NAT Sci 102: Radioactivity and Age Determinations. Due Date: April 27

Name: NAT Sci 102: Radioactivity and Age Determinations Due Date: April 27 During this lab session you are going to witness how radioactive decay works, and you will see firsthand what "half life" means. You will also explore the character of randomness in nature as radioactive decay is a random process. This means one that in any given one second time interval, you won't be able to predict whether an atom will decay or not, but you can predict the results of averaging what happens in many one second intervals. The first part of the assignment needs to be turned in on paper (including this handout with the requested information filled in). It is worth 40 points. The 20 questions at the end should be turned in on a Scantron and are worth 2 points each, or 40 points in total. Radioactivity Recall that atoms are comprised of electrons, neutrons, and protons with the neutrons and protons being concentrated in their centers, called "atomic nuclei". The number of protons equals the number of electrons, and the electrons orbit the nucleus in a fashion similar to planets orbiting the Sun. The protons and electrons are electrically charged with protons having a positive charge and electrons having a negative charge. The equal numbers of each ensures that the atom itself has no net electrical charge. The number of protons (or equivalently, the number of electrons) determines the element of the atom -- for example, carbon atoms have 6 protons while iron atoms have 26 protons. There are about 100 elements, that is, atoms can be reasonably stable with anywhere from 1 to 100 protons in their nuclei. The number of protons in an atomic nucleus is not the whole story. Atomic nuclei also contain neutrons which have nearly the same mass as protons but have no electrical charge. For light elements (those with relatively few protons), the number of protons and neutrons in the nucleus is roughly equal. For example, the most abundant form of carbon has 6 protons and 6 neutrons in its nucleus. As more and more protons are added to the nucleus, larger numbers of neutrons tend to be added also. The uranium nucleus has 92 protons and 146 neutrons. The sum of the number of protons and neutrons in a nucleus is equal to the atomic weight of the atom (and gives the actual mass of the atom if the atomic weight is multiplied by 1.6x10-24 grams = mass of a proton or neutron). The atomic weight is usually written as a superscript to the letter symbol denoting the element so you will see C 12 and U 238 for the common types of carbon and uranium. The number of neutrons in a nucleus is not fixed so other variants of an element can exist, such as C 13 indicating a carbon atom with 6 protons and 7 neutrons. Most elements have two or three variants called isotopes, with differing numbers of neutrons. Some isotopes are not stable and can spontaneously eject particles with the isotope being transformed into a different type of atom (either different element, or different isotope of the same element). This process is called radioactive decay, and the process of ejecting a particle is radioactivity. Isotopes with larger than average numbers of neutrons or very large atoms like uranium are the most likely to be radioactive. Radioactive decay can involve ejection of alpha ( ) particles made of 2 protons and 2 neutrons or beta ( ) particles which are just electrons. Radioactive decays may also involve the emission of very energetic photons called gamma ( ) rays (and other particles also). Look at the following two reactions, which are examples of radioactive decays: C 14 N 14 + e- U 238 Th 234 + He 4 (e- = electron = particle) (He 4 = particle) In the first case, a neutron in the C 14 nucleus changes into a proton so the atom becomes nitrogen (which

