Plasmon Charge Density Probed By Ultrafast Electron Microscopy
e = electron charge = 1.6x10-19 C me = electron mass = 9.1x10-31 ...
Transcript of e = electron charge = 1.6x10-19 C me = electron mass = 9.1x10-31 ...
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Exam3DEX1:Physicsofnewenergy26-1-2016van9:00-12:00
PLEASEREADTHESEINSTRUCTIONSFIRST!Inthisexamwewouldliketosummarizewithyoutheissuesdiscussedinthisclassby4exercises:
1. Energyingeneral:theenergyproblem,consumptionandstorage.2. Thermodynamics:howthermodynamicsimposesalimitonthemaximumefficiency
toconvertheatintowork.3. Fusionpower:thebasicprinciplesanditschallengestorealiseafusionpowerplant4. Solarcells:itsstructureandtheoperatingprinciple
AllfourquestionsareposedinEnglish.YoucanchooseyourselftoanswerineitherEnglishorDutch.Foreachofthesubquestionsthenumberofpointsthatcanbescoredisindicated.Thetotalnumberofpointsis100.ThefinalresultFiscalculatedaccordingtoF=1.0+0.09x(numberofpointsscore)androundedto1decimal.Wealsoprovideanindicativetimeneededtocompletetheexercise(justourownestimate,maybeithelpsyoutocheckwhetheryourpaceissufficient). Theuseofcalculatorsisallowed,butanyotherbooks,phones,laptops,internetaccess,formularyisstrictlyprohibited.Belowyoufindsomeconstants,whichyoumightneedforsolvingsomeexercises.(Notethatyoudonotnecessarilyneedallofthem,itisjustastandardlist).Constants e= electroncharge = 1.6x10-19 Cme= electronmass = 9.1x10-31 kgmp protonmass = 1.67x10-27 kgc= speedoflight = 2.99x108 m/sε0= vacuumpermittivity = 8.85x10-12 F/mμ0= magneticpermeability= 1.26x10-6 Vs/Amh= Planckconstant = 6.63x10-34 JskB= Boltzmannconstant = 1.38x10-23 J/Kg= gravitationofEarth = 9.81 m/s2
NA=Avogadro’snumber = 6.02x1023 mol-1R= Gasconstant = 8.31 J/(molK)(=8.31Pam3/(molK)atm= atmosphere = 1.01x105 Paρair= densityofair = 1.3 kg/m3
ρwater=densityofwater = 1000 kg/m3kw= thermalconductivitywood= 0.1 W/(mK)kr= heatconductivityrubber= 0.15 W/(mK)atomicmass(amu):hydrogen=1,helium=4,carbon=12,oxygen=16
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1. Energy–General–15pts–estimatedtime:25minutesa) [5pts]EnergyProblem:Whatarethethreemainfactorsdeterminingtheenergy
problemwhichwewillfaceinthenearfuture?Giveonesolutiontothis.Inwhichwaycansciencecontributetothis?Provideoneexample.
Answer:1pt:growthofpopulation1pt:energyuseindevelopingcountriesisincreasing1pt:fossilefuelsupplyisdecreasing1pt:CO2à4points,butmax3forthispart.Solution:1pt:userenewableenergysources,increasedefficiency,decreaseenergyconsumptionExampleScience:(manypossibilities,ownjudgement)à1pt.(e.q.developnewsources.Likenuclearfusion,newhighefficiencysolarcells,developnewbatterieselectricalcars….)b) [5pts]EnergyUse:Thelargestenergyconsumptionbyhouseholdsisfortransport
(cars)andheating.Let’shavealookatthelatterone.Whatismostefficientenergyuseforcooking:anelectricalcookingplateoragascooker(gasstove).Why?Calculatethepowerneededtoraisethetemperatureofa5litrepanfilledwithwaterfrom5°Cto100°C(theboilingtemperature)in3minutes.Whatisthepriceforthisifitisdoneelectrically(25ct/kWh,40%efficiency)?
1pt:gasisbestchoice:,efficiencytoproduceelectricityislow(40%),atleastforfossilpowerplant:a)chemicalenergyàb)thermalenergyàc)electricalenergyàd)thermalenergy:stepcàdveryinefficient 2pt:P=E/t=(c.m.dT)/t=4.18x5x95kJ/180s=11kW2pt:E=P.t/eff=c.m.dT=4.18x5x95/0.4=5MJ=(5/3.6)kWhàprice=5/3.6*0.25euro=0.35euroc) [5pts]Energystorage:Inanelectriccartheenergyisstoredinabattery.Typicallythis
carcandriveonly100kmonfullelectricenergy,whereasaconventionalcaronpetrolcandriveeasily500kmonasingletank.Whatistheessentialdifferenceintheenergystorage?Howisthisdifferentforfuelcellsbasedonhydrogen?Explain
2pt:Batterieshavealowerenergydensityperkgcomparedtopetrol 3pt:Fuel(H)hasevenmuchhigherenergydensityperkgthanpetrol,butthequestionishowhydrogencanbetransported:asafluid(alotofmassneededforkeepingthefuelcool)asagas(heavytankneededtokeepitathighpressure),orashydride(alsoheavy).Sofinallystillnotahighenergydensityperkg.
