0+=c E E eV; 4610614.2⨯=m 3-J 1- 4.0+=c E E eV; 4610018.3⨯=m 3-J 1-(b) ()E E hm g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4______________________________________________________________________________________For υE E =; 0=υg 1.0-=υE E eV; 4510634.5⨯=υg m 3-J 1-2.0-=υE E eV;4510968.7⨯=m 3-J 1-3.0-=υE E eV; 4510758.9⨯=m 3-J 1-4.0-=υE E eV;4610127.1⨯=m 3-J 1-_______________________________________ 3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pncmm g g υ_______________________________________3.30 Plot_______________________________________ 3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66= (ii)()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________ 3.32()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F exp 11(a) kT E E F =-, ()()⇒+=1exp 11E f ()269.0=E f(b) kT E E F 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f (c) kT E E F 10=-, ()()⇒+=10exp 11E f()51054.4-⨯=E f_______________________________________ 3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F exp 1111or()⎪⎪⎭⎫⎝⎛-+=-kT E E E f F exp 111(a) kT E E F =-, ()269.01=-E f (b) kT E E F 5=-, ()31069.61-⨯=-E f (c) kT E E F 10=-, ()51054.41-⨯=-E f_______________________________________ 3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp F f 61066.5-⨯=kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +; ()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+; ()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=______________________________________________________________________________________(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =; ⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kT E -υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯= kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=_______________________________________ 3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So ()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c +-=-+υOr midgap c F E E E E =+=2υ_______________________________________ 3.3622222man E n πη= For 6=n , Filled state()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eVFor 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eVTherefore 8.1240.9<<F E eV_______________________________________ 3.37(a) For a 3-D infinite potential well ()222222⎪⎭⎫⎝⎛++=a n n n mE z y x πη For 5 electrons, the 5thelectron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x πη()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π 1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--E eV For the next quantum state, which is empty, the quantum state is 2,2,1===z y x n n n . This quantum state is at the same energy, so 35.2=F E eV(b) For 13 electrons, the 13th electronoccupies the quantum state______________________________________________________________________________________3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE 1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eVThe 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is at the same energy, so 746.5=F E eV_______________________________________ 3.38The probability of a state at E E E F ∆+=1 being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111 or()⎪⎭⎫ ⎝⎛∆+=-kT E E f exp 11122 so ()()22111E f E f -= Q.E.D._______________________________________ 3.39(a) At energy 1E , we want01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F FThis expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F For()⎪⎪⎭⎫⎝⎛-=kT E E F 1exp 01.01Then()100ln 1kT E E F += orkT E E F 6.41+= (b)At kT E E F 6.4+=,()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F which yields()01.000990.01≅=E f_______________________________________ 3.40 (a)()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0______________________________________________________________________________________or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kTwhich yields 740=T K_______________________________________ 3.41 (a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b) At 1000=T K, 08633.0=kT eV Then()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫⎝⎛-+=E for 99.7% (d)At F E E =, ()21=E f for alltemperatures_______________________________________ 3.42(a) For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =,82.030.012.12=-=-E E F eV Then()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b) For 4.02=-E E F eV, 72.01=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E f_______________________________________ 3.43(a) At 1E E =()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f F or()61032.9-⨯=E f At 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.012.1exp______________________________________________________________________________________or()191066.11-⨯=-E f(b) For 4.02=-E E F ,02.11=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________ 3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛⨯kT E E kT F exp 1 or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kT E E kT E E kT dE E df F F (a) At 0=T K, For()00exp =⇒=∞-⇒<dE dfE E F()0exp =⇒+∞=∞+⇒>dEdfE E FAt -∞=⇒=dEdfE E F(b) At 300=T K, 0259.0=kT eVFor F E E <<,0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()65.91110259.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1-(c) At 500=T K, 04317.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =, ()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________ 3.45(a) At midgap E E =,()⎪⎪⎭⎫ ⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kTE kTE E E f gF2exp 11exp 11Si: 12.1=g E eV, ()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for()101007.4-⨯=E fGe: 66.0=g E eV______________________________________________________________________________________()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E for()61093.2-⨯=E f GaAs: 42.1=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0242.1exp 11E for()121024.1-⨯=E f(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._______________________________________ 3.46(a) ()⎥⎦⎤⎢⎣⎡--=kT E E f F F exp ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 108 or ()810ln 60.0+=kT()032572.010ln 60.08==kT eV()⎪⎭⎫⎝⎛=3000259.0032572.0Tso 377=T K(b) ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 106()610ln 60.0+=kT()043429.010ln 60.06==kT ()⎪⎭⎫⎝⎛=3000259.0043429.0Tor 503=T K_______________________________________ 3.47(a) At 200=T K,()017267.03002000259.0=⎪⎭⎫⎝⎛=kT eV⎪⎪⎭⎫ ⎝⎛-+==kTE E f FF exp 1105.019105.01exp =-=⎪⎪⎭⎫⎝⎛-kT E E F()()()19ln 017267.019ln ==-kT E E F 05084.0=eV By symmetry, for 95.0=F f , 05084.0-=-F E E eVThen ()1017.005084.02==∆E eV (b) 400=T K, 034533.0=kT eV For 05.0=F f , from part (a),()()()19ln 034533.019ln ==-kT E E F 10168.0=eV Then ()2034.010168.02==∆E eV_______________________________________。

