In physics, radiation is the emission or transmission of energy in the form of waves or particles through space or through a material medium. This includes: electromagnetic radiation, such as radio waves, microwaves,

1. RadiationandHalf-life

2. Half-life:Itisthetimetakenbyaradioactivesourcetodecreaseitsactivitythehalfofits
initialcount/value.
Half-lifecanvaryfromfractionofsecondtomillionsofyears.Forthesamesource,half-life
alwaysremainsthesamebutitisdifferentfordifferentsources.
Therearetwotypesofhalflife: (a)Longhalf-life (b)Shorthalf-life
Shorthalf-life:Ifaradioactivesourcetakesashorttimetodecreaseitsactivitytothehalfof
itsinitialcount,thenitiscalledshorthalf-life.Shorthalf-liferadioactivesourcesareusedin
medicalsciencediagnosisandthetherapy.
Longhalf-life:IfaradioactivesourcetakesaLongtimetodecreaseitsactivitytothehalfof
itsinitialcount,thenitiscalledlonghalf-life.Longhalf-liferadioactivesources,usedin
schoolLaboratorysothatitcangiveaconstantreadingforalongtime.
MeasurementofHalf-Life
Half-lifecanbemeasuredbytwoways:(a)Bythemethodoffraction(b)Byusingagraph
MethodofFraction:Tomeasurethehalf-lifebythemethodoffractionfirstwehavetorecord
thenumberofhalflife.Letthenumberofhalf-lifeben.So,thefractionofthissourcethatis
stillpresentis(1/2)n
.Sothefractionofthesourcethathasbeendecayedis[1-(1/2)n
]
Que.Halflifeofaradioactivesourceis4.5millionyears.Whatfractionofthesourcewill
havebeendecayedin18millionyearstime?
Ans.Noofhalf-liferequired=18/4.5=4
Thefractionofthesourcethatisstillpresent=(1/2)4
=1/16
Thefractionofthesourcedecayed=1-1/16=15/16
Que.Theactivityofasourcedecreasesto1/32inatimeof30hours.Calculatethehalf-life.
Ans.5half-lives=30years,1half-halflife=6years
Que.Initialactivityofasourceis7200countspersecond.Anditshalf-lifeis10hours.
Whattimeisrequiredtodecreasetheactivityto450countspersecond,
Ans:7200/2--10h--3600/2----10h---1800/2---10--900/2----10h--450.Thereis4half-lives.
MeasurementofHalflifeusingagraph:
*
Tomeasurethe
half-lifeusingagraph,firstwehavetorecordtheinitialcount.Thiscountisthendividedby
twotofindthehalfcount.Timecorrespondingtothishalfcountishalf-life. Forexample,
thehalf-lifeinthegraphis30minutes. Nowthe
backgroundradiationisnotnegligible.
Letthebackgroundcountbe20.Thereforetheinitialcountfromthesourcealonebe980

3. (1000-20).Halfcountofthesource=980/2=490+20=510.
Wecannotreadthegraphonlyforthesource,sobackgroundradiationshouldbeaddedto
readthegraph.So,timecorrespondingtothis510istheactualhalf-life.
Backgroundradiation:Itisthelowlevelradiationthatisalwayspresentintheenvironment.
Backgroundradiationvariesfromplacetoplace.Someofthesearenuturalandsomeare
artificial.
Theyaredifferentsourcesbackgroundradiation
butthemostcommonsourcesare
(a)Radongas
(b)Cosmicradiation(Comingfromstars)
(c)Earthcrust(volcaniceruption)
(d)RocksandMountainsetc.
SourcesofbackgroundradiationintheUK.
GMtubeandGMcounter:
GMtubeisadevicewhichisusedtoidentifythe
presenceofradioactivesource.GMcounterisa
devicewhichisusedtorecordtheactivityofa
radioactivesource.
Experimenttoidentifywhethersourceisalpha,
betaorgamma. Theidentifyaradioactivesourcewhether
itisalpha,betaorgamma,thefollowingstepsaretaken:
1.TheGMCounteristurnedonandthebackgroundradiationisrecorded.Thenthesourceis
placedclosetotheGMtube.AsaresultGMcountershowsahigherreading.Athinsheetof
paperisplacedbetweenthesourceandtheGMtube.
Observations:
a.Ifthereadingdoesnotfallsignificantlythesourceisnotalpha.
b.Ifthereadingdropstothebackgroundradiation,thenthesourceispurealpha.Because
alphacannotpassthroughathinsheetofpaper.
c.Ifthereadingfallssignificantlybutstillabovethebackgroundlevelthenthesourceis
alphaandanyothersource.
2.ThisthinsheetofpaperisthenreplacedbyanaluminiumfoilandthereadingoftheGM
counterisobserved.
a.Ifthereadingdoesnotfallsignificantly,thenthesourceisnotbeta,itisgamma.
b.Ifthereadingdropstobackgroundlevel,thenthesourceisbeta(becausebetaconnot
passthroughaluminiumsheet)
c.Ifthereadingfallssignificantlybutstillabovethebackgroundlevelthanthesourceisbeta

