Decay Constant and Radioactivity
The relationship between half-life and the amount of a radionuclide required to give an activity of one curie is shown in the figure. This amount of material can be calculated using λ, which is the decay constant of certain nuclide:
The following figure illustrates the amount of material necessary for 1 curie of radioactivity. It is obvious, that the longer the half-life, the greater the quantity of radionuclide needed to produce the same activity. Of course, the longer lived substance will remain radioactive for a much longer time. As can be seen, the amount of material necessary for 1 curie of radioactivity can vary from an amount too small to be seen (0.00088 gram of cobalt-60), through 1 gram of radium-226, to almost three tons of uranium-238.
Example – Calculation of Radioactivity
A sample of material contains 1 mikrogram of iodine-131. Note that, iodine-131 plays a major role as a radioactive isotope present in nuclear fission products, and it a major contributor to the health hazards when released into the atmosphere during an accident. Iodine-131 has a half-life of 8.02 days.
- The number of iodine-131 atoms initially present.
- The activity of the iodine-131 in curies.
- The number of iodine-131 atoms that will remain in 50 days.
- The time it will take for the activity to reach 0.1 mCi.
- The number of atoms of iodine-131 can be determined using isotopic mass as below.
NI-131 = mI-131 . NA / MI-131
NI-131 = (1 μg) x (6.02×1023 nuclei/mol) / (130.91 g/mol)
NI-131 = 4.6 x 1015 nuclei
- The activity of the iodine-131 in curies can be determined using its decay constant:
The iodine-131 has half-live of 8.02 days (692928 sec) and therefore its decay constant is:
Using this value for the decay constant we can determine the activity of the sample:
3) and 4) The number of iodine-131 atoms that will remain in 50 days (N50d) and the time it will take for the activity to reach 0.1 mCi can be calculated using the decay law:
As can be seen, after 50 days the number of iodine-131 atoms and thus the activity will be about 75 times lower. After 82 days the activity will be approximately 1200 times lower. Therefore, the time of ten half-lives (factor 210 = 1024) is widely used to define residual activity.