半导体物理与器件第四课后习题答案3.doc

Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V xt x m ,,2222ψ⋅+∂ψ∂- ()tt x j ∂ψ∂=, Assume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m exp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j exp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu exp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u kSetting ()()x u x u 1= for region I, the equation becomes:()()()()021221212=--+x u k dx x du jk dxx u d α where222mE=αIn Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu exp This equation can be written as:()()()2222x x u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV OSetting ()()x u x u 2= for region II, this equation becomes()()dx x du jk dxx u d 22222+ ()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O α where again222mE=α_______________________________________3.3We have()()()()021221212=--+x u k dx x du jk dxx u d α Assume the solution is of the form: ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp The first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexp and the second derivative becomes()()[]()[]x k j A k j dxx u d --=ααexp 2212 ()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differential equation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k ()[]0exp =+-⨯x k j B α We find that 00=For the differential equation in ()x u 2 and the proposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2()[]x k j D +-+βexp for 0<<-x b .The first boundary condition is ()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dx dudx du which yields()()()C k B k A k --+--βαα()0=++D k β The third boundary condition is ()()b u a u -=21 which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp()()[]b k j D -+-+βexp and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp()[]0exp =+-b k j D β The fourth boundary condition isbx a x dx dudx du -===21 which yields()()[]a k j A k j --ααexp()()[]a k j B k j +-+-ααexp ()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexp and can be written as ()()[]a k j A k --ααexp()()[]a k j B k +-+-ααexp()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________ 3.5(b) (i) First point: πα=aSecond point: By trial and error, πα729.1=a (ii) First point: πα2=aSecond point: By trial and error, πα617.2=a_______________________________________3.6(b) (i) First point: πα=aSecond point: By trial and error, πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a_______________________________________ 3.7ka a aaP cos cos sin =+'αααLet y ka =, x a =α Theny x x xP cos cos sin =+'Consider dy dof this function.()[]{}y x x x P dyd sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydxx sin sin -=- Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-'For πn ka y ==, ...,2,1,0=n 0sin =⇒y So that, in general,()()dk d ka d a d dy dxαα===0 And22 mE=αSodk dEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221 α This implies thatdk dE dk d ==0α for an k π= _______________________________________ 3.8(a) πα=a 1π=⋅a E m o 212()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J 34E E E -=∆1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________3.9(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o ()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πoE19105172.2-⨯=J o E E E -=∆11919105172.2104114.3--⨯-⨯= 2010942.8-⨯=Jor 559.0106.110942.81920=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka . From Problem 3.5, πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV_______________________________________3.10(a) πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 12E E E -=∆1919104114.310830.7--⨯-⨯= 19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV (b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J 34E E E -=∆1818103646.1109242.1--⨯-⨯= 1910597.5-⨯=Jor 50.3106.110597.51919=⨯⨯=∆--E eV_____________________________________3.11(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα727.0=a oπ727.022=⋅a E m o o()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=Jo E E E -=∆11919108030.1104114.3--⨯-⨯= 19106084.1-⨯=Jor 005.1106.1106084.11919=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6,πα515.12=aπ515.1222=⋅a E m o()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J23E E E -=∆191810830.7103646.1--⨯-⨯= 1910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV_______________________________________3.12For 100=T K, ()()⇒+⨯-=-1006361001073.4170.124gE164.1=g E eV200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV 600=T K, 032.1=g E eV_______________________________________3.13The effective mass is given by1222*1-⎪⎪⎭⎫⎝⎛⋅=dk E d mWe have()()B curve dkE d A curve dk E d 2222> so that ()()B curve m A curve m **<_______________________________________ 3.14The effective mass for a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p We have that()()B curve dkEd A curve dk E d 2222> so that ()()B curve m A curve m p p **<_______________________________________ 3.15Points A,B: ⇒<0dk dEvelocity in -x directionPoints C,D: ⇒>0dk dEvelocity in +x directionPoints A,D: ⇒<022dk Ednegative effective massPoints B,C: ⇒>022dkEd positive effective mass _______________________________________3.16For A: 2k C E i =At 101008.0+⨯=k m 1-, 05.0=E eV Or ()()2119108106.105.0--⨯=⨯=E J So ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4o m m 488.0=* For B: 2k C E i =At 101008.0+⨯=k m 1-, 5.0=E eV Or ()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C 3711025.1-⨯=⇒CNow ()()37234121025.1210054.12--*⨯⨯==C m 321044.4-⨯=kg or o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_______________________________________ 3.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C 3921025.6-⨯=⇒C()()39234221025.6210054.12--*⨯⨯-=-=C m31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C 382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________ 3.18(a) (i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν1410429.3⨯=Hz(ii) 141010429.3103⨯⨯===νλc E hc 51075.8-⨯=cm 875=nm(b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm_______________________________________ 3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0=k , and is negativearound 2π±=k . _______________________________________ 3.20()[]O O k k E E E --=αcos 1 Then()()()[]O k k E dkdE ---=ααsin 1()[]O k k E -+=ααsin 1 and()[]O k k E dk E d -=ααcos 2122Then221222*11 αE dk Ed m o k k =⋅== or212*αE m =_______________________________________ 3.21(a) ()[]3/123/24lt dn m m m =*()()[]3/123/264.1082.04oom m =o dn m m 56.0=*(b)o o l t cnm m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cn m m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lh hh dp m m m +=*()()[]3/22/32/3082.045.0o om m +=[]o m ⋅+=3/202348.030187.0o dp m m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=*()()()()om ⋅++=2/12/12/32/3082.045.0082.045.0 o cp m m 34.0=*_______________________________________ 3.23For the 3-dimensional infinite potential well, ()0=x V when a x <<0, a y <<0, and a z <<0. In this region, the wave equation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψ Use separation of variables technique, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=⋅+XYZ mEDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ mEz Z Z y Y Y x X XLet01222222=+∂∂⇒-=∂∂⋅X k x X k x X X xx The solution is of the form: ()x k B x k A x X x x cos sin +=Since ()0,,=z y x ψ at 0=x , then ()00=X so that 0=B .Also, ()0,,=z y x ψ at a x =, so that ()0=a X . Then πx x n a k = where ...,3,2,1=x n Similarly, we have2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅From the boundary conditions, we find πy y n a k = and πz z n a k =where...,3,2,1=y n and ...,3,2,1=z n From the wave equation, we can write022222=+---mE k k k z y xThe energy can be written as()222222⎪⎭⎫⎝⎛++==a n n n m E E z y x n n n z y x π _______________________________________ 3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by()332a dk k dk k g T ⋅=ππ where222 mEk =We can then writemEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121 Substituting these expressions into the density of states function, we have()dE E mmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233 ππ Noting thatπ2h=this density of states function can be simplified and written as()()dE E m h a dE E g T ⋅⋅=2/33324π Dividing by 3a will yield the density of states so that()()E h m E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k a n E m n ==*π Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n *⋅=21dE Em dk n⋅⋅⋅=*2211 Then()dE Em a dE E g n T ⋅⋅⋅=*2212 π Divide by the "volume" a , so ()Em E g n *⋅=21πSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o n m m 08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E h m 22/332/33224+*-⋅⋅=π()()2/332/323224kT hm n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV()()19106.10259.0-⨯= 2110144.4-⨯=J Then ()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3-or 19100.6⨯=c g cm 3-(ii) At 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯= 21105253.5-⨯=J Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3- or 191024.9⨯=c g cm 3-(b) GaAs, o nm m 067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π ()()2/3542102288.1kT ⨯=(i) At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m 3- or 171027.9⨯=c g cm 3-(ii) At 400=T K, 21105253.5-⨯=kT J ()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3-181043.1⨯=c g cm 3-_______________________________________ 3.27(a) Silicon, o p m m 56.0=* ()()E E h mE g p-=*υυπ32/324()dE E E h mg E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E hm 32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT hmp-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.4310969.2-⨯⨯=υg2510116.4⨯=m3-or 191012.4⨯=υg cm 3- (ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m3-or 191034.6⨯=υg cm 3- (b) GaAs, o p m m 48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3- or 191027.3⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υg cm 3-_______________________________________ 3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1 For c E E =; 0=c g1.0+=c E E eV; 4610509.1⨯=c g m 3-J 1-2.0+=c E E eV; 4610134.2⨯=m 3-J 1-3.0+=c E E eV; 4610614.2⨯=m 3-J 1- 4.0+=c E E eV; 4610018.3⨯=m 3-J 1- (b) ()E E h m g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4 For υE E =; 0=υg1.0-=υE E eV; 4510634.5⨯=υg m 3-J 1-2.0-=υE E eV; 4510968.7⨯=m 3-J 1-3.0-=υE E eV; 4510758.9⨯=m 3-J 1-4.0-=υE E eV; 4610127.1⨯=m 3-J 1-_______________________________________ 3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pncmm g g υ_______________________________________3.30 Plot_______________________________________ 3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66=(ii) ()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________ 3.32()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F exp 11(a) kT E E F =-, ()()⇒+=1exp 11E f()269.0=E f (b) kT E E F 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f(c) kT E E F 10=-, ()()⇒+=10exp 11E f ()51054.4-⨯=E f_______________________________________ 3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F exp 1111or()⎪⎪⎭⎫ ⎝⎛-+=-kT E E E f F exp 111(a) kT E E F =-, ()269.01=-E f (b) kT E E F 5=-, ()31069.61-⨯=-E f(c) kT E E F 10=-, ()51054.41-⨯=-E f_______________________________________ 3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp F f 61066.5-⨯=kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +; ()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+; ()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =; ⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kTE -υ; ()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯= kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=_______________________________________3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1 ()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So ()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp ()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c +-=-+υOr midgap c F E E E E =+=2υ_______________________________________ 3.3622222ma n E n π =For 6=n , Filled state()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eV For 7=n , Empty state ()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eV Therefore 8.1240.9<<F E eV_______________________________________ 3.37(a) For a 3-D infinite potential well()222222⎪⎭⎫ ⎝⎛++=a n n n mE z y x π For 5 electrons, the 5th electron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x π()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--E eV For the next quantum state, which is empty, the quantum state is 2,2,1===z y x n n n . This quantum state is at the same energy, so 35.2=F E eV(b) For 13 electrons, the 13th electronoccupies the quantum state 3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE 1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eVThe 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is at the same energy, so 746.5=F E eV_______________________________________ 3.38The probability of a state at E E E F ∆+=1 being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111or()⎪⎭⎫ ⎝⎛∆+=-kT E E f exp 11122so ()()22111E f E f -=_______________________________________3.39(a) At energy 1E , we want01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F FThis expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F F or()⎪⎪⎭⎫⎝⎛-=kT E E F 1exp 01.01Then()100ln 1kT E E F += orkT E E F 6.41+= (b)At kT E E F 6.4+=, ()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F which yields()01.000990.01≅=E f_______________________________________ 3.40 (a)()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kT which yields 740=T K_______________________________________ 3.41 (a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b) At 1000=T K, 08633.0=kT eV Then()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫ ⎝⎛-+=E for 99.7% (d)At F E E =, ()21=E f for all temperatures_______________________________________ 3.42(a) For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =, 82.030.012.12=-=-E E F eV Then()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b) For 4.02=-E E F eV,72.01=-F E E eVAt 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E f_______________________________________ 3.43(a) At 1E E =()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f F or()61032.9-⨯=E fAt 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.012.1expor()191066.11-⨯=-E f (b) For 4.02=-E E F ,02.11=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________ 3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛⨯kT E E kT F exp 1or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kT E E kT E E kT dE E df F F (a) At 0=T K, For()00exp =⇒=∞-⇒<dE dfE E F()0exp =⇒+∞=∞+⇒>dEdfE E FAt -∞=⇒=dEdfE E F(b) At 300=T K, 0259.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()65.91110259.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1-(c) At 500=T K, 04317.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________ 3.45(a) At midgap E E =,()⎪⎪⎭⎫⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kT E kTE E E f g F2exp 11exp 11Si: 12.1=g E eV, ()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for()101007.4-⨯=E fGe: 66.0=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E for()61093.2-⨯=E f GaAs: 42.1=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0242.1exp 11E for()121024.1-⨯=E f(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._______________________________________3.46(a) ()⎥⎦⎤⎢⎣⎡--=kT E E f F F exp ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 108or()810ln 60.0+=kT()032572.010ln 60.08==kT eV ()⎪⎭⎫⎝⎛=3000259.0032572.0Tso 377=T K(b) ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 106()610ln 60.0+=kT()043429.010ln 60.06==kT ()⎪⎭⎫⎝⎛=3000259.0043429.0Tor 503=T K_______________________________________ 3.47(a) At 200=T K,()017267.03002000259.0=⎪⎭⎫⎝⎛=kT eV⎪⎪⎭⎫ ⎝⎛-+==kT E E f F F exp 1105.019105.01exp =-=⎪⎪⎭⎫ ⎝⎛-kT E E F()()()19ln 017267.019ln ==-kT E E F 05084.0=eV By symmetry, for 95.0=F f , 05084.0-=-F E E eVThen ()1017.005084.02==∆E eV (b) 400=T K, 034533.0=kT eV For 05.0=F f , from part (a),()()()19ln 034533.019ln ==-kT E E F 10168.0=eVThen ()2034.010168.02==∆E eV _______________________________________。

