Nuclear and Particle Physics Lecture 4

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Nuclear and Particle Physics Lecture 4 Dr.$Dan$Protopopescu Kelvin$Building,$room$524 Dan.Protopopescu@glasgow.ac.uk 12 Recap puzzle Concept Equa.on Defini.on Exponen.

1 Nuclear and Particle Physics Lecture 4 Dr.$Dan$Protopopescu Kelvin$Building,$room$524 Dan.Protopopescu@glasgow.ac.uk 12 Recap puzzle Concept Equa.on Defini.on Exponen.al$decay N(t)$=$N 0 Number"of"nuclei"that"have" decayed"in"the"6me"t c.vity (t)$=$ 0 Number"of"nuclei"decaying"per" unit"6me Decay$probability P decay Probability"of"a"single"nucleus" decaying"in"the"interval"t Life.me Half@life τ$=$1/λ t 1/2 $=$ln2/λ Maximum"6me"un6l"an"unstable" nucleus"decays Time"by"which"half"the"radioac6ve" sample"has"not"yet"decayed 13

2 nswers Concept Equa.on Defini.on Exponen.al$decay N(t)$=$N 0 Number"of"nuclei"that"have"not" decayed"by"6me"t c.vity (t)$=$ 0 Number"of"nuclei"decaying"per" unit"6me,"where" 0$ ="λn 0 Decay$probability P decay Probability"of"a"single"nucleus" decaying"in"the"interval"t* *t+ Mean$life.me$or$ simply$life'me Half@life τ$=$1/λ t 1/2 $=$ln2/λ Mean"6me"un6l"an"unstable" nucleus"decays Time"aCer"which"half"the" radioac6ve"sample"has"decayed 14 Simple decay If$a$sample$of$material$consists$of$nucleus$$which$is$unstable$and$ decays$to$nucleus$b$(of$which$there$are$ini.ally$none)$we$have$simply: $$$$$$$$$$$$$$$$$$ $$$$$$$$$ $B Nomenclature: $$$$@$ parent $$$B$@$ daughter The initial number of each nucleus is: (t = 0) = N 0 (t = 0) = 0 s nucleus decays into nucleus B (t) = N 0 e λ t and since N 0 = (t) + (t) (t) = N 0 (1 e λ t ) ("="total"number"of"nuclei) $@$Dan$Protopopescu 15

3 lternative decay modes n$ini.al$nuclide$$that$decays$into$two$products:$$$$$$$ $B$+$C$ We$have$at$any$.me$t:$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$***and$ (t)+ (t)+n C (t) =N 0 d =, d = B, dn C = C with$λ $=$λ B $+$λ C.$The$decay$constants$λ B $and$λ C $only$determine$ the$probabili.es$of$the$decays$to$products$b$or$c and (t) = B N 0(1 e t ) N C (t) = C N 0(1 e t ) (t) =N 0 (t) N C (t) =N 0 e t $@$Dan$Protopopescu 16 Decay series (or chains) Many$heavy$nuclei$decay$via$complicated$series$involving$several$α$and$β$decays.$ Consider$the$simple$case$of$$ $B$ $C,$where$C$is$stable$and$only$$is$present$ ini.ally: The number of nuclei vary according to: $ $B$ $C (t) = N o e λ t The number of nuclei B as a function of time can be found from: d (t) = λ B (t) + λ (t) where the first term is the decay of nuclei B and the second term is due to B being created from the decay of. Integrating, we can get (t) and its activity B (t) : λ (t) = N 0 ( e λ t e λ Bt ) λ B λ B (t) = λ B (t) = λ λ B N 0 ( e λ t e λ Bt ) λ B λ $@$Dan$Protopopescu (1) (2) 17

4 How was equation (2) derived? We multiply both sides of the equation by d (t) = B (t)+ e B t and we rearrange to obtain This can be written as d (e Bt (t)) = where we use (t) =N 0 e e B t to obtain the form d (e Bt (t)) = N 0 e ( B )t t e B t e B t d (t) + B e Bt (t) = e B t 18 How was eq.(2) derived (part II) We multiply by and integrate both sides R t 0 to obtain d(e Bt (t)) = R t 0 N 0 e ( B )t (t)e B t 0= B N 0(e ( B )t 1) which gives us (t) = t 0(e e Bt ) B QED 19

5 B C decay series For$the$stable$element$C$from$such$a$series$one$would$obtain: N C (t) =N 0 h 1 Be t e B t B i which$we$derived$using N 0 = (t)+ (t)+n C (t) Instead$we$will$focus$on$ (t)*and$inves.gate$a$few$special$cases: λα $ $λ Β$$$$$$$$ (parent$decays$quickly) λα $=$λ Β λα $<$λ Β λα $ $λ Β$$$$$$$ (parent$is$long$lived) 20 B C decay series for λα λβ Parent$decays$quickly,$$τ Α * *τ Β The$number$of$daughter$nuclei$rises$to$maximum,$then$decays$with$ constant$λ B.$ (t) = t 0(e e Bt ) B B =I1 λ Α " "λ Β =0! N 0 e Bt `er$a$given$.me,$daughter$nuclei$decay$almost$as$if$there$were$no$ parent$nuclei. 21

