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EARS 5
Exercise #1
Radioactive decay and dating rocks
Due Monday June 30, 2003
Your mission. As the geologists’ version of Indiana Jones, you are searching, not
for the Temple of Doom, but for the Oldest Rock in the World. Right now your
archenemy claims credit for the Oldest Rock, dated at 4.1 billion years.
However, you believe you have found an older rock which you take to the
laboratory to date. In order to be absolutely sure of the age, you decide to
measure two radioactive elements with different half-lives; thus you will have
two independent estimates of the rock’s age. The elements are uranium-238
(238U), which decays to its daughter product lead-206 (206Pb), and potassium-40
(40K), which decays to argon-40 (40Ar). Uranium, lead, and potassium are solid
elements; argon is gaseous. You measure the relative numbers of atoms of these
elements, and you look up the half-lives for 238U and 40Ar. Here are your data:
Dating system
Ratio of number of
Half-life (years)
daughter atoms to parent
atoms (Nd/Np)
238U — 206Pb
40K — 40Ar
0.894
9.7
4.47 x 109
1.19 x 109
Answer questions 1-4. For questions 2-4, limit your answers to a paragraph or
so . Some background material is presented below that may help you.
1) What age(s) do you calculate for the rock? Be sure to consider the precision of
your answer, i.e. how many digits are meaningful (see discussion below).
2) Explain, in your own words, the physical process that the equation describes.
Why are the parent-daughter ratios different for the U-Pb and K-Ar systems?
3) You perceptively notice that the two ages you calculate are slightly different.
Identify and discuss at least 2 plausible hypotheses that might explain the
discrepancy. (Hint: what assumptions have you made in doing the lab
measurements and the calculation?)
4) Can you claim credit for having found the Oldest Rock in the World? Why or
why not?
Background
Radioactive decay. The number of atoms of a radioactive parent element in a
rock is given by
Ê1
N p (t) = N p (0)Ë ˆ¯
2
t /t1 /2
where Np(t) is the number of atoms of parent element at time t, Np(0) is the
number of parent atoms at the time the rock forms, and t1/2 is the half-life for
that particular radioactive species (this was derived in class and the derivation
can also be found at
http://www.dartmouth.edu/~ears5/handouts/raddecay.html) . Similarly, the
number of atoms of the daughter element, Nd(t) is
È Ê 1 t/t1/ 2 ˘
Nd (t ) = N p (0)Í1 - Ë ˆ¯
˙
2
Î
˚
Implicit in these equations are the assumptions that 1) the parent atoms are
locked into the rock when it forms, 2) that there are no daughter atoms initially
present, and 3) that all atoms formed by the decay process remain in the rock at
all times.
Either of these equations could be used to calculate the age of a rock if Np(0), the
number of parent atoms at the time the rock is formed, is known (as well as the
half life and the current number of atoms of either parent or daughter element).
Unfortunately, since no one was around at the time to measure it, Np(0) is not
known. We can get around this problem if we notice that both of the above
equations include the unknown quantity Np(0). If we divide the second equation
by the first, the unknown term will fall out:
Nd (t )
N p ( t)
=
È
1 (t / t1/ 2 ) ˘
N p ( 0) 1- Ê ˆ
Í Ë 2¯
˙
Î
˚
(
t/
t
)
1/
2
1
N p ( 0)Ê ˆ
Ë2¯
( t/ t1/ 2 )
=2
-1
With a little algebra, this can be solved for the age of the rock, t:
Ï È N d (t) ˘ ¸
Ô log ÍÎ N p (t) +1˙˚ Ô
t = t1/ 2 Ì
˝
log 2
Ô
Ô
Ó
˛
Significant figures. A hand calculator can display 10 or more digits in a number,
but are all of the digits meaningful? The answer is probably not. Here’s a trivial
example. Say you wanted to calculate the length of the third side of a right
triangle, and you have measured the length of two sides of the triangle as 6.3 m
and 7.2 m. The length of the third side, according to the Pythagorean theorem, is
the square root of the sum of the squares of the other two sides. Although your
calculator might be able to tell you that the area of the square is (6.32 + 7.22)1/2 =
9.56761312 m, if you reported this as your answer, a scientist would infer that
you claim to know the length of the third side to the nearest 10-8 m (i.e. about
1/10,000 to 1/1000 the diameter of a human hair). Do you really know the
length that precisely--or are you splitting hairs?? Obviously, you don’t know it
that well, and the number of digits in your answer should reflect how well you
think you know your answer---i.e. the precision of your answer.
Significant figures refers to the total number of digits in a number that are
believed to be precise. In the above example, the measured lengths of the two
sides were given to two significant figures. Reporting the figure to two
significant figures, meters and tenths of meters, implies that your measurement
is precise to 1/10 of a meter, or equivalently ± 0.05 m. If the measurement had
been given as 6.38 m, it would imply that you had measured to a centimeter
(1/100 m), or you could say that the uncertainty is 6.38 ±0.005 m.
When carrying numbers through a calculation, you need to keep in mind the
number of significant figures. A calculated number can only be as precise as the
least precise number that went into it. Thus if some measurements were precise
to three significant digits and some to four or more, a calculated number derived
from those measurements should only be reported to three significant figures.
A practical piece of advice: If you are too careful in a calculation and round all
intermediate results to the correct number of significant digits, you might
accumulate roundoff error. Instead, it is better to use more significant figures
than are justified in intermediate stages of a calculation, but then report the results
of the calculation rounded to the correct number of significant figures.
Rounding numbers. To round a number, look at the digit to the right of the
place to which you want to round. If the digit is 5-9, increase the digit by one and
leave off all following digits. If the digit is 0-4, simply leave off all following
digits. A few examples:
Original number
6.55
6.55666
6.55666
8.131
156,445
1.56445 x 105
Rounded to nearest
tenth
tenth
hundredth
hundredth
thousand
thousand
Rounded number
6.6
6.6
6.56
8.13
156,000
1.56 x 105
Logarithms
It is expected that you will be familiar with logarithms, so the following should
serve as a review.
Logarithms are the inverse operation to exponentiation. For example, you
know that 101=10, 102 = 100, and 103 = 1000; the operation performed in these
examples is exponentiation. We can write these examples as the general
expression bp = a where p is the power to which you raise the base b to give the
answer a.
The reverse operation, taking the logarithm, is equivalent to asking the question,
“what is the power p to which I must raise the base b in order to get a certain
number a?” Stated mathematically, this is p = logb a.
Here are some examples of exponentiation and the corresponding reverse
operation of taking the logarithm.
103 = 1000
10-2 = 1/100
100=1
if 10x = a
e3 = 20.09
e0=1
e-1=1/e
if ez=x
log10 1000 = 3
log10 1/100 = -2
log10 1 = 0
then log10 a = x
loge 20.09 = ln 20.09 = 3
ln 1 = 0
ln 1/e = -1
then ln x = z
Rules for algebraic operations with logarithms (these hold for ln (=natural
logarithm), log10 (=logarithm base 10), or any other base b of logarithm):
log(xy) = log x + log y
log(x/y) = log x - log y
logb x = logc x / logc b (change of base)