半导体物理课后习题解答

半导体物理课后习题解答半导体物理习题解答1－1．（P 32）设晶格常数为a 的一维晶格，导带极小值附近能量E c （k ）和价带极大值附近能量E v （k ）分别为： E c (k)=223m k h +22)1(m k k h -和E v (k)=226m k h －223m k h ；m 0为电子惯性质量，k 1＝1/2a ；a ＝0.314nm 。

试求： ①禁带宽度；②导带底电子有效质量； ③价带顶电子有效质量；④价带顶电子跃迁到导带底时准动量的变化。

[解] ①禁带宽度Eg 根据dkk dEc )(＝232m k h ＋12)(2m k k h -＝0；可求出对应导带能量极小值E min 的k 值： k min ＝143k ，由题中E C 式可得：E min ＝E C (K)|k=k min =2104k mh ；由题中E V 式可看出，对应价带能量极大值Emax 的k 值为：k max ＝0；并且E min ＝E V (k)|k=k max ＝2126m k h ；∴Eg ＝E min －E max ＝21212m k h ＝20248a m h＝112828227106.1)1014.3(101.948)1062.6(----⨯⨯⨯⨯⨯⨯⨯＝0.64eV②导带底电子有效质量m n02020********m h m h m h dkE d C =+=；∴ m n＝22283/m dkE d h C =③价带顶电子有效质量m ’02226m h dk E d V -=，∴222'61/m dk E d h m V n-==④准动量的改变量h △k ＝h (k min -k max)= ah kh 83431=[毕]1－2．（P 33）晶格常数为0.25nm 的一维晶格，当外加102V/m ，107V/m 的电场时，试分别计算电子自能带底运动到能带顶所需的时间。