is the element with 7 protons in its nucleus); the atomic weight doesn't change because the numbers of protons and neutrons sum to the same total. In the second case, the number of protons changes by 2 so the element changes from uranium to thorium and the atomic weight also changes by a total of four. These decays can be measured by counting the and particles using a Geiger counter. Half Lives and Measuring Ages Radioactive isotopes decay in a random fashion. The likelihood that isotopes will decay in any particular time interval is expressed by the half life which is equal to the time for 50% of a sample of isotopes to decay. The half life of the C 14 decay is 5,730 years while that of U 238 to Th 234 is 4.45x10 9 years. By measuring the amount of an isotope that has decayed, the age of a sample can be determined. For example, a sample of wood found in an ancient Indian camp site is found to have 12.5% of its original C 14. The age of this site is equal to three half-lives of the C 14 because 12.5% = 1/8 = ½ x ½ x ½. Three half lives equals 17,190 years so this site is an exciting find. Note that in this process it is essential to know what the starting amount of the radioactive substance is. This can be a tricky problem in using the radioactive dating technique. In the case of C 14 this problem is solved by knowing that a living organism incorporates C into its substance by breathing in CO 2 from the atmosphere. As long as the organism is alive, the amount of C 14 relative to C 12 is kept replenished at the average value for the earth s atmosphere. When the organism dies, the replenishment stops. The amount of C 12 stays fixed because C 12 is stable while the C 14 decays away thus making it possible measure the age by comparing the ratio of C 12 to C 14 in the dead organism to that in the atmosphere. However, C 14 is not useful for measuring astronomical time scales because its decay is too rapid. The decay U 238 to Pb 206 with a half life of 4.5x10 9 years and the decay of K 40 to Ar 40 with a half life of 1.3x10 9 years are good choices. The starting values for uranium and potassium measurements get addressed in different manners. The K 40 to Ar 40 measurement can take advantage of argon being a gas that can be driven out of a sample that is heated. The Ar 40 present in a sample is then indicative of the number of K 40 decays occurring since the sample was last heated. Since uranium decays only to Pb 206 and not to the more commonly occurring Pb 204, uranium dating works by measuring the relative amounts of the two lead isotopes, Pb 204 and Pb 206. Random Processes and Counting Statistics Atomic decays occur randomly, so we need to understand a bit about random processes to understand measuring these decays. Randomness plays a large role in what can observed and learned about nature. This stems from several causes including the fact that some things come in indivisible units -- photons and electrons are examples where you cannot have a third of a photon or 80% of an electron. Either you have one or you don't. Therefore, an experiment that expects 3.4 protons to be detected will have to contend with never getting this exact number! You are going to perform an experiment where you count items and then you will look at the distribution of values that you have counted. Some forms of randomness can be quantified because although we can only get an integer number of electrons, for example, we may have an experiment where on average we would get 3.4 electrons. This means that some of the time we might get 0, sometimes we would get 1, and most of the time we get 3 or 4. If we repeat the experiment many times, we can prepare a plot showing how many repetitions of the experiment saw 3 electrons, how many saw 5 and so on. Figure 1 shows such a plot where the horizontal axis shows the number of radioactive decays and the vertical axis shows the number of repetitions of a radioactive experiment which saw a given number of decays. For example, 100 decays were observed in 50 repetitions of the experiment. In an experiment of this type, the spread of values that you will observe depends on the average where average is the average number of counts you get in Figure 1 the spread (sometimes called the error or uncertainty) = counting experiment to see randomness in counting for yourself. 100 =10. You are going to repeat a You need to perform 20 counting experiments. Each experiment consists of tossing a coin 10 times. For each experiment, record the number of times out of 10 that you got "Heads" in the table. You can speed this up, of course, buy tossing ten coins all together and counting the heads. Put the results in the table below.

Figure 1: How to measure the range of values. The range in this figure is 90 to 110. After you have filled in the table, prepare a bar graph where the horizontal axis is labeled from 0 to 10 indicating the possible values for the number of heads in an experiment and with the vertical axis labeled from 0 to 20 indicating the number of experiments that had any of the values for the number of heads. Count how many experiments had 0 heads, 1 head, 2 heads and so on and plot the values on the graph (a spreadsheet like Excel makes this easier). Your graph should look a bit like Figure 1. Question 1: Given that any one coin toss has equal probability of being heads or tails, how many heads do you think will most likely occur in an experiment consisting of 10 tosses? Question 2: On your graph, which bar is highest? Is this the number of heads you expected Question 3: Not all of your experiments gave the same number of heads. If you repeat a counting experiment many times, you will learn what is the most likely outcome or average result for the experiment.. The uncertainty in any one experiment is related to the spread of values observed in repeats of the same experiment. Look at your graph and determine the range of values for the number of heads in an experiment. Figure 1 shows how you should determine this range. Look for the number of heads that occurred most often the most frequently occurring number of heads is the "maximum". Compute the value of the maximum divided by 2. The range of most likely values begins at the lowest number of heads that occurred at least half of the maximum number of times up to the highest number of heads that occurred at least half of the maximum number of times. What is your range? Compare your range to : The uncertainty in an experiment is approximately half the range computed in this manner.

