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- Fallout Fundamentals
- aFalloutconsists of dust particles that have been coated with radioactive
- by-products from atomic explosions.This occurs when the nuclear or atomic
- blast is a ground rather than air-burst (air-burst meaning that the fireball
- isfar enoughfrom the earth's surface that there is no ground material
- uptake into the high temperature portion of the mushroom cloud).In an air-
- burst the bomb products condensate into such very small particles that they are
- aloft for such a long time that they are mostly non-radioactive by the time
- they come down, typically months or years. The fission process gives off
- hundreds of different radioactive elements and isotopes.Also, a certian
- portion of the fission mass does not fission. The fussion portion of nuclear
- bombs is clean and gives off only helium, the atomic bomb trigger (fission)
- which starts the nuclear bomb (fussion) is the portion of the bomb that leaves
- radioactive by-products.
- Theseby-productscan be classified by their characteristics. One
- characteristic is half-life.The half-life is the length of time it takes for
- an element to give off one-half of its total radioactivity. This would also be
- the length of time required for a given amount to change to one-half the
- radioactive level, in other words if something was giving off radiation that
- would yield 3 Rads/hours, after one half-life it would give off 1.5 Rads/hour.
- An unstable isotope only emits radioactivity when one atomdecaysto
- another isotope or element (which may or not be stable,stable being non-
- radioactive). Therefore the portions of the element that are not in the
- process of decaying are not giving off any radioactivity.If you have
- "X" number of atoms of a radioactive element, "X/2" of those atoms will give
- off their radioactivity in the half-life period and become a different element
- or isotope. If an element has a half-life of 1 day, a given amount of it will
- give off 1/2 of its total radiation during the 1st day, 1/4 during the second
- day, 1/8 the 3rd day, 1/16th the 4th, 1/32 the 5th, 1/64 the 6th, 1/128th the
- 7th, et cetra. If you have a short half-life like Iodine 131 of 8, days most
- of the radioactivity (99+%) will be emitted in two months.In a long half-
- life element like plutonium 239 with a 24,400 year half-life, 1,000,000 atoms
- would in 24,400 years give of 1/2 of their radioactivity leaving 500,000 atoms
- of plutonium 239 at the end of those 24,400 years. 500,000 decays over 24,400
- years equals approx. 21 decays per one year.
- Another characteristic is the type of radiation given off, Alpha, Beta, or
- Gamma radiation. Neutron radiation is only given off by the actual blast
- itself and is not given off by the fallout itself. Only neutron radiation
- can MAKE something that is not radioactive become radioactive. This is why
- fallout can not cause something (like food inside a can) to become radioactive.
- Alpha, beta, and gamma radiation can NOT make anything become radioactive.
- Alpha radiation (helium nucleus, 2 protrons and 2 neutrons), like from
- plutonium, can be shielded with one layer of Cellophane or newspaper or several
- inches of air.Beta radiation (an electron) can be shielded by a layer of
- drywall, or several feet of air. Gamma radiation is electromagnetic radiation.
- Neutron radiation is a neutron and is about twice as hard to stop as Gamma.
- Gamma and neutron are harder to stop, you need several feet of dirt or
- concrete to absorb them. See below for specifics for stopping Gamma radiation.
- One factor that most people don't realize about fallout ishow fast it
- decays.Fallout follows the t-1.2 law which states that for every sevenfold
- increase in time after detonation there is a tenfold drop in radiation output.
- Example, a reading of X level of radioactivity at Y hours after detonation
- would indicate a level of radioactivity of .1X at 7Y hours after detonation.
- This is accurate for 2,500 hours (14 weeks) following theexplosion,
- thereafter the doserate is lower than t-1.2 would predict.Example, if a
- dose rate of 100 REM/hr was found at 1 hour after detonation(this assumes all
- significant fallout from the bomb has fallen, therefore starting with the
- seven hour point is probably more realistic) would be 10 REM/hr at 7 hours,
- 1 REM/hr at 49 hours(2 days), .1 REM/hr at 343 hours(2 weeks), .01 REM/hr at
- 2401 hours (14 weeks). A "survival safe" dose of radiation (this being defined
- as no short term effects or disability) is 3 to 12 Rads/day. This dose rate of
- 3-12 Rads/day can only be taken to an accumulated dose of 150-200 rads if done
- day after day. This would occur (assume 6 Rads/day) in this example at 150
- hours for 24 hour exposure, or at 49 hours for a 6 hours per day outside of
- shelter. Note though that since the level of activity is decreasing the time
- spent outside every day would increase. If you increase the radiation by a
- factor of 10 for another example would be where you would have 1,000 Rem/hr at
- 1 hr, 100 Rem/hr at 7 hrs., 1 Rem/hr at 343 hrs., .1 Rem/hr at 2401 hrs. The 24
- hour exposure would be at 1,000 hours(41 days) and 6 hour work day outside of
- shelter at 300 hours(12 days).
