SSAC2005.QE514.CES1.1
Radioactive Decay and Popping Popcorn –
Understanding the Rate Law
Radiometric determination
of age is crucial to
understanding geologic
time. Radiometric age
dating is possible because
radioactive decay follows a
rate law. What is that rate
law?
Core Quantitative Issue
Exponential function
Supporting Quantitative Concepts
Number sense: Geometric progression
Number sense: Dimensions vs. units
Calculus: rate of change
Differential equation for the exponential function
Graph, logarithmic scale
Graph, trend line
Probability: Law of Large Numbers
Prepared for SSAC by
C E Stringer, University of South Florida - Tampa
© The Washington Center for Improving the Quality of Undergraduate Education. All rights reserved. 2005
1
Preview
This module is the first of a series on radioactive decay and how its mathematics is
used to quantify the age of geologic materials. This subject is fundamental to
understanding the magnitude of geologic time, the rate of geologic processes, and
the quantitative history of the Earth. The key concept of the mathematics is that the
rate of decay (the radioactivity) is proportional to the amount of the reactive isotope
present (the “parent” isotope). As a result, the declining amount of the parent
isotope can be expressed by an exponential-decay function. The concept of a
constant half-life is a corollary.
The goal of this module is to introduce the basic mathematics that describes
radioactive decay. The module uses an analogy between a large number of the atoms
of a radioactive isotope and a large number of popping kernels of popping popcorn.
Slides 3 and 4 give background information on radioactive decay, and Slide 5
introduces a problem designed to help you understand the mathematics of decay by
means of the popcorn analogy.
Slides 6-11 introduce Excel spreadsheets and graphs that help you solve the problem
numerically, using a finite time step. Slides 12-14 have you consider a smaller time
step, and Slide 15 illustrates how the standard analytical solution to the problem is
approached with shorter and shorter time steps.
Slides 16 and 17 wrap up with conclusions, final thoughts and references. Slide 18
gives the homework assignment.
2
Radioactive Decay
When a nuclide decomposes (or decays) to form a
different nuclide, it is called a radioisotope. The
phenomenon is called radioactivity.
Terminology: Forms of an element with the
same atomic number but different mass
numbers (meaning they have different numbers
of neutrons) are called isotopes.
When a radioisotope decays to form a
different nuclide, it emits a particle.
The three initial types of particles
recognized were α-particles, βparticles, and γ-radiation.
The radioisotope can also be thought of as
the “parent” and the nuclide it decays to
can be termed the “daughter.”
Here is an example of a decay equation:
Uranium is the
parent nuclide.
92 is the
atomic number
and 238 is the
mass number.
U
238
92
Th He
234
90
4
2
The helium atom is
an a- particle.
Thorium is the daughter nuclide.
Remember that the atomic number is the number of protons in an atom’s nucleus and
3
the mass number is the number of protons plus neutrons!
Radioactive Decay
When a parent decays to a daughter product,
the daughter may decay again to yet another
atom. These transformations take place until
a stable, non-radioactive isotope is formed.
The series of reactions is referred to as a
decay series or decay chain.
There are three naturally-occurring
decay series: the U-238, Th-232, and U235 chains.
Example
The figure on the right shows the
Uranium-238 series. Uranium-238 is the
parent nuclide and Lead-206 is the
stable, final daughter nuclide. The
column on the left tells you what type of
radiation is emitted in each decay
reaction.
4
From www.compumike.com/ science/halflife1.php
Problem
We can’t say when a given
radioactive atom of a parent
isotope will decay to produce a
radiogenic atom of the
daughter isotope. All we can
say is that there is a certain
probability that the atom will
spontaneously convert in a
given amount of time.
In the same way, we
can’t say exactly when
a given kernel of
popping corn will pop
into a piece of
popcorn…
For example, the
probability that any given
atom of Carbon-14 will
emit a beta particle (and
become an atom of
Nitrogen-14) in the next
year is 0.012%.
Let’s say that there is
a 10% probability that
any given unpopped
kernel in a popcorn
popper will pop in the
next ten seconds.
What then? Assume
there are 1000 kernels
in the popper.
The concept at work here is the Law of Large Numbers, one of the cornerstones
of probability theory. If the number of kernels is large then we can safely say that
10% of them will pop in the 10-second interval. Does 1000 seem to be a large
number of Carbon-14 atoms? See End Note 1.
5
Restating the problem; Setting up the spreadsheet
Suppose you put 1000 kernels of popcorn in
a popcorn popper and raise the temperature
to a constant level hot enough for the kernels
to begin popping. Each kernel of popcorn
has the potential to pop, but they don’t all
begin popping at the same time. If the heat
is left at a constant level for a long enough
period of time, most of the kernels will
eventually pop but you won’t know which
one will pop at which time.
