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MarkLundstrom
ECE-656 Fall20171
SOLUTIONS:ECE656Homework(Week4)MarkLundstromPurdueUniversity
1) Repeattheelectron-phononenergy-momentumconservationargumentsdiscussedin
Sec.2.5ofFundamentalsofCarrierTransport,butthistimeassumeelectronsingraphene.
Solution:Beginwithenergyandmomentumconservation: ′E = E ± ω (*) !′p = !p ± "!q (**)Usethedispersionofgraphenetowrite:
E = υF k =υF p
E2 = υF p( )2
E = p2
E υF2( ) =
p2
m* E( ) m* E( ) ≡ E υF
2( ) Usingthisdefinitionofaneffectivemassforgraphene,(*)becomes
′p 2
m* = p2
m* ± ω (*’)
whichissimilartotheparabolicbandresult.
Nowtakethedotproductof(**)withitselftofind:
!′p i!′p = ′p 2 = p2 ± 2"!p i
!q + "2q2 nowdividethroughby m*
′p 2
m* = p2
m* ±2!"p i
"qm* + !
2q2
m*
anduse(*’)
±!ω = ± 2!"p i
"qm* + !
2q2
m* = ± 2!pqcosθm* + !
2q2
m*
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MarkLundstrom
ECE-656 Fall20172
ECE656Homework(Week4)Solutions(continued)Solvingfor !
2q2 m* wefind
!2q2
m* = ∓ 2!pqcosθm* ± !ω
!β = 2 p ∓cosθ ± m*ω
2 pq⎡
⎣⎢
⎤
⎦⎥ ,
whichisastatementofenergyandmomentumconservationforelectronsingraphene.Nowusethedispersionofgrapheneandourdefinitionofeffectivemasstowrite
m*
p=
E υF2
E υF
= 1υF
whichcanbeusedtoexpressourrelationforenergy-momentumconservationas
!q =2 p ∓cosθ ± ω
2qυF
⎡
⎣⎢
⎤
⎦⎥
Notethatthisresultisalmostthesameasfortheparabolicbandresult(exceptforthefactorof2downstairs).Theconclusionswillbesimilar.2) Assumeatransitionrateoftheform:
S!p→ !′p( ) = Cδ E − ′E( )δ !′p − !p ∓ #!q( )
whereCisaconstant.Answerthefollowingquestionsassumingparabolicenergybands.
a) Deriveanexpressionfor
!q = q ,whichexpressesconservationofenergyandmomentum.
b) Usingtheresultsofa),determinetheminimumandmaximummagnitudeof !q .
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MarkLundstrom
ECE-656 Fall20173
ECE656Homework(Week4)Solutions(continued)
Solution:a)Beginwithenergyandmomentumconservation:
′E = E (*) !′p = !p ± "!q (**)Sincethebandareparabolic,(*)canbewrittenas:
′p 2
2m* =p2
2m* ± ω (*’)
Nowtakethedotproductof(**)withitselftofind:
!′p i!′p = ′p 2 = p2 ± 2"!p i
!q + "2q2 nowdividethroughby 2m*
′p 2
2m* =p2
2m* ±2!"p i
"q2m* + !
2q2
2m*
anduse(*’)
′p 2
2m* −p2
2m* = 0 = ± 2!"p i"q
2m* + !2q2
2m* = ± 2!pqcosθ2m* + !
2q2
2m*
Solvingfor !q wefind
!q = ∓2 pcosθ ,whichisastatementofenergyandmomentumconservationonforelasticscattering.b)Notethat q isthemagnitudeof
!q ,soitmustalwaysbegreaterthanorequaltozero.Thelargest q willoccurfor
Absorption: cosθ = −1, θ = π Emission: cosθ = 1, θ = 0
Ineithercase, !qmax = 2 p Thesmallest q occursfor θ = π 2 forbothabsorptionandemission.Ineithercase, !qmin = 0 Weconcludethatforelasticscattering:
0 ≤ !q ≤ 2 p
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MarkLundstrom
ECE-656 Fall20174
ECE656Homework(Week4)Solutions(continued)3) Thisproblemconcernselectronscatteringinbulk(3D)GaAs.Assumethattheoptical
phononenergyis ω0 = 35 meV.RecallthatGaAsisadirectgapsemiconductorandthattheLvalleys(along<111>)haveenergyminimathatare0.3eVabovetheΓ valleyminimum.ThefourXvalleys(along<100>)haveenergyminima0.5eVabovetheΓ valleyminimum.RecallthatΓ valleyelectronshavealighteffectivemassandthattheLandXvalleyelectronshavealarge(Si-like)effectivemass.Answerthefollowingquestions.