[解] 设电场强度为E ，∵F=h dtdk =qE （取绝对值） ∴dt =qE h dk∴t=⎰t dt 0=⎰a qE h 21dk =aqE h 21代入数据得： t ＝E⨯⨯⨯⨯⨯⨯--1019-34105.2106.121062.6＝E6103.8-⨯（s ）当E ＝102V/m 时，t ＝8.3×10－8（s ）；E ＝107V/m 时，t ＝8.3×10－13（s ）。

半导体物理与器件第四版课后习题答案3

Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V x t x m ,,2222ψ⋅+∂ψ∂- ()tt x j ∂ψ∂=, Assume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m exp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j exp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu exp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u kSetting ()()x u x u 1= for region I, the equation becomes:()()()()021221212=--+x u k dx x du jk dxx u d α where222mE=α Q.E.D.In Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu exp This equation can be written as:()()()2222x x u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV OSetting ()()x u x u 2= for region II, this equation becomes()()dx x du jkdx x u d 22222+ ()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O αwhere again222mE=α Q.E.D._______________________________________3.3We have()()()()021221212=--+x u k dx x du jk dxx u d α Assume the solution is of the form: ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp The first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexp and the second derivative becomes()()[]()[]x k j A k j dx x u d --=ααexp 2212()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differential equation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k ()[]0exp =+-⨯x k j B α We find that00= Q.E.D. For the differential equation in ()x u 2 and the proposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions ()()[]x k j A x u -=αexp 1 ()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2 ()[]x k j D +-+βexpfor 0<<-x b .The first boundary condition is()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dx du dx du which yields()()()C k B k A k --+--βαα ()0=++D k β The third boundary condition is ()()b u a u -=21 which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp()()[]b k j D -+-+βexp and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp()[]0exp =+-b k j D β The fourth boundary condition isbx a x dx dudx du -===21 which yields()()[]a k j A k j --ααexp ()()[]a k j B k j +-+-ααexp ()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexp and can be written as ()()[]a k j A k --ααexp ()()[]a k j B k +-+-ααexp ()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________ 3.5(b) (i) First point: πα=aSecond point: By trial and error, πα729.1=a (ii) First point: πα2=aSecond point: By trial and error, πα617.2=a_______________________________________3.6(b) (i) First point: πα=aSecond point: By trial and error, πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a_______________________________________ 3.7ka a aaP cos cos sin =+'αααLet y ka =, x a =α Theny x x xP cos cos sin =+'Consider dydof this function.()[]{}y x x x P dy d sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydxx sin sin -=-Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-'For πn ka y ==, ...,2,1,0=n 0sin =⇒y So that, in general,()()dk d ka d a d dy dxαα===0 And 22 mE=α Sodk dEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221 α This implies thatdk dE dk d ==0α for an k π= _______________________________________3.8(a) πα=a 1π=⋅a E m o 212 ()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV (b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J 34E E E -=∆1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________3.9(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o ()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πoE19105172.2-⨯=J o E E E -=∆11919105172.2104114.3--⨯-⨯= 2010942.8-⨯=Jor 559.0106.110942.81920=⨯⨯=∆--E eV(b) At π2=ka , πα23=a π2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka . From Problem 3.5, πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV _______________________________________3.10(a) πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 12E E E -=∆1919104114.310830.7--⨯-⨯= 19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J 34E E E -=∆1818103646.1109242.1--⨯-⨯= 1910597.5-⨯=Jor 50.3106.110597.51919=⨯⨯=∆--E eV_____________________________________3.11(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα727.0=a oπ727.022=⋅a E m o o()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=J o E E E -=∆11919108030.1104114.3--⨯-⨯= 19106084.1-⨯=Jor 005.1106.1106084.11919=⨯⨯=∆--E eV(b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6, πα515.12=aπ515.1222=⋅a E m o()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J23E E E -=∆191810830.7103646.1--⨯-⨯= 1910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV _______________________________________3.12For 100=T K, ()()⇒+⨯-=-1006361001073.4170.124gE164.1=g E eV200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV 600=T K, 032.1=g E eV_______________________________________ 3.13The effective mass is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d mWe have()()B curve dkE d A curve dk E d 2222>so that ()()B curve m A curve m **<_______________________________________ 3.14The effective mass for a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p We have that()()B curve dkEd A curve dk E d 2222> so that ()()B curve m A curve m p p **<_______________________________________ 3.15Points A,B: ⇒<0dk dEvelocity in -x directionPoints C,D: ⇒>0dk dEvelocity in +x directionPoints A,D: ⇒<022dk Ednegative effective massPoints B,C: ⇒>022dk Edpositive effective mass_______________________________________ 3.16For A: 2k C E i =At 101008.0+⨯=k m 1-, 05.0=E eV Or ()()2119108106.105.0--⨯=⨯=E J So ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4 o m m 488.0=*For B: 2k C E i =At 101008.0+⨯=k m 1-, 5.0=E eV Or ()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C3711025.1-⨯=⇒CNow ()()37234121025.1210054.12--*⨯⨯==C m 321044.4-⨯=kgor o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_______________________________________ 3.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C 3921025.6-⨯=⇒C ()()39234221025.6210054.12--*⨯⨯-=-=C m31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C 382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________ 3.18(a) (i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν 1410429.3⨯=Hz(ii) 141010429.3103⨯⨯===νλc E hc 51075.8-⨯=cm 875=nm(b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm_______________________________________ 3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0=k , and is negativearound 2π±=k ._______________________________________ 3.20()[]O O k k E E E --=αcos 1 Then()()()[]O k k E dkdE ---=ααsin 1()[]O k k E -+=ααsin 1 and()[]O k k E dk E d -=ααcos 2122Then221222*11αE dk Ed m o k k =⋅== or212*αE m = _______________________________________ 3.21(a) ()[]3/123/24l t dnm m m =*()()[]3/123/264.1082.04oo m m =o dnm m 56.0=*(b)oo l t cn m m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cnm m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lhhh dp m m m +=*()()[]3/22/32/3082.045.0oom m +=[]o m ⋅+=3/202348.030187.0o dpm m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=*()()()()o m ⋅++=2/12/12/32/3082.045.0082.045.0 o cpm m 34.0=*_______________________________________3.23For the 3-dimensional infinite potential well, ()0=x V when a x <<0, a y <<0, and a z <<0. In this region, the wave equation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψUse separation of variables technique, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂022=⋅+XYZ mEDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ mEz Z Z y Y Y x X XLet01222222=+∂∂⇒-=∂∂⋅X k x X k x X X x x The solution is of the form: ()x k B x k A x X x x cos sin +=Since ()0,,=z y x ψ at 0=x , then ()00=X so that 0=B .Also, ()0,,=z y x ψ at a x =, so that ()0=a X . Then πx x n a k = where ...,3,2,1=x n Similarly, we have2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅From the boundary conditions, we find πy y n a k = and πz z n a k = where...,3,2,1=y n and ...,3,2,1=z n From the wave equation, we can write022222=+---mE k k k z y xThe energy can be written as()222222⎪⎭⎫ ⎝⎛++==a n n n m E E z y x n n n z y x π _______________________________________ 3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by()332a dk k dk k g T ⋅=ππ where222 mEk =We can then writemEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121 