Cell C3 is the number of kernels you start
with in the popper.
Column B lists the numbers of seconds
that have passed. Remember we are
thinking in 10-second intervals.
Let’s say that each unpopped kernel has a 10%
probability of popping during any 10-second
interval. How many unpopped kernels will there
be after a 10-second time step? Create the Excel
spreadsheet shown below to find out!
B
2
3
4
5
6
7
8
Set up Column C to calculate the number of kernels
remaining unpopped after each 10-second period. Create an
absolute reference in Cell C7 by typing =$C$3. The formula in
Cell C8 should be
=C7-$C$4*C7.
Nstart =
Ppop
Time (t )
(sec)
0
10
C
D
1000
0.1 per 10 sec
Nunpopped
(N )
1000
900
Cell C4 is the probability that a kernel
will pop in a
10-second interval.
6
What happens in the 10-second intervals after the first one?
B
Expand your Excel spreadsheet to chart the
number of remaining kernels through 14 more
time steps.
First, create all fifteen 10-second time steps
in Column B.
Because we assumed that our 10%
probability of popping remains the same, we
can simply copy and paste our formula from
Cell C8 down the column to complete
Column C.
Create Column D to look at the fraction of
the kernels that remain unpopped after
each 10-second time step. Why is this
number (N/N0) of interest?
Change Cell C3 to 2000, or
5000, or 10,000. What do
you observe about N/N0?
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Nstart =
Ppop
sec
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
C
D
1000
0.1 per 10 sec
Nremain
(N )
1000
900
810
729
656
590
531
478
430
387
349
314
282
254
229
206
N /N 0
1.00
0.90
0.81
0.73
0.66
0.59
0.53
0.48
0.43
0.39
0.35
0.31
0.28
0.25
0.23
0.21
Do you notice a pattern in Columns C
and D? See End Note 2.
7
Looking at Popcorn Popping Graphically
Create a graph by plotting the seconds on
the x-axis and number of remaining kernels
(parents) on the y-axis
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Nstart =
Ppop
C
D
F
E
G
Time (t )
(sec)
Nunpopped
(N )
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
1000
900
810
729
656
590
531
478
430
387
349
314
282
254
229
206
1000
N /N 0
1.00
0.90
0.81
0.73
0.66
0.59
0.53
0.48
0.43
0.39
0.35
0.31
0.28
0.25
0.23
0.21
Exponential-decay phenomena
are characterized by a
constant half-life.
I
H
J
This looks like an exponential-decay
function. An exponential function plots
as a straight line when the dependent
variable is plotted on a logarithmic scale.
So, right-click on the y-axis, select
“’format axis,” select the “scale” tab, and
select “logarithmic scale.”
1000
0.1 per 10 sec
900
800
Kernels remaining
B
The half-life is the time that it takes for the
reaction to proceed to where half of the
popcorn remains unpopped (End Note 3).
700
600
500
400
300
200
100
0
0
100
50
150
Time (seconds)
Estimate the half-life (in seconds) from the table and
graph. Is the half-life constant? In other words: How
long is the quarter-life? Is it two half-lives?
8
Looking at Popcorn Popping Graphically, 2
Insert an exponential trend line and use the
option tab to display the equation for the line and
correlation coefficient. Record the equation.
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Nstart =
Ppop
C
D
E
F
G
H
I
J
1000
0.1 per 10 sec
S
Nunpopped
(N )
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
1000
900
810
729
656
590
531
478
430
387
349
314
282
254
229
206
Here y represents
N, the number of
unpopped
kernels, and x
represents time in
seconds.
1000
N /N 0
1.00
0.90
0.81
0.73
0.66
0.59
0.53
0.48
0.43
0.39
0.35
0.31
0.28
0.25
0.23
0.21
900
-0.0105x
y = 1000e
R2 = 1
800
Kernels remaining
B
2
3
4
5
700
600
500
400
300
200
100
0
0
50
100
Time (seconds)
150
R2 = 1 means that
the fit is perfect.
The equation
describes the
listed values of N
vs. time with no
scatter.
So, now, how does the rate of decay (i.e., the
radioactivity) vary with time? Would you say
that the rate of decay is constant? (End Note 4)
9
Looking at Popcorn Popping Graphically, 3
Insert Column E to calculate the number of
kernels that pop in the next 10-second time
step. This is ΔN10-sec, where the subscript
refers to the 10-sec time step.
