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In the October 2004 issue, David W. Ball published a paper entitled “How Radioactive Is Your Banana?” (1). He calculated that the activity of a big banana equals 18.4 Bq (decays per second). It is interesting to calculate the activity of an average human body.
Potassium, one of the elements most abundant in the Earth crust (2.1%), has three isotopes (1,
2):
Nuclide |
Content in Natural
Potassium
(%) |
Relative Atomic Mass
(rounded) |
|
39K |
93.2581 |
38.9637 |
40K |
0.0117 |
39.9640 |
41K |
6.7302 |
40.9618 |
|
The relative atomic mass of natural potassium Ar = 39.0983. One of its isotopes,
40K, is radioactive with half-life of t1/2 = 1.26 × 109 y. A 70-kg person’s
body contains 140 g of potassium (interestingly, the sodium content is less:
100 g)(2). Hence, a person’s body contains approximately 0.0164 g of
40K, which is 230 times more than in a big banana (1). Thus, the activity of
a body will be 230 × 18.4 ≈ 4230 Bq, and during one year, there
will occur about 1.33 × 1011 (130 billion!) decays of 40K.
It is easy to calculate how much “radiogenic” calcium-40 will accumulate during this time (and also during a person’s life time) in a body according to beta-decay: 40K → 40Ca + e. But it is much more interesting to calculate the volume of gaseous argon formed in the body. As a matter of fact, not all 40K atoms decay as in the direction indicated above; approximately 12% of all decays proceed through electron capture (3): 40K + e → 40Ar (by the way, most of atmospheric argon is formed via the decay of 40K). Hence, during one year, in the 70 kg person, 1.33 × 1011 × 0.12 = 1.6 × 1010 atoms of argon will be produced, and for 50 years 8.0 × 1011 atoms or 1.33 × 10–12 mol. At STP, it will account for only 3 × 10–8 mL. Thus, to inflate a 5-L rubber balloon with this argon, 5 × 103/3 × 10–8 ≈ 1.7 × 1011 persons should “deliver” radiogenic argon during almost all their life time, and this number is 30 times greater than the entire population of the Earth!
Radioactive disintegration of 40K also allows solution of another problem: what would be the relative atomic mass of potassium if an imaginary chemist could define it at the moment when our planet was formed? The atomic mass of natural potassium (39.0983) is less than that of 40K (39.9640), therefore, as a result of decay of 40K, the value Ar(K) does not remain constant, but gradually decreases, albeit very slowly. The half-life of 40K equals 1.26 × 109 years (1). Hence, approximately four half-life periods have passed from the time of the Earth formation (approx. 4.6 × 109 years) (3) and the amount of 40K has decreased 16-fold (more precisely, N0/N = exp[(ln2) × 4.6/1.26] = 12.6). Consider a present-day sample of Earth’s crust containing 100 potassium atoms. They contain (on average) 93.2581 atoms of K, 0.0117 atoms of 40K, and 6.7302 atoms of 41K. At the time of Earth’s formation, the quantity of 39K and 41K atoms was the same, whereas the number of 40K atoms was 12.6 times greater, that is, 0.1474 atoms. Therefore, when Earth was formed, this sample contained 100 + (0.1474 – 0.0117) = 100.1357 atoms of potassium, the total weight of these atoms was 93.2581 × 38.9637 + 0.1474 × 39.9640 + 6.7302 × 40.9618 = 3915.2524, and the relative atomic mass of natural potassium was 3915.2524/100.1357 = 39.0995.
Literature Cited
- Ball, D. W. J. Chem. Educ. 2004, 81, 1440.
- Emsley, J. The Elements, 2nd ed.; Clarendon Press: Oxford, 1991.
- Hurley, P. M. How Old Is The Earth? Anchor Books: New York, 1959;
Chapter 4.
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