第46届国际物理奥林匹克竞赛(IPhO2015)理论试题与解答(2)

TheoreticalTask1(T-1):Solutions

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Alternativemethodsleadingtothesameresultareacceptable.η(2.27)=0.457

(A8)ThebandgapofpuresiliconisEg=1.11eV.Calculatetheef?ciency,ηSi,ofasiliconsolarcell

usingthisvalue.

Solution:

=

1.11×1.60×10?19

xg1.38×10?23×5763=2.23ηxgSi=6

(x2g+2xg+2)e?xg

=0.457Inthelatenineteenthcentury,KelvinandHelmholtz(KH)proposedahypothesistoexplainhowthe

Sunshines.Theypostulatedthatstartingasaverylargecloudofmatterofmass,M??,andnegligibledensity,theSunhasbeenshrinkingcontinuously.TheshiningoftheSunwouldthenbeduetothereleaseofgravitationalenergythroughthisslowcontraction.

(A9)LetusassumethatthedensityofmatterisuniforminsidetheSun.Findthetotalgravitational

potentialenergy,?,oftheSunatpresent,intermsofG,M??andR??.

Solution:

M??

ThetotalgravitationalpotentialenergyoftheSun:?=?

??

Gmdm0

r

Forconstantdensity,ρ=

3M??

34πR3m=4

3

πrρ

dm=4πr2ρdr

???=???

R??

G

??

43??????dr5203πrρ4πr2ρ16π2Gρ2R??3GM??r=?35=?5R??

(A10)EstimatethemaximumpossibletimeτKH(inyears),forwhichtheSuncouldhavebeenshin-ing,accordingtotheKHhypothesis.AssumethattheluminosityoftheSunhasbeenconstant

throughoutthisperiod.

Solution:

τ??KH=

L??

2

τKH

=3GM??5R=1.88×107years??L??

TheτKHcalculatedabovedoesnotmatchtheageofthesolarsystemestimatedfromstudiesofmete-orites.ThisshowsthattheenergysourceoftheSuncannotbepurelygravitational.

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[0.3]

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TheoreticalTask1(T-1):Solutions

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B.NeutrinosfromtheSun:

In1938,HansBetheproposedthatnuclearfusionofhydrogenintoheliuminthecoreoftheSunisthesourceofitsenergy.Thenetnuclearreactionis:

41H?→4He+2e++2νe

The“electronneutrinos”,νe,producedinthisreactionmaybetakentobemassless.TheyescapetheSunandtheirdetectiononEarthcon?rmstheoccurrenceofnuclearreactionsinsidetheSun.Energycarriedawaybytheneutrinoscanbeneglectedinthisproblem.

(B1)Calculatethe?uxdensity,Φν,ofthenumberofneutrinosarrivingattheEarth,inunitsof

m?2s?1.Theenergyreleasedintheabovereactionis?E=4.0×10?12J.AssumethattheenergyradiatedbytheSunisalmostentirelyduetothisreaction.

Solution:

4.0×10?12J?2ν

?ΦL??ν=4πd2×2=3.85×10264π×(1.50×1011)2×4.0×10?12

×2=6.8×1014m?2s?1

.??δETravellingfromthecoreoftheSuntotheEarth,someoftheelectronneutrinos,νe,areconvertedto

othertypesofneutrinos,νx.Theef?ciencyofthedetectorfordetectingνxis1/6thofitsef?ciencyfordetectingνe.Ifthereisnoneutrinoconversion,weexpecttodetectanaverageofN1neutrinosinayear.However,duetotheconversion,anaverageofN2neutrinos(νeandνxcombined)areactuallydetectedperyear.

(B2)IntermsofN1andN2,calculatewhatfraction,f,ofνeisconvertedtoνx.

Solution:

N1=??N0

Ne=??N0(1?f)Nx=??N0f/6N2

=

Ne+Nx

OR

(1?f)Nf

1+6N1=N2

?f=6??N2

??51?

N1

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TheoreticalTask1(T-1):Solutions

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Inordertodetectneutrinos,largedetectors?lledwithwaterareconstructed.Althoughtheinteractionsofneutrinoswithmatterareveryrare,occasionallytheyknockoutelectronsfromwatermoleculesinthedetector.Theseenergeticelectronsmovethroughwaterathighspeeds,emittingelectromagneticradiationintheprocess.Aslongasthespeedofsuchanelectronisgreaterthanthespeedoflightinwater(refractiveindex,n),thisradiation,calledCherenkovradiation,isemittedintheshapeofacone.

(B3)Assumethatanelectronknockedoutbyaneutrinolosesenergyataconstantrateofαper

unittime,whileittravelsthroughwater.IfthiselectronemitsCherenkovradiationforatime?t,determinetheenergyimpartedtothiselectron(Eimparted)bytheneutrino,intermsofα,?t,n,me,c.(Assumetheelectrontobeatrestbeforeitsinteractionwiththeneutrino.)

