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Banana dose

Dear reader,

When I consider the banana dose I feel that it is so very silly.

I consulted the reference man and I discovered that a typical human gets 3.3 grams of potassium in their diet each day, also a typical human is 0.2 % potassium by weight. From these two facts I hold a view that it is possible to estimate a biological half life for potassium.

If the typical man weighs 75 kg, then he will contain 150 grams of potassium.

If we assume that all the potassium in the human body is in the same chemical form and is fast to exchange between the different parts of the body then we can make an estimate of how likely an atom of potassium is to be lost from the human body each day.

3.3 / 150 = 0.022

So each day 2.2 % of the potassium in the human body is exchanged for new potassium.

Now we should bear in mind that for a first order reaction, if we were to dose a person with a carrier free sample of potassium-40 then their radioactivity level would go up.

We can use the maths of radioactive decay to consider the problem.

The rate of loss of potassium will be given by the formula

Loss of potassium per day = k . amount in body

We should put some numbers into this equation

0.022 = k  x 1.000

So k = 0.022 days-1

As half life = ln 2 / k

Then the half life for potassium in a human body is normally equal to 31.4 days

We can then used the equation A = Ao exp -kt to deal with the problem, we can then work out how many decays will occur inside a person if they swallow a given activity of potassium-40. Based on the average amount of energy delivered to the person’s tissues per radioactive decay event we can then get a dose.

But this only is true for a person who has been internally contaminated with a carrier free sample of potassium-40.

 The great problem with the banana dose is that if a person was to go on a banana binge then their body would adapt to it by increasing the rate at which potassium is excreted from the person. As a result it is not possible to used the equation A = Ao exp -kt to deal with the problem. The radiation dose which the person will get will be smaller if the potassium-40 is mixed with normal potassium (K-39), than it would be if the potassium-40 was carrier free.

The maths will be much more complex for the potassium-40 mixed with normal non radioactive potassium, if any of my readers have time to do the calculation then please do write in. I will find the results interesting.

A better banana dose would be the radiation dose which a person will get if they hold a banana in their hand for an hour.

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