Question

Reactivity and Decay

A banana in your lunch contains 380 mg of potassium, of which a small fraction is the radioactive isotope 40 19 K. Using the percentage of radioactive potassium to nonradioactive potassium in Appendix B, estimate the activity of this banana due to the radioactive 40 19 K. decays/s

Answer 1

The activity of a radioactive sample consisting of a single radioactive isotope is given by:

A = ?*N

where ? is the decay constant of the isotope and N is the number of atoms of the isotope in the sample.

The decay constant and halflife (T) of a radioactive isotope are related by:

? = ln(2)/T

so,

A = ln(2)*N/T

The atomic mass of K is 39.1 gm/mol, so 380 mg = 0.38 gm of K is 9.72*10^-3 moles of K.

Multiplying by Avogadro's number, this many moles contains:

(9.72*10^-3 moles)*(6.022*10^23 atoms/mol) = 5.85*10^21 atoms.

40K makes up just 0.012% of all the atoms in a modern sample of ordinary potassium, so there are about:

0.00012*5.85*10^21 atoms = 7.02*10^17 atoms of 40K in a typical banana.

The activity of 40K in a typical banana is then given by:

A = (7.02*10^17 atoms)*ln(2)/(4.04*10^16 sec)

A = 12 atoms decaying/second