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Nstart =
Ppop
C
D
E
G
F
J
I
H
K
Here y represents
ΔN10-sec, the
number of
popping kernels
per 10 seconds,
and x represents
time in seconds.
1000
0.1 per 10 sec
Time (t )
(sec)
Nunpopped
(N )
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
1000
900
810
729
656
590
531
478
430
387
349
314
282
254
229
206
N /N 0
1.00
0.90
0.81
0.73
0.66
0.59
0.53
0.48
0.43
0.39
0.35
0.31
0.28
0.25
0.23
0.21
Number
popped
(ΔN 10)
100
90
81
73
66
59
53
48
43
39
35
31
28
25
23
100
90
Kernels popped per 10 secs
B
2
3
4
5
Plot ΔN10-sec vs. time and the exponential trend
line for the variation of reaction rate vs. time.
What can you conclude from the graph and trend
line? When is the reaction rate half of the original
rate? When is it one-fourth of the original rate?
-0.0105x
y = 100e
2
R =1
80
70
60
50
40
30
20
10
0
0
50
100
Time (seconds)
150
R2 = 1 means that
the fit is perfect.
The equation
describes the
listed values of
ΔN vs. time with
no scatter.
So, the rate decreases exponentially.
What besides the half-life is constant?
10
The Constant Ratio
Insert Column F to calculate the ratio of the
change in number of kernels in the 10-second
time step to the number of kernels present at
the beginning of the time step. Note that that
ratio is constant down the column.
B
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Nstart =
Ppop
C
D
E
F
1000
0.1 per 10 sec
Time (t )
(sec)
Nunpopped
(N )
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
1000
900
810
729
656
590
531
478
430
387
349
314
282
254
229
206
N /N 0
1.00
0.90
0.81
0.73
0.66
0.59
0.53
0.48
0.43
0.39
0.35
0.31
0.28
0.25
0.23
0.21
Number
popped
(ΔN 10)
100
90
81
73
66
59
53
48
43
39
35
31
28
25
23
Relative
rate
(ΔN 10/N )
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
Thus the ratio ΔN10-sec/N is
constant. Moreover, it is equal to
the probability that any given
kernel will pop in the next 10
seconds (the value in Cell C4)
Hence the rate law for this case:
the number of kernels that pop in
the next 10 seconds is directly
proportional to the number of
kernels that are present,
ΔN10-sec = k10-secN,
where k10-sec, the constant of
proportionality, is the probability
that any given kernel will pop in
the next 10 seconds.
11
A Shorter Time Increment
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
Nstart =
Ppop
C
D
E
F
G
H
I
J
Now modify the spreadsheet
of Slide 8 to consider a
smaller time increment.
Instead of a 10-second time
step, use a 1-second time
step. Instead of a 10%
probability of popping in 10
seconds, use a 1% probability
of popping in one second.
Plot the number of unpopped
kernels as a function of time
for 30 seconds, and determine
the equation of the trend line.
1000
0.01 per sec
Time (t )
(sec)
Nunpopped
(N )
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1000
990
980
970
961
951
941
932
923
914
904
895
886
878
869
860
851
843
835
826
818
810
802
794
786
778
770
762
755
747
740
N /N 0
1.00
0.99
0.98
0.97
0.96
0.95
0.94
0.93
0.92
0.91
0.90
0.90
0.89
0.88
0.87
0.86
0.85
0.84
0.83
0.83
0.82
0.81
0.80
0.79
0.79
0.78
0.77
0.76
0.75
0.75
0.74
1000
900
800
Kernels remaining
B
2
3
4
5
700
600
y = 1000e -0.0101x
R2 = 1
500
400
300
200
100
0
0
5
10
15
20
25
30
Time (seconds)
Compare the exponent in this
equation to that in Slide 9.
Compare the number of
remaining kernels at 30
seconds to that in Slide 9.
What is going on?
We will come back to that
question. First, what
about ΔN/N for this
shorter increment?
12
A Shorter Time Increment, 2
B
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
Nstart =
Ppop
C
D
E
F
1000
0.01 per sec
Time (t )
(sec)
Nunpopped
(N )
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1000.0
990.0
980.1
970.3
960.6
951.0
941.5
932.1
922.7
913.5
904.4
895.3
886.4
877.5
868.7
860.1
851.5
842.9
834.5
826.2
817.9
809.7
801.6
793.6
785.7
777.8
770.0
762.3
754.7
747.2
739.7
N /N 0
1.000
0.990
0.980
0.970
0.961
0.951
0.941
0.932
0.923
0.914
0.904
0.895
0.886
0.878
0.869
0.860
0.851
0.843
0.835
0.826
0.818
0.810
0.802
0.794
0.786
0.778
0.770
0.762
0.755
0.747
0.740
Number
popped
(ΔN 1sec)
10.0
9.9
9.8
9.7
9.6
9.5
9.4
9.3
9.2
9.1
9.0
9.0
8.9
8.8
8.7
8.6
8.5
8.4
8.3
8.3
8.2
8.1
8.0
7.9
7.9
7.8
7.7
7.6
7.5
7.5
Relative
rate
(ΔN 1sec/N )
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
As in Slides 10 and 11, insert a column (E) to
calculate the number of kernels that pop in the
next time step, and a column (F) to calculate
the ratio of the number of kernels that are
popping to the number that are present. Note
that the ratio is constant through all the time
steps.