Solution:
a) Sketchthetotalelectronscatteringratevs.energyforelectronsintheΓ valley.
Labelallcriticalenergiesandgiveabriefexplanation(labelabsorptionandemissionprocessesseparately).AllenergiesshouldbereferredtothebottomoftheΓ valley,i..e. ECΓ = 0 .
Thescatteringrateisthesumof6differentprocesses;thelastfiveofthesehavethresholdsfortheironset.
1) POPAbsorption(ABS).Γ − Γ intravalleyscattering.2) POPEmission(EMS).Γ − Γ intravalleyscatteringplus1)3) OPABS. Γ − L intervalleyscatteringplus1)and2).4) OPEMS. Γ − L intervalleyscatteringplus1),2),and3).5) OPABS. Γ − X intervalleyscatteringplus1),2),3),and4).6) OPEMS. Γ − X intervalleyscatteringplus1),2),3),4),and5).
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MarkLundstrom
ECE-656 Fall20175
ECE656Homework(Week4)Solutions(continued)
b) SketchtheΓ toΓ electronscatteringratevs.energy.Labelallcriticalenergiesandgiveabriefexplanation(labelabsorptionandemissionprocessesseparately).
1) POPAbsorption(ABS).Γ − Γ intravalleyscattering.2) POPEmission(EMS).Γ − Γ intravalleyscattering
NotethatPOPABSisproportionalto N0 andPOPEMSisproportionalto N0 +1 ,soPOPEMSislargerinmagnitude.
c) SketchtheΓ toLelectronscatteringratevs.energy.Labelallcriticalenergiesand
giveabriefexplanation(labelabsorptionandemissionprocessesseparately).
3) OPABS. Γ − L intervalleyscattering.4) OPEMS. Γ − L intervalleyscattering
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MarkLundstrom
ECE-656 Fall20176
ECE656Homework(Week4)Solutions(continued)
NotethatEMSis,again,strongerthanABS.Alsonotethatthescatteringratesincreaseasthesquarerootofenergybecausetheygoesasthedensityoffinalstates(noPOPintervalleyscatteringbecauselargephononwavevectorsareneededforthistypeofscattering).
d) SketchoftheLtoΓ electronscatteringratevs.energy.Labelallcriticalenergies
andgiveabriefexplanation(labelabsorptionandemissionprocessesseparately).
1) OPABS. L − Γ intervalleyscattering.2) OPEMS. L − Γ intervalleyscattering
NotethatbothABSandEMSbeginatthesameenergybecauseelectronsatthebottomoftheL-valleycanscatterdowntotheΓ valleybyeitherabsorbingoremittingaphonon.Alsonotethatthemagnitudeofthescatteringratesissmallerthaninc)becausethefinalstatesinthiscasehaveavalleydegeneracyof1insteadof4asinc).Mostimportantly,notethatthescatteringratesdonotbeginatzero,becausetheyareproportionaltoaDOSinthefinalvalley,wheretheΓ valleyDOSisquitehigh. 1 τΓ−L ∝ DΓ E ± ω0( ) beginsatanenergy E = 0.3± ω0 ,wherethedensity-of-finalstates(intheΓ valley)isquitehigh.
e) SketchtheLtoLelectronscatteringratevs.energy.Labelallcriticalenergiesand
giveabriefexplanation(labelabsorptionandemissionprocessesseparately).
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MarkLundstrom
ECE-656 Fall20177
ECE656Homework(Week4)Solutions(continued)
AnelectronatthebottomofanL-valleycanabsorbaphononandscattertoanotherL-valley,butanelectronnearthebottomofanL-valleycannotscattertoanotherL-valleybyemission,becausetherearenoL-valleystatesavailableatthatenergy.
4) Thedeformationpotentialscatteringrateforopticalphononemission(ODPemission)
isdescribedby:
1τ= 2πD0
2
2ρω0
⎛
⎝⎜⎞
⎠⎟N0 +1( ) D3D E − ω0( )
2 .
Obtaintheenergyfluxrelaxationrateforthisscatteringprocess.
