Re: Bananas (K-40)

["Tom Johnson" <tjohnson@radtrain.com>]

Assuming that the person of interest is standing on top of an "infinite"
depth of bananas (6000 tons would appear to be so about 1 meter from the
surface), 100 Bq/kg of K-40, and that a banana's density is 1.6 g/cc (I am
not sure of a banana's density, do they float?), assuming no beta dose, and
using ICRP 26 methodology (weighting factors) and FGR 12, I get around
0.0003 mrem/hr to the person. Using ICRP 60, I get about the same answer.

My wife informed me that I have gone bananas.

Tom Johnson

PS: I need to know the isotope to calculate the dose from the surface
contamination on the cask to a person 1 meter away.


>Inspired by the recent discussion we had about BBq (bananabecquerels) I
>wonder if you could help me with a calculation. Going back to the data,
>bananas should contain about 100 Bq (K-40) per kilo. My mind then went
>to boat transports of bananas. A boat for Sweden may typically come with
>6000 tons of bananas in one transport.

>My question concerns the K-40 related dose rate for someone working
>close to the bananas on the boat. The bananas come in boxes so for
>simplicity, I think one can assume that the "critical group/person" is
>standing 1 meter from the wall of a "cube" or "rectangular superbox"
>with bananas (symmetrical position – same distance to the right as to
>the left corner). I realize that some assumptions must be made here
>about density, self-absorption etc –but I still ask: Can anyone simplify
>the problem and give me a realistic estimate of the K-40 dose rate? I
>don't expect any megadoses to come out from this - it is more of
>curiosity in the question.


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