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2. Thermodynamics–15pts–estimatedtime:25minutes
Two moles of an ideal mono-atomic gas (cV= 12.47 J/(mol⋅K), cp=20.78 J/(mol⋅K)) follow athermodynamiccycleaccordingtothepathDgAgBgCgD.ThestepsDAenBCareisochoric.ThestepsABandCDareisothermal.InstateD,thepressureis2⋅105Paandthetemperatureis360K.InstateB,thevolumeisVB=3VDandthepressureisPB=2PC.
a) [2pts]DrawthePVdiagramforthecycleDABCD.b) [8pts]Calculatetheworkandtheheatforeachstepofthecycle.c) [5pts]Calculatetheefficiencyofthecyclicprocess.
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3. Fusion–35pts–estimatedtime50minutesFusionEnergyisaverypromisingnewenergysource:
• Thefuelisabundantlypresentforthousandsofyears• NoCO2emission• Inherentsafe• Largescale
Nevertheless,westilldonothaveaworkingfusionreactoryet.Beforethiswillberealisedmanychallengeshavetobeovercome.Let’shaveacloserlookatafewofthose.
a) (6pts)Challenge1:Thefuel:theeasiestfusionreactionistheonebetween
deuteriumandtritiumnuclei.However,inaplasmamixofdeuteriumandtritiumalsootherfusionreactionsarepossible.Giveatleasttwootherreactionsthatwilloccurinthisplasmaandusethepicturebelowtoestimatetheamountofenergyreleasedinthisreaction.
Differentpossibilities(energiesonlyapproximate,countvaluecorrectifdifferenceislessthen1MeV)D+DàT+p+4MeVD+Dà3He+n+3.3MeVT+Tà4He+2n+11MeV3He+Dà4He+p+18MeV3He+Tà4He+D(orp+n)+12-14MeV3He+3Heà4He+p+13MeV
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b) (3pts)Challenge2:TheBurnCondition.Tohavenetfusionpowerweneedtofulfilthefollowingcondition:Morepowershouldbeproducedbyfusionreactionsthanweneedtoprovidetoheattheplasma.Thisleadstothefusiontripleproduct.Whicharethethreeparameterswhichdeterminethiscondition?
nxTxτE>constantn=densityT=temperatureτE=energyconfinementtime
c) (7pts)Challenge3:Magneticconfinement.Weneedtoconfinethehotplasmawith
amagneticfieldBof5T.• (2pt)calculatetheaveragespeedofadeuteriumionina15keVplasma• (2pt)calculatetheradiusatwhichthisiongyratesaroundthemagneticfield
line.• (3pt)Sketchthemotionofthisioninthefollowingthreesituations:theionis
movingparalleltothemagneticfield,perpendiculartothemagneticfieldandoblique(i.e.underanangle)tothemagneticfield.Indicateclearlythedirectionsofmagneticfieldandvelocity.
Averagevelocity:0.5*md*v2=kTàv=√(2kT/md)
=√(2x15000*1.6.10-19/2x1.6x10-27)=√1011=3.3x105m/s
(Note:kTisenergyunit.Thisisgivenhereas15keV,toconvertthisbacktoJoule,youhavetomultiplybye=1.6x10-19J/eV)Radius:rlarmor=mxv/(qxB)=2x1.6x10-27x3.3x105/(1.6x10-19x5)=1.3x10-3mMotion(1pteach):- parallel:noforce,ioncontinuestomovestraighton- perpendicular:circulatingthemagneticfield.Checkalsodirection(-0.5ptifthisis
notindicatedorincorrect! - oblique:combinationofboth:spiral(alsocheckdirection)
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d) (4pts)Challenge4:Temperature.FusionoftheD-Treactioniseasiestat15keV
(equaltoapproximately165MillionK).Thiscanbedonebyeitherinjectingparticlesorinjectingelectromagneticwaves
• (2pts)Whichparticlesarebesttoinjecttobesttoinjecttoheattheplasma.Explainthisheatingprinciplein2-3sentences.
