1

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?

Prepared for SSAC byC E Stringer, University of South Florida - Tampa

© The Washington Center for Improving the Quality of Undergraduate Education. All rights reserved. 2005

Radioactive Decay and Popping Popcorn – Understanding the Rate Law

SSAC2005.QE514.CES1.1

Supporting Quantitative ConceptsNumber sense: Geometric progressionNumber sense: Dimensions vs. unitsCalculus: rate of changeDifferential equation for the exponential functionGraph, logarithmic scaleGraph, trend lineProbability: Law of Large Numbers

Core Quantitative IssueExponential function

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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.

Preview

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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 nuclide decomposes (or decays) to form a different nuclide, it is called a radioisotope. The

phenomenon is called radioactivity.

The radioisotope can also be thought of as the “parent” and the nuclide it decays to can be termed the “daughter.”

Remember that the atomic number is the number of protons in an atom’s nucleus and the mass number is the number of protons plus neutrons!

HeThU 42

23490

23892 +→

Here is an example of a decay equation:

Uranium is the parent nuclide.

92 is the atomic number and 238 is the mass number.

Thorium is the daughter nuclide.

The helium atom is an particle.

When a radioisotope decays to form a different nuclide, it emits a particle. The three initial types of particles recognized were α-particles, β-particles, and γ-radiation.

Radioactive Decay

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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 U-235 chains.

From www.compumike.com/ science/halflife1.php

ExampleThe 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.

Radioactive Decay

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In the same way, we can’t say exactly when a given kernel of popping corn will pop into a piece of popcorn…

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.

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.

Problem

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.

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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.

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!

Cell C3 is the number of kernels you start with in the popper.

Cell C4 is the probability that a kernel will pop in a 10-second interval.

Column B lists the numbers of seconds that have passed. Remember we are

thinking in 10-second intervals.

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.

Restating the problem; Setting up the spreadsheet

B C D23 Nstart = 10004 Ppop 0.1 per 10 sec5

6Time (t )

(sec)Nunpopped

(N )7 0 10008 10 900

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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?

What happens in the 10-second intervals after the first one?

Do you notice a pattern in Columns C and D? See End Note 2.

B C D23 Nstart = 10004 Ppop 0.1 per 10 sec5

6 sec

Nremain (N ) N /N 0

7 0 1000 1.008 10 900 0.909 20 810 0.81

10 30 729 0.7311 40 656 0.6612 50 590 0.5913 60 531 0.5314 70 478 0.4815 80 430 0.4316 90 387 0.3917 100 349 0.3518 110 314 0.3119 120 282 0.2820 130 254 0.2521 140 229 0.2322 150 206 0.21

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Create a graph by plotting the seconds on the x-axis and number of remaining kernels (parents) on the y-axis

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).

Looking at Popcorn Popping Graphically

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?

Exponential-decay phenomena are characterized by a

constant half-life.

B C D E F G H I J23 Nstart = 10004 Ppop 0.1 per 10 sec5

6

Time ( t ) (sec)

Nunpopped

(N ) N /N 0

7 0 1000 1.008 10 900 0.909 20 810 0.81

10 30 729 0.7311 40 656 0.6612 50 590 0.5913 60 531 0.5314 70 478 0.4815 80 430 0.4316 90 387 0.3917 100 349 0.3518 110 314 0.3119 120 282 0.2820 130 254 0.2521 140 229 0.2322 150 206 0.21

0

100

200

300

400

500

600

700

800

900

1000

0 50 100 150

Time (seconds)

Kernels remaining

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.”

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Looking at Popcorn Popping Graphically, 2

Here y represents N, the number of

unpopped kernels, and x

represents time in seconds.

R2 = 1 means that the fit is perfect.

The equation describes the

listed values of N vs. time with no

scatter.

Insert an exponential trend line and use the option tab to display the equation for the line and correlation coefficient. Record the equation.

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)

B C D E F G H I J23 Nstart = 10004 Ppop 0.1 per 10 sec5

6 S

Nunpopped

(N ) N /N 0

7 0 1000 1.008 10 900 0.909 20 810 0.8110 30 729 0.7311 40 656 0.6612 50 590 0.5913 60 531 0.5314 70 478 0.4815 80 430 0.4316 90 387 0.3917 100 349 0.3518 110 314 0.3119 120 282 0.2820 130 254 0.2521 140 229 0.2322 150 206 0.21

y = 1000e-0.0105 x

R 2 = 1

0

100

200

300

400

500

600

700

800

900

1000

0 50 100 150

Time (seconds)

Kernels remaining

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Looking at Popcorn Popping Graphically, 3

Here y represents ΔN10-sec, the number of

popping kernels per 10 seconds, and x represents time in seconds.

R2 = 1 means that the fit is perfect.

The equation describes the

listed values of ΔN vs. time with

no scatter.

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.

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?

So, the rate decreases exponentially. What besides the half-life is constant?

B C D E F G H I J K23 Nstart = 10004 Ppop 0.1 per 10 sec5

6

Time ( t ) (sec)

Nunpopped

(N ) N /N 0

Number popped (_ N 10 )

7 0 1000 1.00 1008 10 900 0.90 909 20 810 0.81 81

10 30 729 0.73 7311 40 656 0.66 6612 50 590 0.59 5913 60 531 0.53 5314 70 478 0.48 4815 80 430 0.43 4316 90 387 0.39 3917 100 349 0.35 3518 110 314 0.31 3119 120 282 0.28 2820 130 254 0.25 2521 140 229 0.23 2322 150 206 0.21

y = 100e-0.0105 x

R2 = 1

0

10

20

30

40

50

60

70

80

90

100

0 50 100 150

Time (seconds)

Kernels popped per 10 secs

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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.

