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。