A bottle of red wine is thought to have been sealed about `5 years` ago. The wine contain a number of different kinds of atoms, including carbon, oxygen,and hydrogen. Each of these has a radioactive isotope. The radioactive isotope of carbon is the familiar `._6^(14)C` with a half-life of `5730 years`. The radioactive isotope of oxygen is. `._(8)^(15)O` and has a half-life of `122.2 s`. The radioactive isotope of hydrogen `._1^(3)H` and is called tritium, its half-life is `12.33 years`. The activity of each of these isotopes is known at the time the bottle was sealed. However, only one of the isotopes is useful for determining the age of the wine accurately. Which is it?

Explanation

The age of a sample is realted to the change in activity during the time peroid in question. The expected age of `5` years is a small fraction of the `5730` years half-life of `._6^(14)C`, so very few nuclei of `._6^(14)C` would decay during wine's life, the change in avtivity would be very small. Due to short half-life of `._8^(15)O` (122.25) so many half-lives of `122.25` would occur that the activity would decrease to a vanishingly samll level. The expected age of `5` years is long enough relative to the half-life of `._1^(3)H` `(12.33 years)` that a measurable change in activity will occur, so it is the most appropriate isotope. Hence, the radioactive isotope is chosen such that the expected age of the sample should be great enough so that the activity is undetectable. If a sample is about `50` years old and it contained hydrogen,one could measure activity of `._(1)^(3)H` (tritium) a beta emitter with a half- life `12.3 years`.