4. andgamma.
3.Whenthealuminiumfoilisreplacedbyblockoflead,itisobservedthatreadingdropstoa
backgroundradiationbecausegammacannotpassthroughablockoflead.
Tableshowsacountrateofaradioactivesourceforobservations
l
a.Whydothereadingsfluctuate?
Becauseradioactivityisarandomprocess,Italwaysfluctuatesaboutthemeanlevel.
b.Whatisthebackgroundcount?
Backgroundradiationis20Counts.
c.Whyarethreereadingstakenforeachobservation?
Becauseradioactivityisarandomprocess.Sothreereadingsaretakentocalculatethe
averagevaluewhichismorereliable.
d.Isalphapresent?Givereason.
Alphaispresentbecausewhenthethinsheetofpaperisusedreadingsdropsignificantly.
Weknow,alphacannotpassthroughthinsheetofpaper.
e.Isbetapresent?Givereason.
Betaisnotpresent.Becausewhenaluminiumfoilisused,thereisnoSignificantchangeto
itscountrate.
f.Isgammapresent?Givereason.
Grammaispresentbecausewhenablockofleadisplacedbetweenthesourceandthe
detector,readingdropstothebackgroundradiation.
GeigerandMarsden'sexperiment
Object placed
bewtween the
source&detecter
1stcount 2ndcount 3rdcount
Nothing 1202 1199 1200
Thinsheetofpaper 598 600 600
aluminiumfoil 601 600 599
Blockoflead 22 18 22

5. Description:Abeamofalphaparticleswasfiredatagoldleaf.Azincsulfidescreenwas
usedasadetecter.
Observations:
(a)Mostofthealphaparticlespassedstraightthrough.
(b)Onlyatinyminority(approx1outof8000)weredeviatedwithananglemonethan900
.
(c)Someofthealphaparticlesweredeviatedwithansmallangle.
Conclusion:
(a)Mostofthealphaparticlespassed,suggestedthatmostoftheavailablespaceinanatom
isempty.
(b)Onlyatinyminorityweredeviatedwithananglemorethan900
suggestedthateveryatom
hasanucleuswhereallofitspositivechargeandmostofitsmassisconcentrated.
Ifthesizeofanatomis10-10
mthenthesizeofthenucleusis[10-10
/8000]=1.25x10-25
m.The
actualsizeofthenucleusis10-15
whichisveryclosetothis.
(c)SomeoftheAlphaparticlesweredeviatedwithasmallangle,suggestedthatnucleusis
positivelycharged.
EdexelInternationalGCSE(9-1),Studentbookquestionsandanswers.
1aExplainwhatismeantbybackgroundradiation?
BackgroundradiationisradiationproducedbyradioactivematerialintheEarthandinthe
Earth’satmosphere.Itshouldbemeasuredandtakenintoaccountwhenmeasuringthe
activityofradiationfromaparticularsource.
b.Explainthedifferencebetweennaturalbackgroundradiationandartificialbackground
radiation.
Naturalbackgroundradiationisduetothedecayofnaturallyoccurringradioactive
isotopesintheEarththatwereformedwhentheSolarsystemwascreated.Natural
radiationalsoresultsfromhigh-energyparticlesthatbombardtheEarth.Artificial
backgroundradiationcomesfromman-madesources,ratherthanfromtherocks
thatmakeuptheEarth.
c.Givethreedifferentsourcesofbackgroundradiation.Saywhetheryourexamplesare
naturalorartificialsources.
UraniuminrocksintheEarth’scrust(natural).Radioactivematerialsthathave
escapedintotheenvironmentfromnuclearpowerstations(artificial).Falloutfrom
atmosphericnuclearweapontestinginthe1950sand1960s(artificial)
2.a.ExplainsimplytheprincipleoftheGeiger-Müllertube.
Ionisingradiationcausesionisationofalow-pressuregasinsidethetube.The
ionisedgasallowsacurrenttoflowbetweentwoelectrodes,andthecurrentis
detectedbyanelectroniccircuit.
b.TheGeiger-Müllertubeisoftenconnectedtoaratemeter.Explainwhatthisinstrument
measures.
Aratemetergivesanindicationofthenumberofdecaysoccurringpersecond
c.TheratemeteriscalibratedinkBq.Howisthisunitdefined?

