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The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final examination. These solutions are not the official set of solutions used by the examiners of the SACE Board.

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Page 1: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

 

         

        The following pages offer suggested solutions to the  2012 SACE Stage 2 Physics final  examination. These solutions are not the official set  of solutions used by the examiners of the SACE Board.  

Page 2: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Tension

F mv2

r 0.035x2.42

0.32 0.63 N

Page 3: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

vH vsin 25sin 40 16.1 ms1

v2 vo2 2as v 0 max height

0 16.12 2x9.8s

s 16.12

2x9.813.2 m

Total height above the ground = 13.2 + 1.5 = 14.7 m

COMMONERRORThereleaseheightmustbeaddedasthequestionasksforthemaximumheightabovetheground.

Page 4: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Increasingthelaunchheightincreasestherange.Thejavelinspendsmoretimeintheairasitfallstheextraheight.Sincetherangeistheproductofthehorizontalcomponentoftheinitialvelocity(vH)andthetimeofflight.Thehorizontalcomponentoftheinitialvelocityisconstantsoanincreaseinthetimeofflightincreasesintherange.

Page 5: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

FN VerticalcomponentFV

HorizontalcomponentFH

Page 6: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Theverticalcomponentofthenormalisequalinmagnitudebutoppositeindirectiontotheweight,whereweightW=mg.

ThehorizontalcomponentofthenormalforceFH,providesallthecentripetalaccelerationforuniformcircularmotion.

tan FH

FV

mv2

rmg

v2

rg

Tan v2

rg

v tanxrg tan 42x26x9.8 15 ms1

Page 7: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

ForceontheMoonduetotheEarth

FME GmM mE

rmE2

6.67x1011 x7.35x1022 x5.97x1024

(3.85x108 )2

ForceontheMoonduetotheSun

FMS GmM mS

rmS2

6.67x1011 x7.35x1022 x1.99x1030

(1.5x1011)2

Force on the Moon due to the Earth

Force on the Moon due to the Sun

6.67x1011 x7.35x1022 x5.97x1024

(3.85x108)2

6.67x1011 x7.35x1022 x1.99x1030

(1.5x1011)2

5.97x1024

(3.85x108 )2x

(1.5x1011)2

1.99x1030

0.455

Page 8: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Sincespeedofasatellitevisgivenby v GM

Ro

whereMisthemassofEarthand

GistheGravitationalconstantthen v 1

rwhereristheradiusoforbitofthe

satellite.ThisisbecausebothMandGareconstant.TheQuickBirdsatellitehasalargerradiusoforbitandwillthereforehaveaslowerspeedthantheinternationalspacestation.

Asexplainedabove,themassintheequationforthespeedofasatelliteisthemassoftheEarthnotthemasofthesatellite.Thedifferentmasseshavenoeffectontheirspeeds.

Imageswithhigherresolutionareproduced.

MEANINGAlow‐altitudeorbitmeansthatthesatelliteisclosertotheground.Theimagesproducedshowmoredetail.Werefertothisasresolution.

Page 9: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

ThemomentumvectorofBafterthecollision

ThetotalinitialmomentumvectorisjustthatofAasofBwasinitiallystationarybeforethecollision

Page 10: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Usingthelawofconservationofmomentum,thetotalinitialandfinalmomentumisthesame(bothmagnitudeanddirection).Sincepi=pfthenpAi=pAf+pBfThefinalmomentumvectorisshownonthediagramabove.

Page 11: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Thechangeinmomentumpisgivenbythefinalmomentumofthephotonsubtracttheinitialmomentum.Thefinalmomentumofaphotonthatisabsorbediszerobutthefinalmomentumofaphotonthatisreflectedisthesameastheinitialmomentumbutintheoppositedirection.Thechangeinmomentumofaphotonthatisabsorbedisgivenbyp pf p

i 0 p p whichishalfthatofaphotonthatisreflected

p pf pi p p 2 p .

Usingthelawofconservationofmomentum,thechangeinmomentumexperiencedbythesolarsailisequalbutoppositeindirectiontothechangeinmomentumexperiencedbythephoton.

