Measurement of the fission mass yields of Am242 at the ... · PDF fileCharlotte AMOUROUX1...

CharlotteAMOUROUX1
MeasurementofthefissionmassyieldsofAm242attheLohengrin Spectrometer
A.Bidaud2,N.Capellan2,S.Chabod2,H.Faust3,G.Kessedjian2,U.Kster3,
A.Letourneau1,F.Martin2,T.Materna1,S.Panebianco1,Ch.Sage2,O.Serot4
1CEA,DSMSaclay,France2LPSCGrenoble,CNRS/IN2P3,France
3InstitutLaueLangevin,France4CEA,DENCadarache,France

242Am(Z=95)
241Am:90%oftheradiotoxicity ofthenuclearwaste (without plutonium)between 200and1000years >Transmutationof241Am
n+241 Am 242 AmFP1
FP2
242Am:two longlived statesZ=95(odd charge)
2/12CharlotteAMOUROUXWONDER25/09/2012
141Y
16.02H1
5
48.6keV

PLAN
Experimental Setup&AnalysisMethod EnergyandChargeDistributions UncertaintiesDetermination Results
CharlotteAMOUROUXWONDER25/09/2012 3/17

HighneutronfluxReactor
Target
Magnet:Selection A/q
Condenser:Selection:E/q
Detector:E
Experimental setup

Howdowe measure theenergy ofthefragment?
E>E/q>A/q>A
*A.Bail thesis
=>Y(A,E,q)CharlotteAMOUROUXWONDER25/09/2012
EEIonisationChamber
5/17

Energy Distributionforagiven q
ChargeDistributionforagiven E
N(A,q,E)according toqandE
Valid ifnocorrelation between Eandq
Howtocalculate thefissionyields ?
Inrealitywe haveacorrelation butits influenceonY(A)inless than 3%

Example ofQdistribution:two differents casesMeasured Chargeis determined at thelastcrossed material
(Nickel)
With nanosecond isomerWithout nanosecond isomer
QDistribution
A=105
A=136
Target
Target
NickelFoil NickelFoil
Magnet Magnet
Q~2122 Q~2425

EDistribution
A=105Q=21
0.6MeVfordetermination ofKE0.6MeVfor(q,E)correlation

Kinetic energy asafonctionofthefragmentmass

EDistributionStatisticerrors:
~1%
Amplitu
de[a
.u]
Energy [MeV]
~1,5%ofthetotalarea
_
maximalfluctuationof
1,5%between 2Edistribution
~0,8%ofthetotalarea
_
maximalfluctuationof
1,0%between 2Edistribution
A=105Q=21
Systematics errors:
1,5%lowenergypart 1,0%highenergypart

Determination ofthesystematic errorThispointis known twice
Forthesame massasafunction oftime:
Forallmasses:
~3%

Source Contribution
Statistical ~1%
Extrapolationofthelowpartoftheenergydistribution
1.5%
Extrapolationofthehighpartoftheenergydistribution
1%
Discrepanciesbetweenthetwomeasurementsofthe
commonpoint3%
Normalisation ?
Totalofthesystematicerror 3.5%
Sourcesofrelativeuncertaintiesandtheirrespectivecontributions.
CharlotteAMOUROUXWONDER25/09/2012

n+241 Am
242m Am
242
Am
Y
Yfission=2644(281)barn*91,4%
8,6%
Y=Y ?
Objectivesoftheexperiment:
FissionMassYieldsfromAm241(2n,f)
IsthereanydifferencebetweenthefissionyieldsofAm242(n,f)andAm242m(n,f)?
FissionYields ofAm242
(141y)
(16h)
fission=6856(656)barn*
*G.Fioni etal,Nucl Phys.A693(2001)546O.Bringer,Ph.D Thesis,INP Grenoble,
October 2007

Howdowe proceed toobserveapossibledifference ?
B:vaccum problemA:Strong evolution ofthepropertiesofthetarget (largeenergy shift)
A
Shutdow
nofthe
reactor
B

Hypothesis:X=0.
forX=0(0,04forallYand0,07forYm )
Generalcase: ~1
What is themaximumpossibledifference ?
n+241 Am242m Am
242
Am

Conclusions
Nodifference between theyields:quantificationongoing.
Ifyou assumethey areequal
NormalisationY105=6,5%
Onlystatisticalerror

Isotopic fissionyields
Future
239Pu233U
*A.Bail Thesis *MeetingGEDEPEONJan2011G.Kessedjian (F.Martin thesis (ongoing))
241Pu
242AmThank you
for your attention
235U

Backup
18

Comparaisonwith theGEFcode(June 2012)
19

Number offissions
20

Evolutionofthekinetic energy asafunction oftime
21

FWHMoftheenergy distributionasafunction oftime
22
PLANHow do we proceed to observe a possible difference ?Conclusions

Measurement of the fission mass yields of Am242 at the ... · PDF fileCharlotte AMOUROUX1...

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