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>>
View the other Key Equations and Concepts in this
chapter
Dating by Isotopic Analysis
| t = |
|
 ln |
|
(Equation 2.28) |
where t = time or the age of the object, t1/2
= half-life of the isotope. The units of time and half-life will
be the same. The mathematical function of natural log (its opposite
is ex), 1n is available on scientific calculators.
A0 is the initial amount of the isotope; At
is the amount of isotope at the time of measurement. Since it is
a ratio, the units don't matter, as long as they are the same for
both A0 and At. The units can
be mass, concentration, or radioactive activity. It is also possible
to start with the fraction of isotope rather than with the individual
amount of A.
Since logarithms are a different type of mathematical function,
there is actually a different significant figure rule for this function.
However, we will ignore that for this chapter and allow you to use
the multiplication and division rule for the entire problem.
>> Example 1
What is the age of an object that has 31.5% of its initial 14C
activity? (t1/2 = 5730 years)
Solution:
Since the remaining amount of 14C is given as a percentage,
the easiest way to interpret this into the equation is A0
= 1 and At = 0.315. (Note: the 1 is exact;
the 0.315 has three significant figures.)
| t |
= |
|
ln |
|
| |
= |
8268.4 |
ln |
(3.1746) |
| |
= |
8268.4 |
|
1.1552 |
| |
= |
9551.51 |
The units are years. There are three significant figures because
of the value of 0.693 and t1/2.
age = 9.55 x 103 years
>> Example 2
244Cf has a half-life of 17.9 years. How long will
it take for 1.00 mg to be left from a 1.00-g original sample?
Solution:
| t = |
|
ln |
|
From the problem, t1/2 = 17.9 y, A0
= 1.00 g, and At = 1.00 mg. The values for A
must be in the same units. As there are 1000 mg = 1 g, A0
is easily changed to 1000 mg.
| t |
= |
|
ln |
|
| |
= |
25.8297 |
|
6.907755 |
| |
= |
178.425 years |
(Remember time units must also match.)
Note that each value only has three significant figures, so
t = 178 years
>> Example 3
If after 348 days, 1.00 g of an isotope remains from a 5.00 g
sample, what is the half-life of the isotope?
Solution:
| t = |
|
ln |
|
From the problem, t = 348 days, A0 =
5.00 g, and At = 1.00 g
| 348 |
= |
|
ln |
|
| 348 |
= |
2.322 |
|
t1/2 |
|
|
= |
t1/2 |
| t1/2 |
= |
149.84 |
The time unit is days. The answer should have three significant
figures, so the final answer is 150 days.
>> Example 4
210Po has a half-life of 138.38 days. If 4.00 mg of
210Po were allowed to decay for 30.0 days, how much
would remain?
Solution:
| t = |
|
ln |
|
From the problem, t = 30.0 days, t1/2
= 138.38 days, A0 = 4.00 mg.
| 30 |
= |
|
ln |
|
| 30 |
= |
199.68 |
ln |
|
| 0.15024 |
= |
ln |
|
| e0.15024 |
= |
|
| 1.16211At |
= |
4.00 |
| At |
= |
3.44 mg |
The three significant figures were pretty constant throughout
the problem. Since the units on the original sample were milligrams,
the same units were on the final answer.
>> View
the other Key Equations and Concepts in this chapter
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