Solution:

WhentheelectronstopsemittingCherenkovradiation,itsspeedhasreducedtovstop=c/n.Itstotalenergyatthistimeis

E=??mec2

=√nmec2stop

1?v2stop/c

2n2?1Theenergyoftheelectronwhenitwasknockedoutis

E=α?t+√nmec2

start

n2?1Beforeinteracting,theenergyoftheelectronwasequaltomec2.Thus,theenergyimpartedbytheneutrinois

Eimparted=Estart?mec2=α?t+

??

√n?1??

mec2n2?1ThefusionofHintoHeinsidetheSuntakesplaceinseveralsteps.Nucleusof7Be(restmass,mBe)isproducedinoneoftheseintermediatesteps.Subsequently,itcanabsorbanelectron,producinga7

Linucleus(restmassmLi

7

Be+e??→7Li+νe.

WhenaBenucleus(mBe=11.65×10?27kg)isatrestandabsorbsanelectronalsoatrest,theemittedneutrinohasenergyEν=1.44×10?13J.However,theBenucleiareinrandomthermalmotionduetothetemperatureTcatthecoreoftheSun,andactasmovingneutrinosources.Asaresult,theenergyofemittedneutrinos?uctuateswitharootmeansquarevalue?Erms.

(B4)If?Erms=5.54×10?17J,calculatethermsspeedoftheBenuclei,VBeandhenceestimateTc.

(Hint:?Ermsdependsonthermsvalueofthecomponentofvelocityalongthelineofsight.)

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TheoreticalTask1(T-1):Solutions

Solution:

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Moving7BenucleigiverisetoDopplereffectforneutrinos.Sincethefractionalchangeinenergy(?Erms/Eν～10?4)issmall,theDopplershiftmaybeconsideredinthenon-relativisticlimit(arelativistictreatmentgivesalmostsameanswer).Takingthelineofsightalongthez-direction,

?Ermsvz,rms

=Eνc

=3.85×10?4

1VBe=√3c

?VBe=

√

3×3.85×10?4×3.00×108ms?1=2.01×105ms?1.

Theaveragetemperatureisobtainedbyequatingtheaveragekineticenergytothethermalenergy.

312

mBeVBe=kBTc22?Tc=1.13×107K

Q T-2 The Extremum Principle1

?? I A A The Extremum Principle in Mechanics II Page 1 of 2 (Total Marks: 10)

?? Consider a horizontal frictionless plane shown in Fig. 1. It is × divided into two regions, I and II, by a line AB satisfying the ?? equation . The potential energy of a point particle of mass in ?? ?? region I is while it is in region II. The particle is sent ?? from the origin O with speed along a line making an angle with the x-axis. It reaches point P in region II traveling with speed along a ?? B line that makes an angle with the x-axis. Ignore gravity and ?? ?? relativistic effects in this entire task T-2 (all parts). Figure 1 A1 Obtain an expression for in terms of , and . A2 Express in terms of , and . We define a quantity called action ∫ , where is the infinitesimal length along the trajectory of a particle of mass moving with speed . The integral is taken over the path. As an example, for a particle moving with constant speed on a circular path of radius , the action for one revolution will be . For a particle with constant energy , it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA).

0.2 0.3

PLA implies that the trajectory of a particle moving between two fixed points in a region of constant potential will be a straight line. Let the two fixed points and in Fig. 1 have coordinates and A3 respectively and the boundary point where the particle transits from region I to region II have 1.0 coordinates Note that is fixed and the action depends on the coordinate only. State the expression for the action . Use PLA to obtain the relationship between and these coordinates. ?? B The Extremum Principle in Optics II ?? ?? A light ray travels from medium I to medium II with refractive indices ?? ?? and respectively. The two media are separated by a line parallel to the x-axis. The light ray makes an angle with the y-axis in medium I ?? ?? I and in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known ?? ?? as Fermat’s principle of least time. Figure 2 The principle states that between two fixed points, a light ray moves along a path such that time taken B1 between the two points is an extremum. Derive the relation between and on the basis of 0.5 Fermat’s principle. ?? Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident ?? horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height. ?? ?? Figure 3: Tank of Sugar Solution Assume that the refractive index depends only on . Use the equation obtained in B1 to obtain the 1.5 expression for the slope of the beam’s path in terms of refractive index at and . The laser beam is directed horizontally from the origin into the sugar solution at a height from the B3 bottom of the tank as shown in figure 3. Take where and are positive constants. 1.2 Obtain an expression for in terms of and related quantities for the actual trajectory of the laser beam. B2 1

Manoj Harbola (IIT-Kanpur) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the principal

authors of this problem. The contributions of the Academic Committee, Academic Development Group and the International Board are gratefully acknowledged.

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