Substituting these expressions into the density of states function, we have()dE EmmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233 ππ Noting thatπ2h=this density of states function can be simplified and written as()()dE E m h a dE E g T ⋅⋅=2/33324πDividing by 3a will yield the density of states so that()()E hm E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k an E m n ==*π Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n*⋅=21dE Em dk n⋅⋅⋅=*2211Then()dE Em a dE E g n T ⋅⋅⋅=*2212 π Divide by the "volume" a , so ()Em E g n *⋅=21πSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o nm m 08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E h m 22/332/33224+*-⋅⋅=π()()2/332/323224kT h m n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV()()19106.10259.0-⨯=2110144.4-⨯=J Then ()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3- or 19100.6⨯=c g cm 3-(ii) At 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯=21105253.5-⨯=J Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3- or 191024.9⨯=c g cm 3-(b) GaAs, o nm m 067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π()()2/3542102288.1kT ⨯=(i) At 300=T K, 2110144.4-⨯=kT J()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m3-or 171027.9⨯=c g cm 3- (ii) At 400=T K, 21105253.5-⨯=kT J()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3-181043.1⨯=c g cm 3-_______________________________________ 3.27(a) Silicon, o p m m 56.0=* ()()E E h mE g p-=*υυπ32/324()dE E E hm g E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E h m32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT h m p-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.4310969.2-⨯⨯=υg2510116.4⨯=m 3- or 191012.4⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J ()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m 3-or 191034.6⨯=υg cm 3- (b) GaAs, o p m m 48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3- or 191027.3⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υg cm 3-_______________________________________ 3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1 For c E E =; 0=c g1.0+=c E E eV; 4610509.1⨯=c g m 3-J 1-2.0+=c E E eV; 4610134.2⨯=m 3-J 1-3.0+=c E E eV; 4610614.2⨯=m 3-J 1- 4.0+=c E E eV; 4610018.3⨯=m 3-J 1- (b) ()E E hm g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4 For υE E =; 0=υg1.0-=υE E eV; 4510634.5⨯=υg m 3-J 1-2.0-=υE E eV; 4510968.7⨯=m 3-J 1-3.0-=υE E eV; 4510758.9⨯=m 3-J 1-4.0-=υE E eV; 4610127.1⨯=m 3-J 1- _______________________________________ 3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ_______________________________________ 3.30 Plot_______________________________________ 3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W ()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66=(ii) ()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________ 3.32()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F exp 11(a) kT E E F =-, ()()⇒+=1exp 11E f()269.0=E f (b) kT E E F 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f (c) kT E E F 10=-, ()()⇒+=10exp 11E f()51054.4-⨯=E f_______________________________________ 3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F exp 1111or()⎪⎪⎭⎫ ⎝⎛-+=-kT E E E f F exp 111(a) kT E E F =-, ()269.01=-E f (b) kT E E F 5=-, ()31069.61-⨯=-E f(c) kT E E F 10=-, ()51054.41-⨯=-E f _______________________________________ 3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp F f 61066.5-⨯=kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +; ()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+; ()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =; ⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kT E -υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯= kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=_______________________________________ 3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1 ()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So ()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp ()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c +-=-+υOr midgap cF E E E E =+=2υ_______________________________________ 3.3622222ma n E n π =For 6=n , Filled state ()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eV For 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eVTherefore 8.1240.9<<F E eV_______________________________________ 3.37(a) For a 3-D infinite potential well()222222⎪⎭⎫⎝⎛++=a n n n mE z y x π For 5 electrons, the 5th electron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x π()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--E eVFor the next quantum state, which is empty, the quantum state is 2,2,1===z y x n n n . This quantum state is at the same energy, so 35.2=F E eV(b) For 13 electrons, the 13th electronoccupies the quantum state 3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE 1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eVThe 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is at the same energy, so 746.5=F E eV_______________________________________ 3.38The probability of a state at E E E F ∆+=1 being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111or()⎪⎭⎫⎝⎛∆+=-kT E E f exp 11122so ()()22111E f E f -= Q.E.D. _______________________________________ 3.39(a) At energy 1E , we want01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F F This expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F For()⎪⎪⎭⎫⎝⎛-=kT E E F 1exp 01.01Then()100ln 1kT E E F += orkT E E F 6.41+= (b)At kT E E F 6.4+=, ()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F which yields()01.000990.01≅=E f_______________________________________ 3.40 (a)()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kT which yields 740=T K_______________________________________ 3.41 (a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b) At 1000=T K, 08633.0=kT eV Then()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫⎝⎛-+=E for 99.7% (d)At F E E =, ()21=E f for all temperatures_______________________________________ 3.42(a) For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =, 82.030.012.12=-=-E E F eV Then()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b) For 4.02=-E E F eV,72.01=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E f_______________________________________ 3.43(a) At 1E E = ()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f F or()61032.9-⨯=E fAt 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫⎝⎛-=0259.012.1expor()191066.11-⨯=-E f(b) For 4.02=-E E F ,02.11=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________ 3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛⨯kT E E kT F exp 1 or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kT E E kT E E kT dE E df F F (a) At 0=T K, For()00exp =⇒=∞-⇒<dE dfE E F()0exp =⇒+∞=∞+⇒>dEdfE E FAt -∞=⇒=dEdfE E F(b) At 300=T K, 0259.0=kT eVFor F E E <<, 0=dEdfFor F E E >>, 0=dEdfAt F E E =,()()65.91110259.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1-(c) At 500=T K, 04317.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =, ()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________ 3.45(a) At midgap E E =,()⎪⎪⎭⎫ ⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kTE kTE E E f gF2exp 11exp 11Si: 12.1=g E eV, ()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for()101007.4-⨯=E fGe: 66.0=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E for()61093.2-⨯=E fGaAs: 42.1=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0242.1exp 11E for()121024.1-⨯=E f(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._______________________________________ 3.46(a) ()⎥⎦⎤⎢⎣⎡--=kT E E f F F exp ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 108 or ()810ln 60.0+=kT()032572.010ln 60.08==kT eV()⎪⎭⎫⎝⎛=3000259.0032572.0Tso 377=T K(b) ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 106()610ln 60.0+=kT()043429.010ln 60.06==kT()⎪⎭⎫⎝⎛=3000259.0043429.0Tor 503=T K_______________________________________ 3.47(a) At 200=T K,()017267.03002000259.0=⎪⎭⎫⎝⎛=kT eV⎪⎪⎭⎫ ⎝⎛-+==kTE E f FF exp 1105.019105.01exp =-=⎪⎪⎭⎫⎝⎛-kT E E F()()()19ln 017267.019ln ==-kT E E F 05084.0=eV By symmetry, for 95.0=F f , 05084.0-=-F E E eVThen ()1017.005084.02==∆E eV (b) 400=T K, 034533.0=kT eV For 05.0=F f , from part (a),()()()19ln 034533.019ln ==-kT E E F 10168.0=eVThen ()2034.010168.02==∆E eV_______________________________________。