Solution:
Beginwiththedefinitionoftheenergyfluxrelaxationrate:
1τ FW
= S p, ′p( ) ′p ,↑∑
FWp( )− FW
′p( )FWp( )
⎡
⎣⎢⎢
⎤
⎦⎥⎥ (*)
Workontheterminbracketsandassumethattheinitialfluxisorientedalongthez-axis:
FWp( )− FW
′p( )FWp( )
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
Eυz − E − ω0( ) ′υz
Eυz
= 1−E − ω0( )
E′υz
υz
⎛
⎝⎜⎞
⎠⎟= 1−
′υz
υz
⎛
⎝⎜⎞
⎠⎟+ω0
E′υz
υz
⎛
⎝⎜⎞
⎠⎟
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MarkLundstrom
ECE-656 Fall20178
ECE656Homework(Week4)Solutions(continued)
FWp( )− FW
′p( )FWp( )
⎡
⎣⎢⎢
⎤
⎦⎥⎥= 1−
′υz
υz
⎛
⎝⎜⎞
⎠⎟+ω0
E′υz
υz
⎛
⎝⎜⎞
⎠⎟= 1−
′υz
υz
⎛
⎝⎜⎞
⎠⎟−ω0
E1−
′υz
υz
⎛
⎝⎜⎞
⎠⎟+ω0
E
Ifweassumethattheeffectivemassisconstant(parabolicenergybands),then
m*υz = pz andwefind:
FWp( )− FW
′p( )FWp( )
⎡
⎣⎢⎢
⎤
⎦⎥⎥= 1−
′pz
pz
⎛
⎝⎜⎞
⎠⎟+ω0
E−ω0
E1−
′pz
pz
⎛
⎝⎜⎞
⎠⎟
Nowinsertthisexpressionin(*)
1τ FW
= S p, ′p( )′p ,↑∑ 1−
′pz
pz
⎛
⎝⎜⎞
⎠⎟+ω0
E−ω0
E1−
′pz
pz
⎛
⎝⎜⎞
⎠⎟⎡
⎣⎢⎢
⎤
⎦⎥⎥
= S p, ′p( ) 1−′pz
pz
⎛
⎝⎜⎞
⎠⎟+ S p, ′p( ) ω0
E⎛⎝⎜
⎞⎠⎟′p ,↑
∑′p ,↑∑ − S p, ′p( )
′p ,↑∑ ω0
E1−
′pz
pz
⎛
⎝⎜⎞
⎠⎟
1τ FW
= 1τ m
+ω0
E⎛⎝⎜
⎞⎠⎟
1τ−ω0
E⎛⎝⎜
⎞⎠⎟
1τ m
1τ FW
= 1τ m
1−ω0
E⎛⎝⎜
⎞⎠⎟+ω0
E⎛⎝⎜
⎞⎠⎟
1τ
Sinceweareassumingthatphononemissionoccurs,thisonlyappliesfor E > ω0 .
5) TheODPscatteringratefor2Delectronsis:
1τ n, ′n
⎞
⎠⎟
a,e
= πD0
2
ρω0
⎛
⎝⎜⎞
⎠⎟N0 +
12
12
⎛⎝⎜
⎞⎠⎟
D2 D E ± ω0( )2
2 +δn, ′n
2⎛
⎝⎜
⎞
⎠⎟
Let ω0 = 1.1kBT ,assumetwosubbands,andplottheabsorptionandemissionscatteringratesvs.energyforanelectroninsubbandone.
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MarkLundstrom
ECE-656 Fall20179
ECE656Homework(Week4)Solutions(continued)Solution:Let’swritethescatteringrateas
1τ n, ′n
⎞
⎠⎟
a,e
= Γ0 N0 +12
12
⎛⎝⎜
⎞⎠⎟
2 +δn, ′n
2⎛
⎝⎜
⎞
⎠⎟
where
Γ0 =
πD0
2
ρω0
⎛
⎝⎜⎞
⎠⎟m*
2π2
N0 =
1eω0 kBT −1
= 1e1.1 −1
= 13−1
= 12
Nowworkoutthevariousscatteringrates:
1τ n, ′n
⎞
⎠⎟
a,e
= Γ0
12+ 1
2
12
⎛⎝⎜
⎞⎠⎟
2 +δn, ′n
2⎛
⎝⎜
⎞
⎠⎟ = Γ0 1 1
2⎛⎝⎜
⎞⎠⎟
2 +δn, ′n
2⎛
⎝⎜
⎞
⎠⎟
1τ1,1
⎞
⎠⎟
a
= Γ0
12
⎛⎝⎜
⎞⎠⎟
32
⎛⎝⎜
⎞⎠⎟= 3
4Γ0
1τ1,1
⎞
⎠⎟
e
= Γ0
32
⎛⎝⎜
⎞⎠⎟
32
⎛⎝⎜
⎞⎠⎟= 9
4Γ0
1τ1,2
⎞
⎠⎟
a
= Γ0
12
⎛⎝⎜
⎞⎠⎟
1( ) = 24Γ0
1τ1,2
⎞
⎠⎟
e
= Γ0
32
⎛⎝⎜
⎞⎠⎟
1( ) = 64Γ0
Totalscatteringratefor E > ε2
1τ1
⎞
⎠⎟ tot
= Γ0
34+ 9
4+ 2
4+ 6
4⎛⎝⎜
⎞⎠⎟= 5Γ0
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MarkLundstrom
ECE-656 Fall201710
ECE656Homework(Week4)Solutions(continued)
Thetotalscatteringrateforelectronsinsubband1isplottedbelow.Inthisplot,wetakethezeroofenergytobethebottomofthefirstsubband, E = ε1 = 0 .