6. Abequerelisarateofdecayofonedisintegrationpersecond.1kBqis1000
disintegrationspersecond.
3.a.Definewhatismeantbythehalf-lifeofaradioactivematerial.
Thehalf-lifeofaradioactivesampleistheaveragetimetakenforhalftheunstable
atomsinthesampletoundergoradioactivedecay.
bRadioactivedecayisarandomprocess.Explainwhatthismeans.
Randommeansthatthedecayofanindividualatomisunpredictable;wecannotsaywhen
anyparticularatomwillundergodecay.
4.Theactivityofaradioactivesampleismeasured.Theactivity,correctedforbackground
radiation,isfoundtobe240Bq.Theactivityismeasuredagainafter1hour30minutesand
isnow30Bq.Whatisthehalf-lifeofthesample?
In1.5hoursthesamplehashalvedinactivitythreetimes(2401206030)so
threehalf-livesperiodshavepassed;thehalf-lifeis,therefore,30minutes.
5.Inanothermodelofradioactivedecay,astudentfillsaburettewithwaterasshownin
thediagram,andstartsatimerattheinstantthetapatthebottomisopened.Shenotes
theheightofthecolumnofwateratregularIntervals.Ittakes35secondstoemptyfrom50
mlto25ml.Assumingthatthearrangementprovidesagoodmodelofradioactivedecay.
a.Howlongwillittakeforthreequartersofthewaterintheburettetorunaway?
35secondsisthetimetakenforhalfthewatertodrainaway.Ifthearrangementisa
goodmodel,inafurther35secondstheamountofwaterintheburettewillhave
halvedagainandthree-quartersofthewaterwillhavedrainedaway,sotheansweris
70seconds(1minute10seconds).
b.Howmuchwatershouldremainintheburetteafter13/4minutes?
After13/4minutes,three‘half-lives’,afurtherhalvingwillhavetakenplace,leavingtheburette
one-eighthfull,i.e.containing6.25ml.
6.Astudentwantstomeasurethehalf-lifeofaradioactiveisotope.Heistoldtheisotopehasa
half-lifeofbetween10and20minutes.Illustratingyouanswersasappropriate,describe:
a.themeasurementsthatheshouldtake.
Studentsshouldmeasurethebackgroundradiation.Theyshouldthenmeasurethe
radioactivityofthesampleusingaGMtubeandratemeteratregularintervalsof,say,
5minutesforaperiodof30–40minutes.
b.howheshouldusethemeasurementstoarriveatanestimateofthehalf-lifeforthe
isotope.
Thereadingsoftheactivityofthesampleshouldbecorrectedbysubtractingtheaverage
backgroundradiationcount.Thecorrectedreadingsshouldthenbeplottedonagraphof
countrateagainsttime.Thetimetakenfortheinitialactivitytofalltohalfcanthenbe

7. measuredfromthegraph.
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