8.) Our radiological environment

Transcription

1 8.) Our radiological environment Te average effective annual dose Indoor radon Rn-222 is te biggest problem, because of a relatively large abundance of its element of origin, U- 238, and a relatively long lifetime compared to oter Rn-isotopes (Rn-220 and Rn-219). Building sites wit ig concentrations of Ra [ Bq kg ] (Ra 226 Rn 222) and ig gas permeability represent te biggest problem. 1

2 Indoor Rn-concentration dχ Rn dt = u(t) χ Rn (λ Rn + λ v ) - χ Rn,air = te concentration of Rn-222 activity in air. - u(t) =te rate of flow of Rn-222 into te building. [ ] Bq m 3 [ ] Bq m 3 s - λ Rn = te disintegration constant. [ 1 s ] - λ v = te rate of flow out of te building. [ 1 s ] If u and λ v are constant, te equilibrium concentration is: χ Rn,air = u λ Rn + λ v Wat contributes to u(t) is: - Ground conditions - Building materials - Water (ouseold water) - Outdoor air (ventilation) Rn and Rn-daugters get stuck to tiny particles of dust and surfaces (plate-out). 2

3 Rn-concentration in Norwegian ouses Most probable value: Mean value: 20 Bq m 3 88 Bq m 3 in % of te ouses ad values > 400 Bq m 3 in % of te ouses ad values > 200 Bq m 3 in 2001 Rn dosimetry and risk limits (ICRP 50) Contributions from Rn-222 (a gas) and its metal-like daugter nuclides ave to be accounted for separately. 1.) Contributions from Rn: Dose rate in soft tissue excluding te lungs (due to Rn dissolved in te tissue): ngy Ḋ soft tissue = S st χ Rn,air, S st = Bq m 3 For lung tissue, te contribution from Rn in te alveolar air comes in addition to te contribution from dissolved Rn: Dose rate: Ḋ lungs = S l χ Rn,air, S l = 0.04 ngy Bq m 3 Equivalent dose rate: Ḣ T = ω R Ḋ T,R, ω R = 20 for α Effective dose rate: Ė = ω T Ḣ T = ω R (ω l Ḋ l + ω st Ḋ st ), ω l = 0.12, ω st = 0.88 Ė = S tot,rn χ Rn,air, S tot,rn = 0.2 nsv 1 2.) Contributions from sort-lived Rn-daugter nuclei: Cain of disintegrations: 3

4 No Nuclide T 1 2 E α ε pi ε pi Rn 3.82d 5.49 [MeV ] [MeV ] [ MeV Bq ] Po 3.05m Pb 26.8m Bi 19.7m Po 164µ s Pb 19.4s Were numbers 1 troug 4 represent sort-lived daugter nuclei. Equilibrium activity concentration: C act,eq = λ Rn C Rn = C i, i = 1,..4 Were C Rn is te number of Rn atoms per unit volume of air. Now, in a real situation, te activity concentration of Rn-daugters will be lower tan te equilibrium Rn activity concentration. Tis is because of ventilation and plate-out, wic affect te daugters more tan it affects Rn itself. Te potential α-energy per Rn-daugter atom (ε pi, i = 1,..4) is te sum of α-disintegration energies for one atom of te nuclide and its sort-lived daugter nuclei: ε pi = 4 j i E αj Potential α-energy per unit activity: ε pi = Niεpi N i = εpit 12,i ln 2 Potential α-energy concentration: C p = 4 i=1 C act,i εpi [ J m 3 ] Equivalent equilibrium Rn-concentration in air: EEC Rn χ eq,rn = 4 εpi Cact,i i=1 λ i 4 ε pi i=1 = 4 C p i=1 ε pi = K C p EEC Rn is te concentration of activity of Rn in equilibrium wit its sort-lived daugter nuclei, wic would ave te same potential α-energy per unit volume air as te mixture of interest: χ eq,rn = 0.10 C act,po C act,pb C act,bi 214 Note! Contribution from C act,po 214 is extremely small because εpi 1 for Po-214. Empirical value of te equilibrium factor: F = EECRn χ Rn,air 0.5 4