Experiment No. of Heads Experiment No. of Heads 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20 What you have just illustrated is the principle that in an experiment where you count items, there is an uncertainty proportional to the square root of the counts. These means that if you expect to count 100 - particles, your counts will most probably lie in the range from 100 100 to 100 + 100, 90 to 110 (since 100 = 10). It is still possible for an experiment to result in counts outside this range but with lower probability. Measuring the Half Life of Pa 234 We operated a Geiger counter that was projected in the lecture hall so you could see the rate of particles emitted by decaying atoms in a sample of material. The decaying atom, Pa 234 (Pa = protactinium), has a half life measured in minutes. Your job is to determine the value of the half life. To be able to measure radioactivity from a rapidly decaying atom like protactinium requires a fresh source of Pa 234 for each experiment. We have such a renewable source. It works by virtue of Pa 234 dissolving in a different liquid than the Th 234 which decays into Pa 234. When the bottle of liquids is shaken up and then allowed to settle, the liquid on the top has the Pa 234 in it while the bottom has the remaining Th 234. The Geiger counter is arranged to detect the particles emitted from the top, eg., by the Pa 234. Two sets of readouts for the Geiger counter are listed below. Fill out the rest of following tables. The background counts are from the Geiger counter reading long after the radioactivity had decayed. Average them to get a more accurate value. Fill in the average for each background entry. Complete the table by computing the net counts by subtracting the average background counts from each of the counts that was observed during the decay of the protactinium. Then calculate the uncertainties as the square roots of the raw counts for each entry.

Below, we give the results of two independent counting experiments like the one that was run for you in class. Use these numbers to carry out the items below. Determine the background counts from the Geiger counter reading long after the radioactivity has decayed. There is a separate table with these numbers. Average the five entries to get a best value. Although of the values has an error of the square root of the number, because we have averaged five of them we can ignore this type of error in the best value. Enter this value in the background columns in the tables for runs 1 and 2. Compute the net counts by subtracting the background counts from each of the counts that was observed during the decay of the protactinium. Then put in the uncertainties for each reading. Prepare a graph where the horizontal axis represents elapsed time and the vertical axis represents net counts. A very convenient way to make a graph is to put all the numbers into a spreadsheet such as Excel provides. The number of decays will change with time as the quantity of protactinium changes. Use your graph to estimate the half life of protactinium and write it here: 70 seconds is the official number, although our data give a slightly longer value, about 83 seconds Do you see any deviations from a smooth decay with time? yes Give an explanation for these deviations (how do they compare with your estimates of the uncertainties in the counts): The deviations arise because of the statistics of the decays they are random and independent, so the uncertainties are equal to the square root of the number of counts. Run 1 Elapsed Time In sec Raw Counts Background Net Counts (Raw minus Background) Uncertainty (Square Root of Raw) 0 610 30 580 24.1 30 510 30 480 21.9 60 438 30 408 20.2 90 416 30 386 19.6 120 266 30 236 15.4 150 194 30 164 12.8 180 158 30 128 11.3 210 108 30 78 8.8 240 122 30 92 9.6

Background counts (use for both runs) Elapsed Time Background Counts In sec 0 28 30 42 60 18 90 34 120 28 Run 2 Elapsed Time In sec Raw Counts Background Net Counts (Raw minus Background) Uncertainty (Square Root of Raw) 0 712 30 682 26.1 30 608 30 578 24.0 60 522 30 492 22.2 90 476 30 446 21.1 120 366 30 336 18.3 150 256 30 226 15.0 180 210 30 180 13.4 210 166 30 136 11.7 240 140 30 110 10.5

number of trials 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 number of heads The distribution of heads in my coin toss experiment. I have marked my estimate of the range.

counts 800 700 600 500 400 300 run 1 run 2 average 200 100 0 0 50 100 150 200 250 300 time (seconds) A graph of the counts, from which you can estimate the half life. I did it from the average of the two runs, but you might have used either one or the average.

From here, put the answers on a Scantron and turn it in separately (with your name on it!!) 1. Estimate the half life of protactinium: a. about 30 seconds b. about 70 seconds c. about 150 seconds d. very long 2. Do you see any deviations from a smooth decay with time? a. yes b. no 3. Do the two runs agree perfectly, or are there differences? a. agree perfectly b. there are differences 4. Give an explanation for these deviations and differences, if there are any: a. there aren t any b. the must have been outside disturbances that affected the decay rate c. maybe the Geiger counter wasn t working right d. the decay is a random process so no two runs will be identical 5. Making allowance for the expected differences in the runs, is the half life the same for both runs? a. yes b. no More on Random Processes Although complex mathematics can modify the rules a little, the square root law describes normal statistics pretty well in nearly all situations. It states that if I have a likely number of successes of N, then the uncertainty in N is the square root of N. For example, if I toss a coin 100 times, then the likely number of times it will come up heads is 50 times and the uncertainty is 7 (the square root of 50 is very close to 7). What does uncertainty mean? It means that the number of heads will be between 50-7 = 43 and 50+7 = 57 two thirds of the time, so if I were to toss the coin 100 times for sixty experiments (6000 tosses total), then I expect the number of heads to be inside the 43 57 range for 40 times (2/3 X 60) and OUTSIDE that range for 20 (what s left, 1/3 X 60). Suppose I want to be more sure. The result is expected to be within two times the square root of N 19 times out of 20, 95% of the time. So I would expect to get either less than 50 (2 X 7) = 36 heads OR more than 50 + (2 X 7) = 64 heads in three of my 60 experiments. Now let s try some examples. The margin of error in a political poll is based on the square root logic, just a slightly more complex form mathematically. Suppose you took a poll of 1000 people (who did not know each other) and 490 answered the way you expected. 6. You could report that 49% were thinking correctly, but what is the margin of error? a. 3.2% b. 2.2%

