BS 4094-1-1966 Recommendation for data on shielding from ionizing radiation - Shielding from gamma radiation《电离辐射防护屏蔽数据推荐标准 第1部分 γ辐射防护屏蔽》.pdf
《BS 4094-1-1966 Recommendation for data on shielding from ionizing radiation - Shielding from gamma radiation《电离辐射防护屏蔽数据推荐标准 第1部分 γ辐射防护屏蔽》.pdf》由会员分享,可在线阅读,更多相关《BS 4094-1-1966 Recommendation for data on shielding from ionizing radiation - Shielding from gamma radiation《电离辐射防护屏蔽数据推荐标准 第1部分 γ辐射防护屏蔽》.pdf(82页珍藏版)》请在麦多课文档分享上搜索。
1、%5,7,6+ 67$1$5%6 5HFRPPHQGDWLRQ IRUDWD RQVKLHOGLQJIURPLRQLLQJUDGLDWLRQ 3DUW6KLHOGLQJ IURP JDPPD UDGLDWLRQ8 other terms commonly encounteredare thekilocurie(kCi)equalling103curies,millicurie(mCi)equalling103curie,and themicrocurie(Ci)equalling106curie.Theactivity,andhence theradiationoutput of the so
2、urcedecays with timeinanexponentialmanner.Aconvenient measureof therateofdecay is givenby thehalf-lifeofa source, whichrepresents the time takenfor the sourceactivity tofallby afactorof two.Toenablerapidassessment of theactivity ofa source tobedeterminedgivenits half-lifeandoriginalactivity,agraphis
3、 given(Figure1)of thefractionalreductioninsourceactivityas afunctionof thenumberofhalf-lives passed sinceits originalactivity was known.Thus forexample,considering24Na withahalf-lifeofhours,ifit is assumed that theoriginal sourceactivity was Qcuries,and theactivity after84hours is required,proceedas
4、 follows:Numberofhalf-lives passed=FromFigure1 thiscorresponds toafractionalreductionof0.02.Hence the sourceactivity is0.02 Qcuries.Conversely,if theexistingactivity is Pcuries, then theactivity84hours previously wasP/0.02 curies.The values ofhalf-lives aregivenin thedata sheets for theradionuclides
5、 considered.Aknowledgeof thehalf-lifeis necessaryifitisdesired tocalculate theactivityofa sourcefor whichagiven shield wasoriginallydesigned,havingknowledgeof theexposurerateat the timeofcalculation.3 Calculation of exposure ratefrom a point sourceIn this section,considerationis only given to theexp
6、osuredue to theprimary beam.Theeffects of scatterareconsideredinAppendixB.Fora sourceofgammaradiation, theexposurerateat agivenpositiondepends on thefollowing:i) theradionuclideconcerned,andits specificgamma-ray constant;ii) the sourceactivity;iii) thedistancebetween the sourceand thepositionofinter
7、est (theexposurerate willbeinverselyproportional to thedistance squared);iv) the typeand thicknessof shieldmaterialplacedbetween thepositionand the source.a)Unshielded source.For thepurposesof this recommendation, the specificgamma-rayconstant foragiven sourceis takenas beingequal to thegamma-ray ex
8、posurerateinroentgens perhourat adistanceof1mfroma sourceofactivity onecurie.NOTE Otherdefinitions exist,andcare shouldbe takenthat thedatain this recommendationare usedinconjunction withspecificgamma-ray constants in these terms.8415-5.6=BS 4094-1:19661 The British Standards Institution 2015BS 4094
9、-1:19662BSI 09-1999Figure 1T he relation between remaining activity and half-livesBS 4094-1:1966 The British Standards Institution 2015BS 4094-1:1966BSI 09-1999 3Thus fora sourceofactivity Qcuries, theexposurerateat dmfrom the sourcein theabsenceofany shieldingmaterialbetween the sourceandpositionof
10、measurement maybedeterminedas follows:Exposureratefromonecurieat onemetredistance=+ R/hwhere + is the specificgammaray constant for that source.Therefore,exposureratefromQcuries at 1m= +QR/handhence,exposureratefromQcuries at dm=the termarisingfrom theinverse squarelaw.Values of the specificgamma-ra
11、y constant + aregivenin thedata sheets; + is plottedas afunctionofphotonenergy inFigure8.b)Shielded source transmissionfactor.Thepresenceofa shieldbetween the sourceandadefinedposition willreduce theexposureratebelow the unshielded valuedue toabsorptionin the shield.Thus for the sourceconsideredina)
12、andagiven shield, theexposurerateat adistancedmfrom the sourceis givenby:Exposurerate=whereTis the transmissionfactor, whichdepends on thematerialand thickness of the shield,and thetypeof source.Values of the transmissionfactoraregivenin thedata sheets in theformofgraphs showing transmissionas afunc
13、tionof thickness for various shieldingmaterials.Thedatagivenrefer tobroadbeams ofradiation(seeAppendix A,SubsectionsA.2and A.3).4 Data sheetsData sheets areprovidedfor thefollowing sources:Theinformationnecessary tocalculate theradiation transmitted through various shieldingmaterials is givenon thed
14、ata sheets foreachof the sources.Theinformationon the sheets includes:Theradionuclideanddecay schemeRadioactivehalf-lifeSpecificgamma-ray constant(+)Broadbeam transmissioncurves for various materials.Examplesi)Suppose that theexposureratedue toa192Ir sourceat thepoint tobe shieldedis2.5R/hand that i
15、t is required toreduce this to0.75mR/h.The transmissionfactor,T,is givenby0.75 103=T 2.5FromFigure 2h(i), therequired thickness of shieldingis5.7cmPb.ii)Considera60Co sourceof200 curies, shieldedby lead,andassume that becauseof thelayout of thelaboratoryin whichit is being used, thenearest distanceo
16、fapproachofpersonnelis twofeet.It is required todetermine the shielding toreduce theexposurerateat this point to2.5mR/h.Fromdata sheet No. 2, the specificgamma-ray constant (+)for60Co=1.32 R/Ci.hat1m.24Na50Co85Kr124Sb137Cs170Tm182Ta192Ir198Au226Ra+decay products232Th+decay products+Qd2-R/h1d2-+QTd2-
17、 R/hi.e. T0.751032.5- 0.0003= =BS 4094-1:19663 The British Standards Institution 2015BS 4094-1:19664BSI 09-1999Hencefora sourceof200 curies,theexposurerateat2 feet(0.61m)from the source= 7.1 102R/hSince therequiredexposurerateis2.5 103R/h,a transmissionfactorof =3.52106is required.FromFigure 2b(i) t
18、herequired thickness of shieldingis22.5cmPb.iii)If,however,it is required todetermine the shield thickness togiveanexposurerateofXmR/hat theshield surface,as fora transport container, theprocedureis morecomplicatedbecausedistance,andhence thegeometry factor and the transmissionfactor,are variables.I
19、t is recommended that thefollowingprocedurebeadopted:Let Xbe therequiredexposurerateat the shield surface;whereTtis the transmissionfactorfora shieldof thickness t(cm)anddis thedistanceinmetres from thesource totheouter surfaceof theshield.ThusEvaluate using transmissiondatafora series of values ofd
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