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The radioactivity of banana (I did this for cigarettes also)

March 28, 2011

Mohan, mdashf

Bananas Are Radioactive

By Anne Marie Helmenstine, Ph.D., Guide



“Bananas are radioactive for a similar reason. The fruit contains high levels of potassium. Radioactive K-40 has an isotopic abundance of 0.01% and a half-life of 1.25 billion years. The average banana contains around 450 mg of potassium and will experience about 14 decays each second. It’s no big deal. You already have potassium in your body, 0.01% as K-40. You are fine. Your body can handle low levels of radioactivity.”


so the half life is 1.25 billion years and decays 14 times a second?

are you sure you know what your talking about?


@joe: There are many, many, many, many, many molecules of potassium in 450 mg of potassium.


@joe, the math looks a bit off, but not in the way you’re thinking, so let me break it down.

half life of K-40 = 1.248e9 years

natural abundance of K-40 = .012 %

molar mass of K-40 = 39.96399848 g/mol

450 mg * .012% = 5.4 mg

5.4 mg = .0054 g

.0054 g / 39.96399848 g/mol = 1.3512e-4 mol

1.3512e-4 mol * 6.022e23 particles/mol= 8.137e19 atoms K-40

4.0685e19 decay in the first 1.248e9 years

average of 3.26e10 per year

average of 1033 per second



Close, but its .012% which is really 0.00012 which works out to about 0.054 mg.


Good work Chris and Mark. However your estimate for number of decays per second will be off because you didn’t figure them using logs. Radioactivity is an exponential decay function. So the number of decays per time unit starts higher and then slowly gets lower over time. So one second now will have almost twice as many decays as one second 1.248e9 years from now. Of course we are interested in the rate of decays for seconds now not 1 billion years in the future. This is what pushed the number to 14 per second.


Hi Chris, Mark:

you laid the ground work. Thanks. I will complete the calculation and show you this is exactly rounded to 14 decays per second (if you change your fraction of radioactive K-40 to 0.01 as in the article rather than 0.012 etc I think it will give the exact answer)

Here is the answer:

I take the calculation you two have done correctly, i.e. the 450 milligram banana contains 8.137×10^17 nuclei of K-40 (not 10^19 as corrected by Mark)

Now the decay constant or transition probability as one may call it which gives the probability at any time of a single nucleus to decay at any time is w = 0.693/T_half_life. This is obtained by using Heisenberg Uncertainty principle for the nuclear decay. Here decay width which is given as a width of energy and mean life time which is given as a width or uncertainity in time must multiply to h-cross (time – energy uncertainty)

Now Mean life is related to half life by the factor 0.693 (69.3% of nucleus would decay in the mean life where as 50% would decay in a half life)

SO the decay probability = w gives the probability or likelihood that any single nucleus will decay or not. SO we know this probability because we know the half life to be 1.25 billion years that is 1.25X365X24X3600X10^9 seconds = 1.25X3.15X10^16 seconds.

Since the banana has 450 milligrams of Potassium it has as many radioactive K-40 as we just counted, 8.137X10^17 nuclei, at the time of eating. SO the probability of decay is w and you know how many nuclei you got. Multiply them to get the exact number of radioactivity of your banana.

Needless to say this product is 14.3.

(8.137X0.693)/(1.25X3.15)X10^(17-16) = 1.43X10 = 14.3

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