6) UseargumentssimilartothoseinSec.2.2ofFCTandevaluatethemomentumrelaxationtimeforpiezoelectricscattering.
a) Showthatthescatteringpotentialis
U PZ =
eePZ
κ Sε0
u (2.29)ofFCT
where ePZ isthepiezoelectricconstantand u istheelasticwavedisplacement.Wehavereplacedthechargeonaelectron,q,withetoavoidconfusionwiththephononwavevector,
!q .
HINT:Beginwith D =κ Sε0E + ePZ
∂u∂x
b) Usethescatteringpotentialofparta)andevaluatethematrixelementforPZ
scattering.Showthattheresultis
H !′p , !p
2=
eePZ
κ Sε0
⎛
⎝⎜⎞
⎠⎟
2kBT
2cℓq2Ω
δ !′p − !p ∓ $!q( ) = Kq
2Aq
2δ !′p − !p ∓ $!q( )
c) Writeanexpressionforthetransitionrate, S
p, ′p( ) ,anddetermine Cq .
d) Evaluate 1 τ m assumingthatthescatteringiselastic.
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MarkLundstrom
ECE-656 Fall201711
ECE656Homework(Week4)Solutions(continued)
Solution:
a)Beginwith D =κ Sε0E + ePZ
∂u∂x (*)
Assumespacechargeneutrality: ∇iD =κ Sε0E + ePZ
∂u∂x
= 0
If, D ∝ eiqx ,thisimpliesthatD=0,sowefindtheelectricfieldas:
E = −
ePZ
κ Sε0
∂u∂x
= −iqePZ
κ Sε0
u
Thescatteringpotentialis
US = −e E dx∫ = e
ePZ
κ Sε0
u
U PZ = e
ePZ
κ Sε0
u
b)Evaluatethematrixelement
H ′p , p =
e−i ′p ir
Ω∫ US
ei ′p ir
Ωd 3r
Usetheresultsfromparta)
U PZ =
eePZ
κ Sε0
u = Kq Aqei!qi!r
Kq =
eePZ
κ Sε0 (eqn.2.59dofFCT)
H !′p , !p = Kq Aq
1Ω
ei !p− !′p ±"!qi!r( ) "∫ d 3r
Followingthetext,FCT,wefind
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MarkLundstrom
ECE-656 Fall201712
ECE656Homework(Week4)Solutions(continued)
H !′p , !p
2= Kq
2Aq
2δ !p − !′p ∓ #!q( ) eqn.(2.60ofFCT) (**)
Nowquantizethelatticevibrationsaccordingtoeqn.(2.71c)
Aq
2→ !
2ρΩω q
Nω + 12∓
12
⎛⎝⎜
⎞⎠⎟
andwefinallywritethematrixelement(**)as
H !′p , !p
2= Kq
2Aq
2δ !p − !′p ∓ #!q( ) = eePZ
κ Sε0
⎛
⎝⎜⎞
⎠⎟
2#
2ρΩω q
Nω + 12∓
12
⎛⎝⎜
⎞⎠⎟δδ !p − !′p ∓ #!q( ) (***)
c)Beginwiththetransitionrate
S !p, !′p( ) = 2π
"H !′p , !p
2δ !p − !′p ∓ "!q( )δ ′E − E ∓ "ω( )
Nowuse(***)forthematrixelementandeqn.(2.66)FCTforthetwodeltafunctionstofind
S !p, !′p( ) = Cq Nω + 1
2∓
12
⎛⎝⎜
⎞⎠⎟δ ′E − E ∓ #ω( )
where
Cq =
2π!2
eePZ
κ Sε0
⎛
⎝⎜⎞
⎠⎟
2!
