[ RadSafe ] Banana Equivalent Dose

[Thompson, Dewey L]

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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