SinceFsail psail

t,thenitfollowsthatsolarsailsthatreflectphotonswill

experienceagreaterforce(ifthecollisionoccursoverthesameamountoftime)

andhencegreateraccelerationsince asail F

msail

p 2 p

p p

Photonisabsorbed

p

p

p

Page 12: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

HINTTheelectricfieldisuniform.Theelectricfieldlinesneedtobeevenlyspaced.Theelectricfieldlinespointawayfromtheconductorastheyindicatethedirectionoftheforceonapositivetestcharge.

Page 13: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

F

B 1

40

QqB

r2 9x109 x8x1019 x2.56x1018

0.082 2.9x1024 N (south)

Page 14: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

ForceduetoqA 2.9x10‐24Neast ForceduetoqB 2.9x10‐24Nsouth

F FA

FB

F (2.9x1024 )2 (2.9x1024 )2 4.1x1024 N

tan1(2.9x1024 )

2.9x1024 450

F 4.1x1024 N SE

2.9x10‐24 N

2.9x10‐24 N

F

HINTThetriangleisrightangled–Pythagorasapplies.

Page 15: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

ThemagneticfieldduetoI1actsintothepageinthepositionwhereconductor2lies.Usingtherighthandruletheforceonconductor2duetoconductor1isuptheplaneofthepage.ThemagneticfieldduetoI2actsoutofthepageinthepositionwhereconductor1lies.Usingtherighthandruletheforceonconductor1duetoconductor2isdowntheplaneofthepage.Theconductorsattracteachother.

HINTUsetherighthandruleforthemagneticfieldaroundastraightconductor.

HINTAgainusetherighthandruleforthemagneticfieldaroundastraightconductor.

Page 16: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

F Bqvsin 2.5x102 x1.6x1019 x1.45x106 xsin 90 5.8x1015 N

Page 17: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

E V

d 6x104

0.451.3x105 Vm1

F Eq 1.3x105 x(1.6x1019 ) 2.1x1014 N

Page 18: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

vv sv

t t sH

vH

0.8

2 0.40 s

Maximumdeflectioninthedirectionoftheelectricfieldis 0.45

2 0.225 m

Thiscanbeusedtofindthemaximumacceleration.

s vot 1

2at 2 vo 0

a 2s

t2 2x0.225

0.42 2.81 ms2

a Eq

m m Eq

a 1.3x105 x1.6x1019

2.81 7.4x1015 kg

REASONINGThiscomponentofvelocityisperpendiculartotheuniformelectricfieldbetweentheplatesandisconstant.Theequationforconstantspeedapplies.

Page 19: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Vertical

v f v

f 3x108

64.25x106 4.67 m

COMMONERRORSIunitsarerequired.MHztoHzx106

REASONINGTheplaneofpolarisationisdefinedastheplaneoftheelectricfield.Theplaneofthemagneticfieldisperpendiculartotheplaneoftheelectricfield.

Page 20: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Anincandescentglobeemitswhitelight.Whitelightconsistsofallcolours/wavelengthsrangingfromredthroughtoviolet(ROYGBIV).Monochromaticlightconsistsofonecolour/wavelengthonly.Thelightfromanincandescentglobeisnotmonochromatic.

Page 21: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

kmax hf W 6.63x1034 x2x1015 (3.66x1.6x1019 ) 7.4x1019 J

Kmax 1

2mv2 v 2K

m 2x7.4x1019

9.11x10311.3x106 ms1

HINT/COMMONERROReVneedstobeconvertedtoJ(x1.6x10‐19)

Page 22: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

SincethedistancebetweenthedoubleslitsandthescreenissomuchlargerthatthedistancebetweenthedoubleslitsthentheangleS1XS2isapproximately90o.UsingthetriangleS1XS2then

sin path difference

d

Foramaxima,thepathdifferenceism

Itfollowsthat

sin path difference

d m

d

ie d sin m

Pathdifference

X

Page 23: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

d sin m

sin1(md

) sin1(3x4.7x107

1.8x104)

0.450

Thelaserlightiscoherentwhereasthebluelightsourcewasn’t.Thismeansthatthesingleslitisn’tneeded.

Sincethefringeseparationy L

d,thenthefringeseparation(distance

betweenadjacentmaxima)isproportionaltothewavelength.Changingthewavelengthfrombluetoredmeansthatthewavelengthisgreater.Agreaterwavelengthmeansagreaterfringeseparation/distancebetweenadjacentmaxima.Thecolourofthebrightfringeswillalsochangefrombluetored.