- For various levels of contamination a "no short term effects" dose of 6 Rads
- per day would be something like this: (for 80 col. printout)(measurements at
- boundries of the oval shaped pattern)
- Hours fromDose rateHours of "safe" work outside per day, medical effect
- explosion
- EXAMPLE AAn area 10 miles wide by 30 miles downwind directly downwind
- from of a missle field that gets dozens of hits
- 1 hr.10,000 R/hr None, 100% dead at 6 minutes of exposure
- 7 hrs.1,000 R/hr None, 100% dead at 1 hour of exposure
- 2 days 100 R/hr None, 50% dead within 3-4 hour continuous exposure
- 2 weeks 10 R/hr 36 minutes. 50% dead for 2 day continuous exposure.
- 14 wks(3 mo) 1 R/hr 6 hours/day. 50% dead for 1 month continuous exposure
- 5% dead for 15 day continuous exposure, no medical care
- and no deaths for 1 week continuous exposure.
- EXAMPLE B An area 10 miles wide by 30 miles downwind of a single 1 MT
- ground burst
- 1 hr 1,000 R/hr None, 100% dead at 1 hour of exposure
- 7 hrs. 100 R/hr None, 50% dead within 7-8 hour of continuous exposure
- 2 days10 R/hr 36 minutes. 50% dead for 5 days of continuous exposure.
- 2 week 1 R/hr 6 hours/day. 50% dead for 1 month continuous exposure.
- 14 weeks0.1 R/hr All day. 0% deaths from radiation hereafter.
- EXAMPLE C An area 12 miles wide by 95 miles downwind for a single 1 MT
- ground burst
- 1 hrradiation has not arrived yet.
- 7 hrs.50 R/hr 12 minutes, 50% dead for 18 hour continuous exposure
- 2 days 5 R/hr. 1 hour, 5% dead for 2 week continuous exposure
- 2 weeks 0.5 R/hr 12 hours/day.
- 14 weeks 0.05 R/hr Unlimited.
- The above three examples indicate conditions and exposures that would only
- be acceptable in wartime. In these examples the wind is continuous in
- direction and velocity. A real wind would not make such nice neat ovals. It
- should be noted that even in real wind conditions, marching perpendicular to
- the depositing wind will remove you from a individual fallout zone.
- Here is an example of the levels of contamination from a single 1 MT ground
- burst with a 15 MPH wind
- Area downwind Arrival Accumulated total radiation doseDose Rate in Rads/hr
- (boundries)time forat
- in milesfallout 1 week 4 weeks 15 weeks 100 yrs 7 hrs. 2 days(14 hrs)
- 33 x 7 1.5 hrs 3000 R 3300 R3600 R 4600 R 100 R/hr10 R/hr
- 95 x 12 5 hrs.900 R1200 R1400 R 1700 R ~50 R/hr 5 R/hr
- 160 x 1810 hrs. 300 R400 R 460 R 650 R not there yet 2 R/hr
- 245 x 2016 hrs90 R 120 R 150 R 240 R not there yet 0.7 R/hr
- For shelter from Gamma radiation the standard rule of thumb is 150 pounds of
- mass per square foot of cross section of shelter wall yields a PF, protection
- factor, of 40. This means if you had two shelters on a flat contaminated field
- with one having walls of one layer of cellophane and the other of walls and
- ceiling of something that had for its thickness 150 lbs/sq. ft.( note this
- would be a thickness of 2.5" of lead, 4" of steel, 12" of concrete, 18" of
- soil, 30" of water, 200' of air) you would recieve 1/40th the dose in the 150
- lb/sq.ft. walled shelter. This effect can be multiplied. If the sq. ft. cross
- section was 300 lbs. that would be 1/40th of 1/40th or 1/1,600th of the
- unprotected dose.Take for example a dose rate starting at 100 Rem/hr at 1
- hr.,10Rem/hr at 7 hrs.,1 Rem/hr at 49 hours, etc. If exposure started at 1
- hour the total dose would be 240 R in 1 day, 310 R in 1 week, 350 R in 4 weeks,
- 390 R in 15 weeks. The same in a PF 40 shelter would be 6 R in 1 day, 7.7 R
- in 1 week,8.7 R in 4 weeks. The difference would be 5% fatalities-most
- others suffering from nausea and taking about 1 month to recover without the
- protection versus 0% fatalities-0% sickness with protection of PF40 in this
- case.
- Another example with a dose rate starting at 1,000 Rem/hr at 1 hr.,100
- Rem/hr at 7 hrs., 10 Rem/hr at 49 hours, etc. If exposure started at 1 hour the
- total dose would be 2,400 R in 1 day, 3,100 R in 1 week, 3,500 R in 4 weeks,
- 3,900 R in 15 weeks. This in a 40 PF shelter would be 60 R in 1 day, 77 R in a
- week, 87 R in 4 weeks. In a 1,600 PF shelter this would be 1.5 R in 1 day,
- about 2 R in 2 weeks, about 2.5 R in 15 weeks. The differences here would be -
- no protection = 100% fatalities in several hours - PF 40 = 0% fatalities, 25%
- suffer nausea(at the most) with total recovery in 7 days, - PF 1600 no effects.
- Please note that protection factor increases as a multiple. If 150 lbs/ft.
- sq. = a PF of 40(1/40th or 2.5%), 300 lbs/ft sq. = a PF of 1,600(1/1,600th or
- 0.0625%), and 450 lbs/ft. sq. = a PF of 64,000(1/64,000th or 0.0015625%)
- Typical Swiss domestic shelters have a PF of 16,000 to over 2,500,000.
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