2ρΩω1!υq
= 2π!2
eePZ
κ Sε0
⎛
⎝⎜⎞
⎠⎟
2!
2ρΩωm*
!pq× q
q
Cq =
eePZ
κ Sε0
⎛
⎝⎜⎞
⎠⎟
2πm*
!ρυSq2 pΩ (eqn.(2.73d)ofFCT
Here,wehaveused: ω q =υS and p = m*υ forthemomentumoftheelectron.
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MarkLundstrom
ECE-656 Fall201713
ECE656Homework(Week4)Solutions(continued)d)Beginwitheqn.(2.80)inFCT.Assume
ωυq
=υS
υ<<1 (averageelectronvelocitymuchlessthanthephononvelocity,so
(2.80)becomes
1τ m
= Ω4π 2 Nω + 1
2∓
12
⎛⎝⎜
⎞⎠⎟qmin
qmax
∫ Cq
"q2 p"q3
pdq
Nowassumeequipartition:
N ≈ N +1=
kBTω
1τ m
=πm*e2ePZ
2 kBTcℓκ Sε0 2 p3( ) q dq
qmin
qmax
∫
Acousticphononscatteringiselastic: qmax = 2 p and qmin = 0
q dq
qmin
qmax
∫ =qmax
2
2−
qmin2
2= 2 p2
Finally,thescatteringratebecomes:
1τ m
=πm*e2ePZ
2 kBTcℓκ Sε0 p
∝ 1E
Themomentumrelaxationratedecreaseswithenergy–asexpectedforanelectrostaticscatteringmechanisms.
7) Foralloysofcompoundsemiconductorslike AlxGa1-xAs ,microscopicfluctuationsinthe
alloycomposition,x,produceperturbationsinthebandedges.Thetransitionrateforallowscatteringis
S p, ′p( ) = 2π
2
3π 2
16⎛⎝⎜
⎞⎠⎟ΔU
2
NΩδ ′E − E( )
whereNistheconcentrationofatomsand
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MarkLundstrom
ECE-656 Fall201714
ECE656Homework(Week4)Solutions(continued)
ΔU = x 1− x( ) χGaAs − χAlAs( ) with χ beingtheelectronaffinity.a) Explainwhythealloyscatteringratevanishesatx=0andatx=1.
b) Deriveanexpressionfor τ m foralloyscattering.
Solution:a)Forx=0,wehaveGaAs,whichisnotanalloy,sothereisnoalloyscattering.Forx=1,wehaveAlAs,whichusnotanalloy,sothereisnoalloyscattering.Thestrongestallowscatteringwilloccurwhenx=½.b)
S p, ′p( ) = C
Ωδ ′E − E( )
where
C = 2π2
3π 2
16⎛
⎝⎜⎞
⎠⎟ΔU
2
N
Sincethereisnodependenceonphononwavevector,thescatteringrateandmomentumrelaxationratesshouldbeequal.Sincethereisnodependentonβ ,wedonotneedtoworryaboutenergy-momentumconservationandcandothesumsimply.
1τ= 1τm
= S p, ′p( ) ′p ,↑∑ = C 1
Ωδ ′E − E( )
′p ,↑∑
Werecognizethesumasone-halfofthedensity-of-states,
1τ E( ) =
1τm E( ) = C
D3D E( )2
τm E( )∝ E−1/2
Whichispowerlawscatteringwithacharacteristicexponentof s = −1/ 2 .
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MarkLundstrom
ECE-656 Fall201715
ECE656Homework(Week4)Solutions(continued)
8) Acousticphononscatteringwasassumedtobeelasticwhenworkingoutthemomentumrelaxationrateineqn.(2.84)ofFCT.Repeatthecalculationbutdonotassumeelasticscattering.Showthattheresultisnearlyequaltoeqn.(2.84)nearroomtemperature.
Solution:Thetransitionrateforphononscatteringis:
S !p, !′p( ) = 2π
"Kq
2Aq
2δ !p − !′p ∓ "!q( )δ ′E − E ∓ "ω( )
Thetwodelta-functionscanbereplacedbyonethatexpressesenergyandmomentumconservation(eqn.(2.66)ofFCT):
δ !p − !′p ∓ #!q( )δ ′E − E ∓ #ω( )→ 1
#υqδ ±cosθ + #q
2 p∓ω q
υq⎛
⎝⎜
⎞
⎠⎟
sothetransitionratebecomes:
S !p, !′p( ) = 2π
"2υqKq
2Aq
2δ ±cosθ + "q
2 p∓ω q
υq⎛
⎝⎜
⎞
⎠⎟,
whichiseqn.(2.67)ofFCT.ForADPscattering:
Kq
2= q2DA
2 (eqn.(2.59a)ofFCT)
Aq
2→ !