c. 4.5% d. Not enough information to tell 7. Your friend, who disagrees with you all the time, took a poll on the same question of 1000 different people and 530 answered the way she wanted. a. Can she claim to have more people on her side, for sure? b. Or would you say she can t make any claims at all the difference is still totally undecided c. Or can she claim that it is pretty likely that more people agree with her, but still there is a small chance it isn t true? 8. Suppose that your polls were taken of groups still a total of 1000 people, but they were organized into groups of five and the view of each group was recorded. Would this change your answers to questions 1 and 2? a. Yes b. No 9. Defend your answer to question 3: a. There would be no change because the poll is still taken with 1000 people b. But they agreed based on groups of 5, so there were only 200 independent views! c. There isn t any way you could learn anything useful from groups 10. To have the same margin or error as for question 1, how many people in groups of five would need to be polled? a. 1000 b. 200 c. 5000 d. You could never get the same margin of error 11. Brandon Lavender took 77 three-point shots in the basketball season, and made 29. MoMo Jones took 79 and made only 25. Who is the better three-point shooter? a. Lavender, his percentage is 37.7% vs. only 31.6% b. It is too close to tell 12. Derrick Williams took 74 three-point shots and made 42. Compared with MoMo Jones, Williams is a. It is too close to tell b. almost for sure better 13. The Gallup Poll has evaluated the number of people approving President Obama s job performance every week. Here are the results in terms of number of people approving for five weeks from late February to early March: 1620 1655 1690 1584 1647 (these numbers average to about 47% of the total) Compute the average over the five weeks, compare the weekly results, and then choose from among the following: a. Obama s approval rating never changed at all b. there were fluctuations in his approval at a barely significant level c. he had a big upswing in the last week 14. The number of homicides in Tucson for each year from 2004 to 2008 was: 54 56 52 51 68. Can we conclude that 2008 was a much worse year for homicides than the previous ones?

a. Not at all b. Almost, but not for sure c. Definitely 15. There have been a lot of newspaper stories in the past weeks about the reduction in the number of auto accidents in recent years. In 2007, 2008, and 2009 the number of auto accidents in the US was 37,435 34,172 30,797. a. was there a significant reduction in the number of accidents b. or is it too close to tell? 16. If an election ends up with 733,459 votes for Brown and 732,854 votes for Jones, and there are no problems with the count, then a. it is too close to know who won, so there should be a runoff b. Brown definitely won c. it should be declared a tie and Brown and Jones should share the office The following table shows the number of home runs Babe Ruth hit in each year (excluding two when he missed a lot of games) 1920 54 1921 59 1922 reduced time 1923 41 1924 46 1925 reduced time 1926 47 1927 60 1928 54 1929 46 1930 49 1931 46 1932 41 17. How many home runs did he hit on average? a. 45.52 b. 47.91 c. 49.36 d. 51.75 e. 53.02 18. According to the rules about uncertainties, how many years would Ruth have hit a number of homers MORE THAN ONE TIMES the uncertainty off the average? a. 2/3 x 11 ~ 7 b. 1/3 x 11 ~ 4 c. Not at all d. All the time

Fill in the following table. The homer difference column is the number of homers he hit minus the average. The homer uncertainty is the square root of the number of homers he hit. The relative difference is the homer difference divided by the homer uncertainty, so it lets you judge quickly how large the difference is compared with the uncertainty. 1920 1921 homer difference homer uncertainty relative difference 1922 reduced time 1923 1924 1925 reduced time 1926 1927 1928 1929 1930 1931 1932 19. How many times did Babe Ruth actually come up more than one times the uncertainty off the average? a. 7 b. 4 c. Not at all d. All the time 20. Was hitting 60 homers a. Just normal luck, given how well he was hitting for the whole period 1920 1932? b. Something that took an exceptional boost in his hitting ability?