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Strategy

Dose in rem is defined by 1 rad = 0 . 01 J/kg size 12{"1 rad"` \( r \) =0 "." "01"`"J/k"} {} and rem = rad × RBE . The energy deposited is divided by the mass of tissue affected and then multiplied by the RBE. The latter two quantities are given, and so the main task in this example will be to find the energy deposited in one year. Since the activity of the source is given, we can calculate the number of decays, multiply by the energy per decay, and convert MeV to joules to get the total energy.

Solution

The activity R = 1 . 00 μCi = 3 . 70 × 10 4 Bq = 3 . 70 × 10 4 size 12{R=1 "." "00"`"μCi "=" 3" "." "70" times "10" rSup { size 8{4} } `"Bq"=3 "." "70" times "10" rSup { size 8{4} } } {} decays/s. So, the number of decays per year is obtained by multiplying by the number of seconds in a year:

3 . 70 × 10 4 decays/s 3 . 16 × 10 7 s = 1 . 17 × 10 12 decays.

Thus, the ionizing energy deposited per year is

E = 1 . 17 × 10 12 decays 5 . 23 MeV/decay × 1.60 × 10 13 J MeV = 0 . 978 J.

Dividing by the mass of the affected tissue gives

E mass = 0 . 978 J 2 . 00 kg = 0 . 489 J/kg. size 12{ { {E} over {"mass"} } = { {0 "." "978" `J} over {2 "." "00"`"kg"} } =0 "." "489"`"J/kg."} {}

One Gray is 1.00 J/kg, and so the dose in Gy is

dose in Gy = 0.489 J/kg 1.00 (J/kg)/Gy = 0.489 Gy.

Now, the dose in Sv is

dose  in  Sv = Gy × RBE size 12{"dose"`"in"`"Sv"=" Gy " times " RBE"} {}
= 0.489 Gy 20 = 9.8 Sv. size 12{ {}= left (0 "." "49"`"Gy" right ) left ("20" right )="10"`" Sv."} {}

Discussion

First note that the dose is given to two digits, because the RBE is (at best) known only to two digits. By any standard, this yearly radiation dose is high and will have a devastating effect on the health of the worker. Worse yet, plutonium has a long radioactive half-life and is not readily eliminated by the body, and so it will remain in the lungs. Being an α emitter makes the effects 10 to 20 times worse than the same ionization produced by β s, γ rays, or x-rays. An activity of 1.00 μ Ci is created by only 16 μ g of 239 Pu (left as an end-of-chapter problem to verify), partly justifying claims that plutonium is the most toxic substance known. Its actual hazard depends on how likely it is to be spread out among a large population and then ingested. The Chernobyl disaster's deadly legacy, for example, has nothing to do with the plutonium it put into the environment.

Risk versus benefit

Medical doses of radiation are also limited. Diagnostic doses are generally low and have further lowered with improved techniques and faster films. With the possible exception of routine dental x-rays, radiation is used diagnostically only when needed so that the low risk is justified by the benefit of the diagnosis. Chest x-rays give the lowest doses—about 0.1 mSv to the tissue affected, with less than 5 percent scattering into tissues that are not directly imaged. Other x-ray procedures range upward to about 10 mSv in a CT scan, and about 5 mSv (0.5 rem) per dental x-ray, again both only affecting the tissue imaged. Medical images with radiopharmaceuticals give doses ranging from 1 to 5 mSv, usually localized. One exception is the thyroid scan using 131 I size 12{"" lSup { size 8{"131"} } I} {} . Because of its relatively long half-life, it exposes the thyroid to about 0.75 Sv. The isotope 123 I is more difficult to produce, but its short half-life limits thyroid exposure to about 15 mSv.

Phet explorations: alpha decay

Watch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half life.

Alpha Decay

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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