半导体物理学第四版答案

半导体物理学第四版答案【篇一：半导体物理学第四章答案】. 300k时，ge的本征电阻率为47?cm，如电子和空穴迁移率分别为3900cm2/( v.s)和1900cm/( v.s)。

试求ge 的载流子浓度。

解：在本征情况下，n?p?ni,由??1/??211知 ?nqun?pqupniq(un?up)ni?1113?3??2.29?10cm?19?q(un?up)47?1.602?10?(3900?1900)2. 试计算本征si在室温时的电导率，设电子和空穴迁移率分别为1350cm2/( v.s)和500cm2/( v.s)。

当掺入百万分之一的as后，设杂质全部电离，试计算其电导率。

比本征si的电导率增大了多少倍？解：300k时，un?1350cm2/(v?s),up?500cm2/(v?s)，查表3-2或图3-7可知，室温下si的本征载流子浓度约为ni?1.0?1010cm?3。

本征情况下，??nqun?pqup?niq(un?up)?1?1010?1.602?10-19?(1350+500)?3.0?10?6s/cm11金钢石结构一个原胞内的等效原子个数为8??6??4?8个，查看附录b知si的晶格常数为820.543102nm，则其原子密度为822?3。

?5?10cm?73(0.543102?10)1?5?1016cm?3，杂质全部电离后，nd??ni，1000000掺入百万分之一的as,杂质的浓度为nd?5?1022?这种情况下，查图4-14（a）可知其多子的迁移率为800 cm2/( v.s) ??ndqun?5?1016?1.602?10-19?800?6.4s/cm?6.4??2.1?106倍比本征情况下增大了?6?3?103. 电阻率为10?.m的p型si样品，试计算室温时多数载流子和少数载流子浓度。

解：查表4-15(b)可知，室温下，10?.m的p型si样品的掺杂浓度na约为1.5?1015cm?3,查表3-2或图3-7可知，室温下si的本征载流子浓度约为ni?1.0?1010cm?3,na??nip?na?1.5?1015cm?3ni(1.0?1010)24?3n???6.7?10cm15p1.5?104. 0.1kg的ge单晶，掺有3.2?10-9kg的sb，设杂质全部电离，试求该材料的电阻率??n=0.38m2/( v.s),ge的单晶密度为5.32g/cm3,sb原子量为121.8?。