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.

B C D E F23 Nstart = 10004 Ppop 0.1 per 10 sec5

6

Time ( t ) (sec)

Nunpopped

(N ) N /N 0

Number popped (_ N 10)

Relative rate

(_ N 10 /N )

7 0 1000 1.00 100 0.18 10 900 0.90 90 0.19 20 810 0.81 81 0.1

10 30 729 0.73 73 0.111 40 656 0.66 66 0.112 50 590 0.59 59 0.113 60 531 0.53 53 0.114 70 478 0.48 48 0.115 80 430 0.43 43 0.116 90 387 0.39 39 0.117 100 349 0.35 35 0.118 110 314 0.31 31 0.119 120 282 0.28 28 0.120 130 254 0.25 25 0.121 140 229 0.23 23 0.122 150 206 0.21

The Constant Ratio

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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.

A Shorter Time Increment

We will come back to that question. First, what about ΔN/N for this shorter increment?

B C D E F G H I J23 Nstart = 10004 Ppop 0.01 per sec5

6

Time ( t ) (sec)

Nunpopped

(N ) N /N 0

7 0 1000 1.008 1 990 0.999 2 980 0.9810 3 970 0.9711 4 961 0.9612 5 951 0.9513 6 941 0.9414 7 932 0.9315 8 923 0.9216 9 914 0.9117 10 904 0.9018 11 895 0.9019 12 886 0.8920 13 878 0.8821 14 869 0.8722 15 860 0.8623 16 851 0.8524 17 843 0.8425 18 835 0.8326 19 826 0.8327 20 818 0.8228 21 810 0.8129 22 802 0.8030 23 794 0.7931 24 786 0.7932 25 778 0.7833 26 770 0.7734 27 762 0.7635 28 755 0.7536 29 747 0.7537 30 740 0.74

y = 1000e-0.0101 x

R2 = 1

0

100

200

300

400

500

600

700

800

900

1000

0 5 10 15 20 25 30

Time (seconds)

Kernels remaining

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?

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A Shorter Time Increment, 2

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.

B C D E F23 Nstart = 10004 Ppop 0.01 per sec5

6

Time ( t ) (sec)

Nunpopped

(N ) N /N 0

Number popped (_ N 1sec )

Relative rate

(_ N 1sec /N )

7 0 1000.0 1.000 10.0 0.0108 1 990.0 0.990 9.9 0.0109 2 980.1 0.980 9.8 0.01010 3 970.3 0.970 9.7 0.01011 4 960.6 0.961 9.6 0.01012 5 951.0 0.951 9.5 0.01013 6 941.5 0.941 9.4 0.01014 7 932.1 0.932 9.3 0.01015 8 922.7 0.923 9.2 0.01016 9 913.5 0.914 9.1 0.01017 10 904.4 0.904 9.0 0.01018 11 895.3 0.895 9.0 0.01019 12 886.4 0.886 8.9 0.01020 13 877.5 0.878 8.8 0.01021 14 868.7 0.869 8.7 0.01022 15 860.1 0.860 8.6 0.01023 16 851.5 0.851 8.5 0.01024 17 842.9 0.843 8.4 0.01025 18 834.5 0.835 8.3 0.01026 19 826.2 0.826 8.3 0.01027 20 817.9 0.818 8.2 0.01028 21 809.7 0.810 8.1 0.01029 22 801.6 0.802 8.0 0.01030 23 793.6 0.794 7.9 0.01031 24 785.7 0.786 7.9 0.01032 25 777.8 0.778 7.8 0.01033 26 770.0 0.770 7.7 0.01034 27 762.3 0.762 7.6 0.01035 28 754.7 0.755 7.5 0.01036 29 747.2 0.747 7.5 0.01037 30 739.7 0.740

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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.

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.

Is there a limit? What if Δt 0?

Comparing the two time steps

y = 1000e -0.0105x

R 2 = 1

0

100

200

300

400

500

600

700

800

900

1000

0 50 100 150

Time (seconds)

Kernels remaining

y = 1000 e -0.0101 x

R2 = 1

0

100

200

300

400

500

600

700

800

900

1000

0 5 10 15 20 25 30

Time (seconds)

Kernels remaining

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Approaching a limit….This spreadsheet calculates the number of kernels that remain unpopped at t = 30 seconds, for ever-smaller 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)

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 unpopped kernels. That’s the rate law.

The Rate Law

B C D2 k 0.01 per second

3 N 0 1000 kernels

4 t 30 sec5

6time step

(sec)unpopped

kernels7 10 7298 1 739.7009 0.1 740.707

10 0.01 740.80711 0.001 740.81712 0.0001 740.81813 0.00001 740.818

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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.

Conclusions and Further Thoughts

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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.”

http://serc.carleton.edu/quantskills/methods/quantlit/RadDecay.html

See Also –

http://serc.carleton.edu/quantskills/activities/popcorn.html

References

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1. Answer the question on Slide 8: determine the half-life and quarter-life 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?

End-of-module assignments

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1. 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)

2. 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)

3. 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)

4. 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)

5. 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, semi-annually, 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)

6. 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 http://www.engineeringtoolbox.com/terminology-units-d_963.html. (Return to Slide 16)

End Notes

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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?

Pretest