1. Radioactive Decay and Popping Popcorn – Understanding the Rate Law
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

2. 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 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.

3. The helium atom is an a- particle. Thorium is the daughter nuclide.
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, β-particles, 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:
The helium atom is an a- particle.
Uranium is the parent nuclide. 92 is the atomic number and 238 is the mass number.
Thorium is the daughter nuclide.
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!

4. Radioactive Decay Example
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.
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.
From science/halflife1.php

5. 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.
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.
In the same way, we can’t say exactly when a given kernel of popping corn will pop into a piece of popcorn…
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.

6. 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.
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.
Column B lists the numbers of seconds that have passed. Remember we are thinking in 10-second intervals.
Cell C4 is the probability that a kernel will pop in a
10-second interval.
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.

7. What happens in the 10-second intervals after the first one?
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?
Do you notice a pattern in Columns C and D? See End Note 2.

8. Exponential-decay phenomena are characterized by a constant half-life.
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
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).
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.”
Exponential-decay phenomena are characterized by a constant half-life.
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?

9. 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.
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.
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)

10. 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.
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?
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.
So, the rate decreases exponentially. What besides the half-life is constant?

11. The Constant Ratio ΔN10-sec = k10-secN,
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.

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

13. 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.

14. Comparing the two time steps
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 t, 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 t, where t is in seconds. The predicted number of kernels at 30 seconds is
Is there a limit? What if Δt  0?

15. The Rate Law In the limit: N = N0e-kt
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; 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.

16. Conclusions and Further Thoughts
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.
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.)
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.
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.

17. 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 –

18. End-of-module assignments
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.
What is the rate of popping at t = 0, t = t1/2 and t = t1/4 for the example in Slides 8 and 9?
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.
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?
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.
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?

19. End Notes
For an explanation of the Law of Large Numbers see: 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: For more about Carbon-14 see: (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, 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)
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 (Return to Slide 16)

20. Which is longer, the half-life or the third life?
Pretest
Given that the half-life is constant in radioactive decay, is the third-life constant too?
Which is longer, the half-life or the third life?
How does the rate of reaction (radioactivity) vary with time in radioactive decay?
What does the equation dN/dt = -kN mean?
In the equation, N = N0e-kt, what are the dimensions of k?
What is the Law of Large Numbers?