REASONINGThepurposeofthesingleslitistoproducetwocoherentlightsourcesatthedoubleslits.

Page 24: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Increasingthefilamentcurrentwillreducetheexposuretime.ThisisbecausethefilamentcurrentreleaseselectronsthatcollidewithatargetmetaltoproduceX‐rays.Ifmoreelectronsarereleased,thenmorecollisionsoccurandmoreX‐raysareproducedreducingtheoverallexposuretime.

Exposuretoionisingradiationcanbereducedbywearingappropriateshieldingsuchasaleadlinedapron.

Page 25: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

h

p h

mv 6.63x1034

9.11x1031 x4.36x1061.67x1010 m

Lowenergyelectronswerefiredtowardsacrystallatticeandtheywerediffractedatpreferredanglesratherthanbeingscatteredrandomly.Diffractionisawavephenomena.Inadditiontheequation d sin mwasusedtocalculatethewavelengthoftheelectronsandmatchedthatfoundusingthede

Broglierelationship h

p.

Page 26: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Thepositionoflinesinthespectraofhydrogenandlithiummatchthelinesinthespectrumofthemixture.Iehydrogenandlithiummustbepresent.

Page 27: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Thereisnoenergygapbetweenthegroundstateandthehigherenergylevelsthatmatches12.50eV.Thephotonwillnotbeabsorbed.

1.89eV

10.2eV

12.09eV

12.75eV

(b)(i)

(a)

REASONINGThesmallestjumpfromE3toE2willemitthesmallest‐energyphoton.

E3 E2 1.513.4 1.89eV

Page 28: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Boththenuclei(UandTh)havediscretenuclearenergylevels.WhenthenucleusdecaystoTh,itmaydecaytothegroundstateoroneofthehigherexcitedstates.Itfollowsthatasthenucleusdecaystheemittedalphaparticleswillhavearangeofdiscreteenergiessothattheoverall(discrete)energyofthisreactionisconserved.

After8alpha(8 2

4 )decaystheatomicnumberofproductnucleusbecomes92‐16=76andthemassnumberbecomes238–32=206.Lead(Pb)hasatomicnumber82.Usingthelawofconservationofcharge,itfollowsthat6betaminusdecaysmustoccur.

92238U 82

206 X 6 10e82

4 6 00

92238U 90

234Th 24

Page 29: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Thereexistsastrongnuclearforcebetweenthenucleons(protonsandneutrons)inthenucleusthatisattractiveandstrongerthantherepulsiveelectrostaticforcesbetweenthe82protons.

Thisforceactsovershortdistancesandquicklybecomesnegligibleatseparationsofmorethanafewnucleondiameters.Thisishowthenucleons(whichareveryclose)canbeheldtogetherinastablenucleus.

Page 30: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

neutrino

Theemittedpositroncancollidewithanearbyelectronandannihilate.Themassisconvertedintoenergyintheformoftwoidenticalphotonsthattravelinoppositedirectionssothatmomentumisconserved.

HINTThehalflifeis2days.Thisisconstant.

Page 31: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

REASONINGThephotonstravelinoppositedirections.

Page 32: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Uranium‐238doesnotreadilyundergoinducedfissionwhereasuranium‐235does.Enrichingthesourcemeansthatalargerpercentageofuranium‐235isaddedtothefuel.Thisensuresthatenoughofthenucleiundergoinducedfissionandachainreactioncanoccur.

E hf hc

6.63x1034 x3x108

502x109 3.96x1019 J

COMMONERRORWavelengthneedstobeinSIunitsnmtommultiplyby10‐9

Page 33: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Massofproducts mcMg+mn= 3.8172x10-26 + 1.6749x10-27 = 3.98469x10-26 kgMassofreactant 2mC= 2x1.9921x10-26 = 3.9842x10-26 kg Themassoftheproductsisgreaterthanthemassofthereactants–energyisabsorbed.

m mproducts mreactants 3.98469x1026 3.9842x1026 4.9x1030 kg

E mc2 4.9x1030 x(3x108)2 4.41x1013 J

Page 34: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Lengthof

pendulum/stringL(cm)

PeriodT

(s)T2(s2)

20 0.94 0.8830 1.08 1.1740 1.36 1.8550 1.48 2.1960 1.54 2.37

Thelengthofthependulum/string.

REASONING:Theindependentvariableisthevariableintentionallychangedbytheexperimenter.