• (2pts)toheatthedeuteriumnucleiintheplasma,whichfrequencyofelectromagneticwaveisbesttoinjectinareactorwithamagneticfieldof5T?
Besttoinjectfuelparticles(0.5pt)DeuteriumortritiumTheseshouldbeneutral(0.5pt)withanenergymuchhigherthantheplasmatemperature(0.5pt).Neutralparticlesareinjectedintotheplasma,particlesgetionizedandthenfollowthemagneticfieldlines.Thentheycollidewiththeplasmaparticlesandtransfertheirkineticenergytotheplasma(0.5pt)Frequencytoheatdeuterium:f=eB/(2pi.Md)=1.6x10-19x5/(2x3.14x2x1.67x10-27)Hz=38MHz(note:ifangularfrequencyisgiven,ie..withoutfactor2pithisisalsocorrectiftherightunitsisgiven)
e) (7pts)Challenge5:Wallpowerload.Theaimofafusionpowerplantistoproduceelectricity.Assumewehaveafusionpowerreactorofthetypetokamak,producing4GWoffusionpower(fromtheD-Treaction).Themajorradiusis6meter,theminorradius=2meter.Assumethetorushasacircularcrosssection.
• (2pt)Describehowthisfusionpowerisconvertedtoelectricityandgiveacoarseestimateofthetheelectricoutputpowerofthisreactor.
• (3pt)Thefusionpowerisdistributedbetweentheneutronsandthealphaparticles.Let’sconcentrateontheneutrons.Whatisthepoweroftheseneutrons?Whereisitdeposited?Besidesthepowerproduction,whatotherusedotheneutronshave?
• (2pt)Calculatethepowerwallload(inMW/m2)asaresultoftheneutrons.Fusionpoweriskineticenergyofneutrons(80%)andalphaparticles(20%).Thiswillbedepositedinthereactorwall(neutrons)orinthedivertor(plasmalossesincludingalphaparticlepower).Coolingpipesinreactorwallandivertorwillheatupthewater,generatingsteam,rotatingturbine,producingelectricityinconventionalmanner.The4GWoffusionpowerwillleadwithaconversionefficiencyofabout25%to1GWofelectricpower.PowerofNeutrons:80%of4GW=3.2GWWheredeposited:inreactorwallOtherreaction:produceTritium:n+6,7LiàHe+T+(n)Powerwallload:Totalarea=4π2Rxa=474m2àpowerload=3.2GW/474m2=6.8MW/m2
f) (4pts)Challenge6:control.Tocontroltheplasmaweneedtomeasureinrealtime
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someplasmaparameters,liketheplasmatemperature.• (2pt)Describeameasurementtechnique(physicsprinciple)todothis.• (2pt)Useasketchtoillustratethisandindicatethemainhardware
component.Temperature:a) ThomsonScattering:injectionoflaser.Laserphotonsscatteron
electrons.PhotonsaredetectedataDopplershiftedfrequency.ThisDopplershiftisameasureoftheelectronvelocity.Fullspectrumrepresentsvelocitydistribution.Itwidthdeterminesthetemperature.
b) Components:laser,lensestodetect,spectrometer.(AlternativelyECEcanbeexplained:microwaves,blackbodyradiation,intensityproportionaltotemperature)g) (4pts)Challenge7:Fusionisonlycompetitivewithalternativeenergysourcesifthe
costs/kWharecomparable.Afusionreactorisextremelyexpensive(about10BillionEuro’s)becauseoftheinfrastructure.
• (2pt)Giveoneargumentwhythecosts/kWhcanstillbereasonable• (2pt)Alsogiveanargumentwhythecosts/kWhareatriskofrisingtoohigh
(assumethatwewillbeabletoreachtherequiredfusioncriteriontoproducenetenergy).
Investmentscostareverylarge,butrunningcostverylow,soeffectivelythecost/kwHcanstillbereasonable.Systemiscomplex.Ifthereactorisdown(lowavailability)thenthecosts/kwHrise.
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4. SolarCells–35pts–estimatedtime50minutes
Analyseandcarefullydescribeinyourownwordsthestructureofacrystallinesiliconsolarcellbyaddressingthefollowingquestions:a) (4pts)Makeacompletesketchofac-Sisolarcellandindicateitscomponents.
b) (3 pts) The p-n junction in a solar cell is essential because it takes care of the
separationof theelectronsandholes,once theyhavebeengenerated bysunlightabsorption in the semiconductor. Make an accurate sketch of the energy banddiagram of a p-n junction in a solar cell and indicate in which directions theelectrons andholeswill be transportedwhen they experience the electric fieldofthejunction.
c) (6 pts) Explainwhy it is not possible to convert all the photocurrent (i.e. all thephoto-generatedcharges)inusefulelectricalcurrentfromasolarcell.Makeuseof
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theelectricalcircuitofasolarcell.