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Chapter 33.1Ifoa were to increase, the bandgap energywould decrease and the material would begin to behave less like a semiconductor and morelike a metal. Ifoa were to decrease, thebandgap energy would increase and the material would begin to behave morelike aninsulator.________________________ _______________3.2Schrodinger's wave equation is:()()()txxVxtxm,,2222ψ⋅+∂ψ∂-()ttxj∂ψ∂=,Assume the solution is of the form:AHA12GAGGAGAGGAFFFFAFAFAHA12GAGGAGAGGAFFFFAFAF ()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Region I: ()0=x V . Substituting the assumed solution into the wave equation,weobtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m exp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j exp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu exp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mEx x u x x u jk x u kSetting ()()x u x u 1= forregion I, the equationbecomes:()()()()021221212=--+x u k dxx du jkdx x u d α where 222 mE=αQ.E.D.In Region II, ()OV x V =.Assume the sameAHA12GAGGAGAGGAFFFFAFAF form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu exp This equation can be written as:()()()2222x x u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV OSetting ()()x u x u 2= forregion II, thisequation becomes()()dxx du jkdx x u d 22222+()022222=⎪⎪⎭⎫⎝⎛+--x u mV k O α where againAHA12GAGGAGAGGAFFFFAFAF 222 mE=α Q.E.D._______________________________________3.3We have()()()()021221212=--+x u k dxx du jkdx x u d α Assume the solution is of the form:()()[]x k j A x u -=αexp 1()[]x k j B +-+αexpThe first derivative is()()()[]x k j A k j dx x du --=ααexp 1()()[]x k j B k j +-+-ααexpand the second derivative becomes()()[]()[]x k j A k j dxx u d --=ααexp 2212()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differentialequation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B αCombining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k kAHA12GAGGAGAGGAFFFFAFAF ()[]0exp =+-⨯x k j B αWe find that00=Q.E.D.For thedifferential equation in()x u 2 and theproposed solution, the procedure is exactly the same as above. _______________________________________3.4We have thesolutions()()[]x k j A x u -=αexp 1()[]x k j B +-+αexpfor a x <<0 and()()[]x k j C x u -=βexp 2()[]x k j D +-+βexpfor 0<<-x b .The first boundary condition is()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dxdu dxduwhich yieldsAHA12GAGGAGAGGAFFFFAFAF ()()()C k B k A k --+--βαα()0=++D k βThe third boundary condition is()()b u a u -=21which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp()()[]b k j D -+-+βexpand can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp()[]0exp =+-b k j D βThe fourth boundary condition isbx ax dxdu dxdu -===21which yields()()[]a k j A k j --ααexp()()[]a k j B k j +-+-ααexp()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexpand can be written as()()[]a k j A k --ααexp()()[]a k j B k +-+-ααexp ()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________3.5(b) (i) First point:πα=aSecondpoint: By trial and error,πα729.1=a(ii) Firstpoint: πα2=aSecond point: By trial and error,πα617.2=a________________________ _______________3.6(b) (i) First point:πα=aSecond point: By trial and error,πα515.1=a(ii) Firstpoint: πα2=aSecond point: By trial and error,πα375.2=aAHA12GAGGAGAGGAFFFFAFAFAHA12GAGGAGAGGAFFFFAFAF _______________________________________3.7ka a aaP cos cos sin =+'ααα Let y ka =, x a =α Theny x xxP cos cos sin =+'Consider dydof thisfunction.()[]{}y x x x P dyd sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydxx sin sin -=- Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-'For πn ka y ==,...,2,1,0=n 0sin =⇒ySo that, in general,()()dkd ka d a d dy dxαα===0 And22mE=α SodkdEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫⎝⎛=-22/122221 α This implies thatdkdE dkd ==0α for an k π=_______________________________________AHA12GAGGAGAGGAFFFFAFAF 3.8(a) πα=a 1π=⋅a E m o 212 ()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o19104114.3-⨯=JFrom Problem 3.5πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J12E E E -=∆1918104114.3100198.1--⨯-⨯=19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πEAHA12GAGGAGAGGAFFFFAFAF 18103646.1-⨯=JFrom Problem 3.5,πα617.24=aπ617.2224=⋅a E m o()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J34E E E -=∆1818103646.1103364.2--⨯-⨯=1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________3.9(a)At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , Bytrial and error,πα859.0=a o()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πo E19105172.2-⨯=Jo E E E -=∆11919105172.2104114.3--⨯-⨯=2010942.8-⨯=Jor559.0106.110942.81920=⨯⨯=∆--E eV(b)At π2=ka , πα23=aπ2223=⋅a E m oAHA12GAGGAGAGGAFFFFAFAF ()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka . From Problem 3.5,πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J23E E E -=∆1818100198.1103646.1--⨯-⨯=19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV _______________________________________3.10(a) πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6,πα515.12=aπ515.1222=⋅a E m o()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J12E E E -=∆1919104114.310830.7--⨯-⨯=19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m oAHA12GAGGAGAGGAFFFFAFAF ()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6,πα375.24=aπ375.2224=⋅a E m o()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J34E E E -=∆1818103646.1109242.1--⨯-⨯=1910597.5-⨯=Jor50.3106.110597.51919=⨯⨯=∆--E eV _____________________________________3.11(a)At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error,πα727.0=a oπ727.022=⋅a E m o o()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=Jo E E E -=∆11919108030.1104114.3--⨯-⨯=AHA12GAGGAGAGGAFFFFAFAF19106084.1-⨯=Jor005.1106.1106084.11919=⨯⨯=∆--E eV(b)At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , FromProblem 3.6,πα515.12=aπ515.1222=⋅a E m o()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J23E E E -=∆191810830.7103646.1--⨯-⨯=1910816.5-⨯=Jor635.3106.110816.51919=⨯⨯=∆--E eV_______________________________________3.12For 100=T K,()()⇒+⨯-=-1006361001073.4170.124gE164.1=g E eV200=T K,147.1=g E eV300=T K, 125.1=gE eV400=T K, 097.1=gE eV500=T K, 066.1=gE eV600=T K, 032.1=gE eV________________________ _______________3.13The effective mass is given by1222*1-⎪⎪⎭⎫⎝⎛⋅=dkEdmWe have()()BcurvedkEdAcurvedkEd2222>AHA12GAGGAGAGGAFFFFAFAFAHA12GAGGAGAGGAFFFFAFAF so that()()B curve m A curve m **<_______________________________________3.14The effective massfor a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p We have that()()B curve dkEd A curve dk E d 2222> so that()()B curve m A curve m p p **<_______________________________________3.15Points A,B:⇒<0dkdEvelocity in -x directionPoints C,D:⇒>0dkdEvelocity in +x directionPoints A,D: ⇒<022dkEdnegative effective massPoints B,C:⇒>022dk EdAHA12GAGGAGAGGAFFFFAFAF positive effective mass _______________________________________3.16For A: 2kC E i=At 101008.0+⨯=k m 1-,05.0=E eVOr()()2119108106.105.0--⨯=⨯=E JSo ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m311044.4-⨯=kgoro m m ⋅⨯⨯=--*31311011.9104437.4o m m 488.0=*For B: 2kC E i=At 101008.0+⨯=k m 1-,5.0=E eVOr()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C3711025.1-⨯=⇒CAHA12GAGGAGAGGAFFFFAFAF Now()()37234121025.1210054.12--*⨯⨯==C m321044.4-⨯=kgoro m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_______________________________________3.17For A: 22k C EE -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C3921025.6-⨯=⇒C ()()39234221025.6210054.12--*⨯⨯-=-=C m31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=*For B: 22k C EE -=-υ()()()2102191008.0106.13.0⨯-=⨯--C382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________3.18AHA12GAGGAGAGGAFFFFAFAF (a)(i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν 1410429.3⨯=Hz(ii)141010429.3103⨯⨯===νλc E hc51075.8-⨯=cm 875=nmAHA12GAGGAGAGGAFFFFAFAF (b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm_______________________________________3.19(c)Curve A: Effective mass is a constant Curve B: Effective mass is positivearound 0=k , and is negativearound 2π±=k ._______________________________________3.20()[]O O k k E E E --=αcos 1Then()()()[]O k k E dkdE---=ααsin 1 ()[]O k k E -+=ααsin 1and()[]O k k E dk E d -=ααcos 2122AHA12GAGGAGAGGAFFFFAFAFThen221222*11 αE dk Ed m ok k =⋅==or212*αE m=_______________________________________3.21(a) ()[]3/123/24l t dnm m m =*()()[]3/123/264.1082.04oo m m =o dnm m 56.0=*(b)o o l t cnm m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cnm m 12.0=*_______________________________________3.22(a) ()()[]3/22/32/3lh hh dp m m m +=*()()[]3/22/32/3082.045.0o om m +=[]o m ⋅+=3/202348.030187.0o dpm m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=* ()()()()om ⋅++=2/12/12/32/3082.045.0082.045.0 o cpm m 34.0=*_______________________________________3.23For the 3-dimensional infinitepotential