Hence the rate law for this case:
the number of kernels that pop in
the next 1 second is directly
proportional to the number of
kernels that are present,
ΔN1-second = k1-secondN,
where k1-second, the constant of
proportionality, is the probability
that any given kernel will pop in
the next second.
13
Comparing the two time steps
1000
900
y = 1000e -0.0105x
R2=1
Kernels remaining
800
700
600
500
400
300
200
100
0
0
50
100
150
For the 10-second time step:
The ratio of ΔN10-sec, the number of
kernels that popped in the 10-second
Δt to the number, N, that is present at
the beginning of the time step is 0.01
per second, and the exponent in the
exponential function is -0.0105t, where
t is in seconds. The predicted number
of kernels at 30 seconds is 729.
Tim e (seconds)
For the 1-second time step:
The ratio of ΔN1-second, the number of
kernels that popped in the 1-second Δt
to the number, N, that is present at the
beginning of the time step is 0.01 per
second, and the exponent in the
exponential function is -0.0101t, where
t is in seconds. The predicted number
of kernels at 30 seconds is 739.7.
1000
900
Kernels remaining
800
700
600
y = 1000e -0.0101x
R2 = 1
500
400
300
200
100
0
0
5
10
15
20
Tim e (seconds)
25
30
Is there a limit? What if Δt  0?
14
The Rate Law
Approaching a limit….
This spreadsheet calculates the
number of kernels that remain
unpopped at t = 30 seconds, for eversmaller time steps, given an initial
1000 kernels and a probability of 1%
per second that any given unpopped
kernel will pop. Row 9 is the result if
one uses a 0.1-second time step; Row
10 is the result for a 0.01-second time
step. The cell equation for C8 is
=$C$3*(1-B8*$C$2)^($C$4/B8). The
equations in the rest of Column C (C7
through C13) are comparable.
It is worth analyzing the cell equation
for C8. The quantity in the first set of
parentheses is 0.9 in Row 7; 0.99 in
Row 8; 0.999 in Row 9, etc. The
quantity in second set of parentheses
is the number of time steps to reach t
= 30 seconds. (End Note 5)
B
C
D
0.01 per second
1000 kernels
30 sec
2 k
3 N0
4 t
5
time step unpopped
(sec)
kernels
6
7
10
729
8
1 739.700
9
0.1 740.707
10
0.01 740.807
11
0.001 740.817
12
0.0001 740.818
13
0.00001 740.818
In the limit:
N = N0e-kt
Find N for N0 = 1000, k = 0.01 per second,
and t = 30 seconds. From your calculus
show that
dN/dt = -kN.
In other words, the rate of popping is
directly proportional to the amount of
15
unpopped kernels. That’s the rate law.
Conclusions and Further Thoughts
1. Popping popcorn is analogous to radioactive decay in that both are
governed by the following rate law: the rate of change (decay) is directly
proportional to the amount of the changing (decaying) material present:
dN/dt = -kN.
2. The parameter, k, is the decay constant. It has dimensions of time-1 (End
Note 6). The decay constant is the probability that any given unpopped
kernel (radioactive atom) will pop (decay) in the unit of time specified by
the units of k. This probability does not change with t. (In the case of
radioactive decay, it does not change with temperature, pressure, or
chemical environment; hence, the decay constant is constant.)
3. The integrated form of the rate law is N = N0e-kt. From this equation one
can show that the half-life (t1/2) is given by t1/2 = ln(2)/k. Because k is
constant, t1/2 is also constant.
4. For modeling the decay phenomenon with a succession of equal, finite
time steps, Δt, the rate law is ΔN/Δt = -kN. Such modeling produces a list
of N vs. t. The same list can be generated by N = N0(1-kΔt)t/Δt. As Δt
approaches zero, this equation approaches N = N0e-kt.
16
References
There are many excellent references on radiometric dating and its context. We
particularly recommend G. Brent Dalrymple (2004), Ancient Earth, Ancient Skies: The
Age of the Earth and its Cosmic Surroundings, Stanford University Press, 247 pp.
