32 Ordinary Differential Equations
in the past), the time that a particular sample of the same substance died can
be calculated from equation (10). For example, in 1950 the rate of radioactive
disintegration of^14 C from a piece of charcoal found in t he Lascaux cave in
France was .97 disintegrations per minute per gram. Tissue from li ving wood
has a disintegration rate of 6.68 disintegrations per minute per gram. So the
tree from which the charcoal came died
t - to = -.^1 ( .97)
00012449
ln
6.
68= 15,500 years before 1950.We have only discussed radioactive substances with relative long half-lives.
However, the reader should be aware that there are radioactive substances
with half-lives on the order of a few years, a year, a day, a second, and even
a very small fraction of a second.
EXERCISES 1.4
l. If 53 of a radioactive substance decomposes in 50 years, what percentage
will be present at the end of 500 years? 1000 years? What is the half-
life of the substance?- If the half-life of a radioactive substance is 1800 years, what percentage
is present at the end of 100 years? In how many years does only 103 of
the substance remain? - If 100 grams of a radioactive substance is present 1 year after the sub-
stance was produced and 75 grams is present 2 years after the substance
was produced, how much was produced and what is the half-life of the
substance? - In 1977 the rate of carbon 14 radioactivity of a piece of charcoal found
at Stonehenge in southern England was 4.16 disintegrations per minute
per gram. Given that the rate of carbon 14 radioactivity of a living tree
is 6.68 disintegrations per minute per gram and assuming the tree which
was burned to produce the charcoal was cut during the construction of
Stonehenge, estimate the date of the construction of Stonehenge.