well,()0=x V when a x <<0,a y <<0, andAHA12GAGGAGAGGAFFFFAFAF a z <<0. In this region, the waveequation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψUse separation of variables technique, so let()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=⋅+XYZ mEDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅mEz Z Z y Y Y x X XLetAHA12GAGGAGAGGAFFFFAFAF 01222222=+∂∂⇒-=∂∂⋅X k xX k x X X x x The solution is of the form:()x k B x k A x X x x cos sin +=Since ()0,,=z y x ψ at0=x , then ()00=Xso that 0=B . Also, ()0,,=z y x ψ atax =, so that()0=a X . Thenπx x n a k = where...,3,2,1=x nSimilarly, we have2221yk y YY-=∂∂⋅ and 2221z k zZZ -=∂∂⋅ From the boundary conditions, we findπy yn a k= andπz z n a k =where ...,3,2,1=ynand...,3,2,1=z nFrom the waveequation, we can write022222=+--- mEk k k z y x The energy can be written as()222222⎪⎭⎫ ⎝⎛++==a n n n m E E z y x n n n zy x π _______________________________________AHA12GAGGAGAGGAFFFFAFAF3.24The total number ofquantum states in the3-dimensionalpotential well is given(in k-space) by()332a dk k dk k g T ⋅=ππwhere222 mEk =We can then writemEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121 Substituting these expressions into thedensityof states function,we have()dE EmmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233 ππ Noting thatπ2h =this density of states function can besimplified and written as()()dE E m ha dE E g T ⋅⋅=2/33324πAHA12GAGGAGAGGAFFFFAFAF Dividing by 3a willyield the density of states so that()()E h m E g ⋅=32/324π_______________________________________3.25For a one-dimensional infinite potential well,222222ka n Em n ==*π Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2 NowE m k n *⋅=21dE Em dk n⋅⋅⋅=*2211Then()dE Em a dE E g n T ⋅⋅⋅=*2212 πDivide by the "volume" a , so()Em E g n *⋅=21πSoAHA12GAGGAGAGGAFFFFAFAF ()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J1-_______________________________________3.26(a) Silicon, o nm m08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E hm 22/332/33224+*-⋅⋅=π()()2/332/323224kT hm n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV()()19106.10259.0-⨯=2110144.4-⨯=JThen()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3-or 19100.6⨯=cgcm3-(ii) At 400=T K,()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯=21105253.5-⨯=JAHA12GAGGAGAGGAFFFFAFAF Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3-or 191024.9⨯=cgcm3-(b) GaAs, o nm m067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π ()()2/3542102288.1kT ⨯=(i) At 300=T K,2110144.4-⨯=kT J()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m3-or 171027.9⨯=cgcm3-(ii) At 400=T K,21105253.5-⨯=kT J()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3-181043.1⨯=cgcm3-_______________________________________3.27(a)Silicon, o pm m56.0=*()()E E h m E g p-=*υυπ32/324()dE E E hm g E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E h m 32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT h m p-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯= (i)At 300=T K,2110144.4-⨯=kT J()()[]2/3215510144.4310969.2-⨯⨯=υgAHA12GAGGAGAGGAFFFFAFAF 2510116.4⨯=m3-or 191012.4⨯=υgcm3-(ii)At 400=T K,21105253.5-⨯=kT J()()[]2/32155105253.5310969.2-⨯⨯=υg 2510337.6⨯=m3-or 191034.6⨯=υgcm3-(b)GaAs, o pm m48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=AHA12GAGGAGAGGAFFFFAFAF (i)At 300=T K,2110144.4-⨯=kT J()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3-or 191027.3⨯=υgcm3-(ii)At 400=T K,21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υgcm3-_______________________________________3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1For cE E =;0=c g1.0+=cEE eV;4610509.1⨯=c g m 3-J1-2.0+=cEE eV;4610134.2⨯=m 3-J 1-3.0+=cEE eV;4610614.2⨯=m 3-J 1-4.0+=cEE eV;4610018.3⨯=m 3-J 1-AHA12GAGGAGAGGAFFFFAFAF (b) ()E E h m g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4For υE E =;0=υg1.0-=υEE eV;4510634.5⨯=υg m 3-J1-2.0-=υEE eV;4510968.7⨯=m 3-J 1-3.0-=υEE eV;4510758.9⨯=m 3-J 1-4.0-=υEE eV;4610127.1⨯=m 3-J 1-_______________________________________3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pncmm g g υ_______________________________________3.30Plot_______________________________________3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66=AHA12GAGGAGAGGAFFFFAFAF (ii)()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________3.32()⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fexp 11(a) kTE EF =-, ()()⇒+=1exp 11E f()269.0=E f (b) kTE EF 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f (c) kTE EF 10=-,()()⇒+=10exp 11E f()51054.4-⨯=E f_______________________________________3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kTE E E f Fexp 1111or()⎪⎪⎭⎫ ⎝⎛-+=-kT E E E f F exp 111(a) kTE EF =-, ()269.01=-E f (b) kTE EF 5=-, ()31069.61-⨯=-E f(c) kTE EF 10=-, ()51054.41-⨯=-E fAHA12GAGGAGAGGAFFFFAFAF _______________________________________3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kTE c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp Ff61066.5-⨯=kT E c +;()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +;()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+;()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =;⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ;()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ;()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kT E -υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯=kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=AHA12GAGGAGAGGAFFFFAFAF _______________________________________3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1 ()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp ()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c+-=-+υOr midgap c FE E E E=+=2υ_______________________________________3.3622222man E n π = For 6=n , Filledstate()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor40.9106.1105044.119186=⨯⨯=--E eV For 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=JAHA12GAGGAGAGGAFFFFAFAF or8.12106.110048.219187=⨯⨯=--E eV Therefore8.1240.9<<F E eV_______________________________________3.37(a)For a 3-D infinite potential wellAHA12GAGGAGAGGAFFFFAFAF ()222222⎪⎭⎫ ⎝⎛++=a n n n mE z y x π For 5 electrons,the 5thelectronoccupies the quantumstate 1,2,2===z y xn n n; so()2222252⎪⎭⎫⎝⎛++=a n n n m E z y x π()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--EeV For the next quantum state, which is empty, the quantum state is2,2,1===z y x n n n . Thisquantum state is at thesame energy, so35.2=FEeV(b)For 13 electrons, the 13thelectron occupiesthe quantum state 3,2,3===z y xn n n; so()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eV The 14thelectron would occupy the quantumAHA12GAGGAGAGGAFFFFAFAF state 3,3,2===z y xn n n.This state is at the same energy, so 746.5=FEeV_______________________________________3.38The probability of a state at E E EF ∆+=1being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kTE E E f Fexp 11exp 11111 The probability of a state at E E EF ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111or()⎪⎭⎫⎝⎛∆+=-kT E E f exp 11122 so ()()22111E f E f -= Q.E.D._______________________________________3.39(a)At energy 1E , we wantAHA12GAGGAGAGGAFFFFAFAF01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F FThis expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F For()⎪⎪⎭⎫ ⎝⎛-=kTE E F1exp 01.01 Then()100ln 1kT E E F +=orkT E E F 6.41+=(b)At kTEE F6.4+=,()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fwhich yields()01.000990.01≅=E f_______________________________________3.40 (a)AHA12GAGGAGAGGAFFFFAFAF ()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫ ⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0or5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kT which yields 740=T K _______________________________________3.41(a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b)At 1000=T K, 08633.0=kT eVThen()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫ ⎝⎛-+=E for 99.7% (d)At FE E =, ()21=E f for all temperaturesAHA12GAGGAGAGGAFFFFAFAF _______________________________________3.42(a)For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =,82.030.012.12=-=-E E F eVThen()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b)For 4.02=-E EFeV,72.01=-F E E eVAt 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E fAt 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E fAHA12GAGGAGAGGAFFFFAFAF_______________________________________3.43(a)At 1E E =()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f FAHA12GAGGAGAGGAFFFFAFAF or()61032.9-⨯=E fAt 2E E =,12.13.042.12=-=-E E F eVSo()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.012.1expor()191066.11-⨯=-E f (b)For 4.02=-E EF,02.11=-F E E eVAt 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E fAt 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫⎝⎛-⎪⎭⎫⎝⎛⨯kTE E kTF exp 1 or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kTE E kT E E kT dE E df FF(a)At 0=T K, For。