Page 35: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Therelationshipbetweenperiodsquared(T2)andlengthofthependulum.T2(s2)

0

0.5

1

1.5

2

2.5

3

0 10 20 30 40 50 60 70

LengthofpendulumL(cm)

line ofbestfit

rise

run

Page 36: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

gradient rise

run 1.7 0.2

40 2 0.039 s2cm1

T 2 4 2 L

g

SincethestraightlinehasanequationT2=0.039L

Itfollowsthattheslope0.039=4 2

g

g 4 2

0.0391010 cms2 10.1ms2

Accuracyishowclosetheexperimentalvalueistotheactualvalueandcanbeimprovedbyreducingsystematicerrors.Anexampleofthiswouldbetocheckthatthelightgateiscorrectlycalibratedbyrepeatingtheexperimentwithadifferentlightgate.Precisionishowmuchscatterthereisinthedatacollected.Thiscanbereduced,byreducinganyrandomerrorsintheexperiment.Anexampleofthiswouldbetorepeatthemeasurementsofperiodforeachlengthatleastthreetimes.

Page 37: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final
Page 38: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Astheionsenterthedeestheydosoat90otoauniformmagneticfield.

Aconstantmagneticforcealwaysactsat90otothevelocityofthecharges.

Eventhoughthespeedremainsthesame,thedirectionofmotionchanges.

Bydefinitionthereisachangeinvelocityv

vf

v

iandhenceacceleration

givenbya

v

t.

Themagneticforcethereforeprovidesthecentripetalaccelerationforuniformcircularmotion.

X

X

X

X

X

X X

X

X

X

X

X

XX X

XX

X

X

X

X

XX

X

dees

Page 39: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Eachtimetheionscrossthegapbetweenthedees,theyareacceleratedbyauniformelectricfield.Everytimetheionscrosstheelectricfield,theworkdone(W)bytheelectricfieldisconvertedintokineticenergy.

Kineticenergyeachtimetheionscrossthegap=W=qV 1

2mv2 .Theionsdo

notgainenergyorspeedwhiletheyareinthedees.Itcanbeshownthatthefinalkineticenergyoftheionsastheyemergefromthe

cyclotronisgivenbyK q2B2r 2

2mwhereristheradiusofthecyclotron.

Itthereforefollowsthatthefinalkineticenergyisdirectlyproportionaltothesquareofthecyclotron’sradius.Thismeansthatiftheradiusofthecyclotronisdoubled,thekineticenergybecomesfourtimeslarger.Alternatively,iftheradiusis10timeslargerthekineticenergywillbe100timeslarger.Iftheradiusis3timessmaller,thekineticenergywillbe9timesetc

HINT22=4102=10032=9

Page 40: 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf · The following pages offer suggested solutions to the 2012 SACE Stage 2 Physics final

Ifmonochromaticlightisincidentonametalsurfaceitcontainsasinglefrequency(f)only.Thelightconsistsofmanyphotons.Photonsarebundlesofdiscreteenergy.EachphotonhasadiscreteenergygivenbyE hf wherehisPlanck’sconstant(6.63x10‐34Js).Electronswithinthemetalareboundbydifferingamountsofenergy(dependingonhowdeepwithinthemetaltheyarepositioned).Usingthelawofconservationofenergy,theenergyofthephotonisusedtoreleaseandelectronandany‘leftover’energyisgivenupasthekineticenergyoftheemittedelectron.Thisproducesarangeofkineticenergiesuptomaximumgivenby:Kmax hf W whereWistheworkfunctionofthemetal.Theworkfunctionisdefinedastheenergyneededtoreleasetheleastboundorsurfaceelectrons.ElectronsinanX‐raytubearereleasedfromaheatedfilament.EachelectrongainsadiscreteenergygivenbyK qV astheyaccelerateacrossthepotentialdifferenceV .Whentheelectronscollidewiththetargetmetaltheydecelerateandlosekineticenergy.Thelawofconservationofenergyapplies.MostofthecollisionsproduceheatbutapproximatelyonepercentofthecollisionsproduceX‐rayphotonsofenergyequaltotheenergy‘lost’bytheelectronsduringthecollision.Theamountofkineticenergylostbythecollidingelectronsdependsonhowcloselytheycollidewiththenucleus.ThisproducesarangeofX‐rayphotonswitharangeofenergies.