2ρΩω q
Nωq+ 1
2∓
12
⎛⎝⎜
⎞⎠⎟ (eqn.(2.71c)ofFCT)
Sowefind:
S !p, !′p( ) = πm*DA
2
"ρ pυS
1Ω
Nωq+ 1
2∓
12
⎛⎝⎜
⎞⎠⎟δ ±cosθ + "q
2 p∓ω q
υq⎛
⎝⎜
⎞
⎠⎟
or
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MarkLundstrom
ECE-656 Fall201716
ECE656Homework(Week4)Solutions(continued)
S !p,
!′p( ) = Cq Nωq+ 1
2∓
12
⎛⎝⎜
⎞⎠⎟δ ±cosθ + #q
2 p∓ω q
υq⎛
⎝⎜
⎞
⎠⎟ (eqn.(2.72FCT)
Cq =
πm*DA2
!ρ pυS
1Ω (eqn.(2.73a)FCT)
FollowingFCT,wewritethemomentumrelaxationrateas:
1τ m
= S!p,!′p( )
!q ,↑∑ ∓#qcosθ( )
p= Ω
8π 3 dφo
2π
∫ S!p,!′p( ) ∓#qcosθ( )
p−1
+1
∫0
∞
∫ d cosθ( )q2dq
or
1τ m
= Ω4π 2 Cq Nωq
+ 12∓
12
⎛⎝⎜
⎞⎠⎟δ ±cosθ + "q
2 p∓ω q
υq⎛
⎝⎜
⎞
⎠⎟∓"qcosθ( )
p−1
+1
∫0
∞
∫ d cosθ( )q2dq
Nowassumeequipartition: Nωq
≈ Nωq+1≈
kBT!ω q
andusetheexpressionfor Cβ .
1τ m
=m*DA
2kBT4π!ρ p2υS
2 δ ±cosθ + !q2 p∓ω q
υq⎛
⎝⎜
⎞
⎠⎟ ∓cosθ( )
−1
+1
∫0
∞
∫ d cosθ( )q2dq
Integrateoverβ first:
1τ m
=m*DA
2kBT4π!ρ p2υS
2
!q2 p∓ω q
υq⎛
⎝⎜
⎞
⎠⎟
qmin
qmax
∫ q2dq =m*DA
2kBT4π!ρ p2υS
2
!q2 p∓υS
υ⎛⎝⎜
⎞⎠⎟qmin
qmax
∫ q2dq
wherewehaveusedtheacousticphonondispersionforthelaststep.
1τ m
=m*DA
2kBT4π!ρ p2υS
2
!2 p
qmax4
4−
qmin4
4
⎛
⎝⎜⎞
⎠⎟∓υS
υqmax
3
3−
qmin3
3
⎛
⎝⎜⎞
⎠⎟⎡
⎣⎢⎢
⎤
⎦⎥⎥
Page 17
MarkLundstrom
ECE-656 Fall201717
ECE656Homework(Week4)Solutions(continued)Assume qmax >> qmin and qmax
4 >> qmax3
1τ m
=m*DA
2kBT8πρ p3υS
2
qmax4
4
⎛
⎝⎜⎞
⎠⎟
Energyandmomentumconservationgives(seeFCT,eqn.2.54)
!qmax = 2 p 1±
υS
υ⎛⎝⎜
⎞⎠⎟
qmax4
4= 4 p4
!4 1±υS
υ⎛⎝⎜
⎞⎠⎟
4
sothemomentumrelaxationratebecomes:
1τm
=m* pDA
2kBT2π4ρυS
2 1±υS
υ⎛
⎝⎜⎞
⎠⎟
4
=m* pDA
2kBT2π4c
1±υS
υ⎛
⎝⎜⎞
⎠⎟
4
Recallthatthe3Ddensity-of-statesis:
D3D E( ) = 2m*( )3/2
4π 23 E1/2
whichcanbeusedtore-expressthemomentumrelaxationrateas:
1τ m E( ) =
m*DA2kBTc
D3D E( )2
1±υS
υ⎛⎝⎜
⎞⎠⎟
4
,
whichisalmostexactlytheresultforelasticscattering(eqn.(2.84)ofFCT)sinceforatypicalelectron υ <<υS .