Thephotocurrent(Iph)anddiodecurrent(Id)areinoppositedirectionssincethefirstrelateswiththecurrentofthechargesgeneratedbysunlightabsorptionandbeingsweptacrossthejunctionviadriftmechanism,whilethediodecurrentisarecombinationcurrentduetothediffusionofelectronsfromntopandholesfrompton.Thisrecombinationdiodecurrentisalwayspresentbecausetheseparationofchargesinducesaforwardbias(p-typesiliconbecomesp-polarizedandn-typesiliconbecomesn-polarized)whichisthenresponsibleforthedevelopmentofthediffusion(diode)recombinationcurrent.ThisisthereasonwhyapartofIphwillbealwayslostintheformofdiodecurrentandtheusefulcurrentIwillbealwayslessthanIph.
d) (4 pts) According to the Matlab simulation, a crystalline silicon (c-Si) wafer with a
thickness of just 200-250 µm is sufficient to quantitatively absorb the solar light.However,thisisindiscrepancywiththevalueofthepenetrationdepthofthesunlightinto c-Si,which suggests thatwewouldneedat least a thicknessof1000µmto takeadvantageofallphotonswithenergyabovethec-Siband-gap(1.1eV).Canyouexplainthisdiscrepancybytakingintoconsiderationthestructureofthesolarcell?
Thediscrepancybetweenthetworesultsarisesfromthefactthatthefirstansweristheresultoftheoptimizationinthedesignofthesolarcell.Specifically,wecanuse“only”200-250µmofc-Si because the optimized structure of the cell includes ametal (preferably of Au, Al or Cu)back contact which serves also as back-reflector and reflects back into the c-Si absorber allthosephotonswhichhavenotbeenabsorbedyetfromc-Si.
e) (5pts)Theefficiencyofthesunlight-to-electricityconversionprocess is limitedbythe2nd law of thermodynamics. Why? Furthermore, provide an estimate of thethermodynamic limit values, according to the twomodelswhichwedescribedduringclasses.Whydothetwomodelsprovidedifferentthermodynamicefficiencyvalues?
Thethermodynamiclimitisduetothefactthatwecannotconvertwith100%efficiencytheheat(photons)fromthesunintousefulelectricalwork,otherwisewewouldgoagainstthe2ndlawofthermodynamics.AccordingtotheShockley-Queissermodel,thislimitisequalto44%.TheDetailedBalancemodelmodelmakesamoreaccuratecalculation:31%.TheDetailedBalancemodelcorrectstheShockleyQueissermodelby:1)consideringthatthecurrentisalwayslowerbecausefundamentalchargerecombinationoccurs;2)themaximumvoltagewecangainfromasolarcellwillbeneverequaltothebandgapvalueofthesemiconductor,butequaltotheVoc(Voc<Bg).
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f) (8pts)Providealistofalllossmechanismsoccurringinacommercialsolarcellandexplainthemindetailbymakinguseofsketches.
Spectralmismatchisduetothemismatchbetweenthesunlightspectrumandthebandgapofthesemiconductor.PhotonswithlowerenergythanBgwillnotbeabsorbedwhilephotonsabovetheBgwillbeabsorbedbuttherestoftheenergywillbelostasheat.Shadowingandreflectionlossesarerelatedtotheopticalpathofthelightwhenhittingacell.Lightreflectingonthemetalcontactswillnotbeabsorbedandlightcanbereflectedalsoattheglasssurfacetoo.Chargescanrecombinebecauseelectronscanfallbackintothevalencebands.Furthermorethepresenceofdefectsatsurfacesandinterfacestrapchargestoo.Afewsketchesareherepresented:
g) (5pts)Whichisthemajorlossmechanism?Canyouprovideandexplainindetailone
approachtoaddressthisspecificlossmechanism?Makealsouseofasketchtopresent
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yourapproach.Spectralmismatchisthemajorlossinsolarcells.Oneapproachistotakeadvantageofsolarcellsmadeoutofmultiplejunctions(tandem,triple)wheretwoorthreesemiconductorswithdifferentbandgapvaluesareusedtobetterexploitthesolarspectrum.Anotherapproachistoconvertphotonsnotusefulforthespecificband-gaptoenergyrangeswhichhaveabettermatchwiththebandgapvalue:wetalkthenaboutconvertors(upconvertersanddownconverters).