See particularly, Chapter 4: “Clocks in Rocks: How Radiometric Dating Works.”
See Also –
http://serc.carleton.edu/quantskills/methods/quantlit/RadDecay.html
http://serc.carleton.edu/quantskills/activities/popcorn.html
17
End-of-module assignments
1.
Answer the question on Slide 8: determine the half-life and quarterlife of our popcorn example by interpolating the spreadsheet. Test
your answer by using the trendline equation.
2.
What is the rate of popping at t = 0, t = t1/2 and t = t1/4 for the example
in Slides 8 and 9?
3.
How would your answers to Questions 1 and 2 be different if you
started with a 37,420 kernels and a popping probability of 6% per 10
seconds? Modify the spreadsheets in Slides 8 and 9 for this new
example and hand them in.
4.
What is the third-life of the popcorn in Slide 8 and of the case in
Question 3? What is the ratio of half-life to third-life in each of the
two cases?
5.
Recreate the spreadsheet in Slide 15, modify it by adding one more
decimal place to the number of unpopped kernels (Col. D), and hand
in the new spreadsheet. What is the value of N from the equation for
the exponential function.
6.
Suppose you have a population of 2000 radon-222 (222Rn) atoms.
The probability that 222Rn will decay in a one-day period is .211 or
21.1%. How many atoms of 222Rn will remain after 30 days? What is
the half-life of 222Rn?
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End Notes
1.
2.
3.
4.
5.
6.
For an explanation of the Law of Large Numbers see:
http://en.wikipedia.org/wiki/Law_of_large_numbers. For the number of Carbon-14 atoms, consider a
gram of plant carbon. Using Avogadro’s number, there are 6.022×1023 carbon atoms in 12 grams (one
mole) of carbon. The abundance of Carbon-14 is 1 atom per 1.0×1010 atoms of carbon. Carrying out
the arithmetic: there are 5×1012 Carbon-14 atoms in a gram of plant carbon (and so the thousand
kernels is a ridiculously small number to compare to the number of parent atoms; we use it only to
simplify the appearance of the spreadsheets). (For Avogadro’s number, see:
http://en.wikipedia.org/wiki/Avogadro's_number; For more about Carbon-14 see:
http://www.c14dating.com.) (Return to Slide 5)
The value in each row is 0.9 times the value in the preceding row. Columns C and D are geometric
progressions (common factor = 0.9, in both cases). By contrast, Column B is an arithmetic progression
(common difference = 10 seconds). That means we are dealing with an exponential function. An
exponential function is produced when a geometric progression is paired against an arithmetic
progression. The succession of times in Column B is an arithmetic progression. (Return to Slide 7)
Notice we are going to great pains in the convoluted wording to avoid saying that a half-life is when half
of the parents have decayed. That is because half-life is defined to be when N/N0 = ½, where N and N0
are the remaining and original parents respectively. Similarly a third-life is when N/N0 = 1/3, or when the
decay has proceeded to where only 1/3 of the original parents remain. According to these definitions,
the time for 1/3 of the parents to decay would be called a two-thirds life. (Return to Slide 8)
When asked about the rate law that applies to radioactive decay, and hence the underlying reason that
radiometric dating works, novice geology students commonly say that the rate of radioactive decay is
constant. That assertion is clearly false as shown in this graph. If the rate or decay were constant, then
the graph of remaining parents vs. time would be a straight line. As shown in this graph, the rate of
decay diminishes with time. (Return to Slide 9)
This example is analogous (but with opposite sign) to the case of compound interest on a savings
account. The path of dollars as a function of time differs if the compounding is done annually, semiannually, monthly, daily or continuously. The continuously compounded case corresponds to the
analytical expression in the right-hand box of Slide 15. The periodically compounded cases correspond
to the spreadsheet examples with the finite time steps. (Return to Slide 15)
Dimensions refer to different kinds of quantities. Length (L), time (T), and mass (M) are the principal
dimensions of mechanics. Units refer to the size of the quantity. Seconds, hours, days, years, and
millions of years are different units of the dimension time. For more on dimensions and units, see
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http://www.engineeringtoolbox.com/terminology-units-d_963.html. (Return to Slide 16)
Pretest
1. Given that the half-life is constant in radioactive decay, is
the third-life constant too?
2. Which is longer, the half-life or the third life?
3. How does the rate of reaction (radioactivity) vary with
time in radioactive decay?
4. What does the equation dN/dt = -kN mean?
5. In the equation, N = N0e-kt, what are the dimensions of k?
6. What is the Law of Large Numbers?
20