5 Heavy duty ventilation results in a smaller F-value. Intake of potential α-energy during a time interval T: I pot = E A V in T χ eq,rn, E A = 4 ε pi i=1 = J/Bq Assume te inalation rate to be: Vin = 0.8 m3 during te time interval T. Te tracea-broncial region: D T B I pot = 1.5 Gy J K T B Pulmonal region: D P I pot = 0.2 Gy J K P T-B dose rate: Ḋ T B = DT B T = S T B χ eq,rn = F S T B χ Rn,air were E S T B = K V T B A in = 7 ngy 1 P dose rate: Ḋ P = DP T = S P χ eq,rn = F S P χ Rn,air were S P = K P E A V in = 0.9 ngy 1 Equivalent dose rate: Ḣ T = ω R Ḋ T,R, for α ω R = 20 Effective dose rate: Ė = ω T Ḣ T = ω R [ω T B Ḋ T B + ω p Ḋ p ] Wen taking into account te tissue weigting factors, it is assumed tat T-B and P contribute equally to te total lung tissue weigting factor. ω T B = ω P = ω l = Ė = S tot,rn daugters χ eq,rn = F S tot,rn daugters χ Rn,air S tot,rn daugters = 9.5 nsv 1 Finally, te total contribution from bot Rn and its daugters becomes: Ė tot = [S tot,rn + F S tot,rn daugters ]χ Rn,air Ė tot = S χ Rn,air If one assumes tat F=0.5 S = [ ] nsv 1 Bqm = 5 nsv 1 3 Yearly, one can assume tat an average person stays indoors about 80% of te time. Tis becomes about 7000 ours per year. Furter on, assuming tat te Rn exposure outdoors can be neglected: S = 5 nsv µsv = 35 Te mean Rn-concentration in Norwegian ouses was χ Rn,air = 88 Bq m 3 in From tis, it follows tat te effective dose rate becomes: 5

6 Ė tot = µsv msv = 3.0 Cancer risk due to Rn exposure Concentration limits in Norway χ Rn,air < 200: It is not necessary to take action. 200 < χ Rn,air < 400: Simple actions required. χ Rn,air > 400: Expensive actions required Measuring te amount of Rn and Rn-daugters Important conditions to take into account 1.) Te measuring device must not be affected by deposited Rn-daugters. For example on te surface of te detector. 2.) It must be known to wic degree it measures Rn, and to wic degree it measures Rn-daugters. 3.) Integration over long time is necessary to obtain good accuracy. 6

7 Measuring metods for air-borne Rn 1.) Te CB (Coal Box)-metod, consists of a box containing active coal, wic absorbs Rn-gas. Te box is open only during exposure wen Rn gas is adsorbed to te active coal. Measuring te activity of Rn-daugters, originating from te absorbed Rn-gas, is done using a NaI scintillation crystal via γ spectroscopy. A problem ere is tat te coal adsorbs air umidity more efficiently tan Rn-gas. Tis means tat te measuring results are more accurate in dry places. Anoter inaccuracy of tis metod is tat it does not integrate over very long time. Oter sources of radiation Cosmic radiation Particle radiation (85% protons, 15% α-particles) from space, and particle radiation as well as γ- radiation from te sun. Tese primary particles are transformed into secondary cosmic radiation consisting of various particle types and some γ-radiation, due to interactions and reactions in te atmospere. Cosmic radiation increases wit altitude above sea level. 7

8 Ground-level: Air-traffic altitude: (10.000m) 0.35 msv 5 µsv i.e µsv External γ-radiation External γ-radiation is mostly due to te existence of radioactive minerals in te ground. Te following nuclides constitute te main contributions to dose: 40 K 40% 232 T 40% 226 Ra 20% Te average effective dose from external γ is around 0.55 msv Naturally occuring internal radiation Natural internal radiation is mainly due to radiation from 40 K (β-emitter, T1 2 =109 s). Natural K consists of about a fraction of K. Te amount of K inside our bodies is regulated by te metabolism. Tis again implies tat te dose contribution is kept at a constant level. Average effective dose from internal radiation is about 0.37 msv. Te Tsjernobyl accident Te outburst resulted in a release of about 3.5% of te total amount of activity contained in te reactor. All te gaseous nuclei ( 85 Kr and 133 Xe) were released. Te fall-out consisted mainly of 137 Cs and 134 Cs. Te mean 137 Cs fall-out in Norway was 7 kbq m, but some places ad more tan 2 80 kbq m. 2 Dosimetry Total transfer factor for effective dose due to all te nuclides, based on 137 Cs ground deposition. For nortern areas, in units of µsv kbqm 2 : 8

9 Year 1 Total External Internal Sum Internal dosimetry is based on averaged transfer coefficients, assumptions concerning diet-composition and biokinetical models for up-take of radioactive substances. Average dose based on Norwegian conditions: Year 1 Total 40 µsv kbqm 2 7 kbq m 2 = 0.28mSv 150 µsv kbqm 2 7 kbq m 2 = 1.0mSv Dose-reducing measures Use of Cs-binders (as feed-admixture and tablets) reduces te up-take of Cs in domestic animals by up to 50-80%. 9