Of the 600 mg of potassium in a large banana, 0.0117% is radioactive K-40 ...

Of the 600 mg of potassium in a large banana, 0.0117% is radioactive K-40, which has a half-life, </2 of 1.25 × 109 y. What is the activity of the banana? MK = 39.102 g/mol


Solution

The activity of a radioactive substance is measured in Becquerels (Bq), which is the number of radioactive decays per second. To calculate the activity of the radioactive potassium in a banana, you need to find the number of radioactive potassium atoms and then multiply by the decay constant, which is the reciprocal of the half-life.

Here is the calculation:

Mass of radioactive potassium: 

600 mg * 0.000117 = 0.0702 mg

Number of radioactive potassium atoms: 

0.0702 mg / 39.102 g/mol = 1.8 x 10^17 atoms

Decay constant: 

ln(2) / 1.25 x 10^9 years = 5.52 x 10^-10 per second

Activity: 

1.8 x 10^17 atoms * 5.52 x 10^-10 per second = 9.9 x 10^6 Bq

So the activity of a large banana is about 9.9 million Becquerels.

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