6 B C decay series for λα = λβ The$solu.on$of$eq.(1)$when$$λ Α =$λ Β =$λ$is: (t) = N 0 te t λ Α = λ Β = λ 22 B C decay series for λα < λβ Parent$$decays$slower$than$the$daughter$B. Ra.o$of$ac.vi.es$becomes$constant$a`er$a$sufficiently$long$.me: B = B = B B (1 e ( B )t ) B B when t!1 23

7 B C decay series for λ λ Β Parent$nucleus$is$long$lived:$λ λ B $or$τ τ B $so: e λ t 1 N 0 λ λ B ( 1 e λ Bt ) `er$a$sufficiently$long$.me$ 1 e λ Bt ( ) 1 λ = λ B d / = 0 c6vity"of"""=""c6vity"of"b in$eq.$(1) This$is$known$a$secular*equilibrium,$i.e.$at$large$.mes$B$is$ decaying$at$the$same$rate$as$it$is$produced. 24 Secular equilibrium (λ λ Β ) n$example$of$secular$equilibrium$is:$ $$$$$$$$$$$$$$$$$$$ 132 Te(12hrs)$ 132 I(2.28hrs)$ 132 Xe 25

lpha decay chains Because$α@decay$always$decreases$ the$atomic$mass$number$$of$the$ nucleus$by$4,$almost$any$decay$will$ result$in$a$nucleus$with$an$atomic$ mass$ $such$that$ $$$$$$$$$$$$$**mod*4*=*

8 lpha decay chains the$atomic$mass$number$$of$the$ nucleus$by$4,$almost$any$decay$will$ result$in$a$nucleus$with$an$atomic$ mass$ $such$that$ $$$$$$$$$$$$$**mod*4*=* *mod*4 s$a$result,$there$are$four$ radioac.ve$decay$chains$known$as$ the$thorium$(4n),$neptunium$(4n +1),$Radium$(4n+2)$and$c.nium$ (4n+3)$series. 26 Image$credits:$Wikipedia Thorium series and the age of the Earth 232 Th$has$a$very$long$half$life$ (t 1/2$ =$14Gyr)$and$goes$through$ a$long$decay$chain$to$stable$ 208 Pb. Image$credits:$Wikipedia It$effec.vely$behaves$as$if$$$$ 232 Th 232 Pb By$measuring$the$rela.ve$ abundance$of$ 208 Pb: N( 208 Pb) N( 232 Th) = N (1 0 e λtht ) N 0 e λ Tht one$can$es.mate$of$the$age$of$ the$earth$at$ yr. 27

Radiometric dating Technique$used$to$date$geological$materials$(rocks)$or$man@made$materials Based$on$a$comparison$between$the$observed$abundance$of$a$naturally$ occurring$radioac.

$ There$are$3$isotopes$of$carbon$@$ 12 C,$ 13 C$and$ 14 C$@$in$the$atmosphere,$from$where$they$ are$absorbed$by$living$organisms. The$ra.o$of$ 14 C/ 12 C$is$known$to$be$γ 0$=$ 1.

9 Radiometric dating Based$on$a$comparison$between$the$observed$abundance$of$a$naturally$ occurring$radioac.ve$isotope$and$its$decay$products,$using$known$decay$ rates. Image$credits:$earthsci.unimelb.edu.au 28 Radiocarbon dating Carbon$is$a$fundamental$part$of$living$.ssue.$ 12 C,$ 13 C$and$ 14 are$absorbed$by$living$organisms. The$ra.o$of$ 14 C/ 12 C$is$known$to$be$γ 0$=$ C$is$permanently$created$by$cosmic$ rays,$i.e.$this$isotopic$ra.o$is$constant$in$ nature The$concentra.on$of$ 14 C$in$living$organisms$ is$the$same$as$that$in$the$environment When$the$organism$dies$it$no$longer$ absorbs$ 14 C.$The$ 14 C$in$the$organism$decays$ but$the$amount$of$ 12 C$remains$constant $$$$$$$$$$$$$$$$$$$$$$$$$ 14 C/ 12 C$=$γ$=$ $ γ 0 Βy$measuring$the$ra.o$of$ 14 C/ 12 C$one$can$ find$out$how$much$.me$has$passed $$$$$$$$$$$$$$$$$$$$$$t$=$ln(γ 0 /γ)/λ $@$Dan$Protopopescu 29

Natural Radiation K 40

Natural Radiation K 40 Natural Radiation There are a few radioisotopes that exist in our environment. Isotopes that were present when the earth was formed and isotopes that are continuously produced by cosmic rays can exist