[ RadSafe ] Banana Equivalent Dose
Thompson, Dewey L
DThompson3 at ameren.com
Fri Mar 25 17:17:44 CDT 2011
Some musings about bananas.
The radioactivity concern is K-40. 0.0118% of all potassium is K-40. So let us do a napkin calculation:
Activity = XN, where X = decay constant, or 0.693/Half Life. N= Number of atoms.
Half Life for K-40 = 1.27 E+9 years. Therefore X = 0.693/1.27E9 = 5.46E-10/yr Convert to Seconds, divide by 3.15 E+7, results in 1.73E-17/sec
N: Gram Atomic Weight for K = 39.1 gm/mole. Therefore, each gram of K has (6.022 E+23 x 1.18 E-4)/ 39.1 = 1.82 E+18 atoms of K-40.
Specific Activity K, therefore = 1.73 E-17 x 1.82 E+18 =31.4 Bq, or 8.4 nci
The average recommended intake per day for Potassium is 4.7 gm.
The average banana has 422 mg K. Therefore, each banana you eat has 0.422 x 31.4 = 13.25 Bq.
The recommended daily intake of bananas therefore, is 11.1. You should eat 11.1 banana equivalents per day, which would be 4.7 x 31.4, or 147.5 Bq/day.
This would be 365 x 11.1, or 4052 banana equivalents per year. This would be 53,838 Bq/yr.
The Occupational ALI for ingestion of K-40 (USA Regulations) is 300 micro curies per year. (11.2 MBq?)
Therefore, you should limit your occupational ingestion of bananas to 3 E+5 /8.4, or 35,714 per year.
Math or logic check anyone?
Grin
Dewey
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