Lab12 Geologic Time Doezie.pdf (1)

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Geologic Time Lab Jacob Doezie 11/27/2023 In geology we use different methods to determine the ages of rocks and more importantly the geologic events which alter the Earth’s lithosphere. Absolute time utilizes the decay of radioactive elements to determine ages which are an actual number. Elements occur in differing in differing isotopes. Atomic number : The number of protons in the nucleus of the atom. Atomic weight: The sum of both the protons and neutrons in the nucleus of the Atom Isotope: An element having the same atomic number but a differing atomic weight. All uranium will have the same atomic number 92, but may vary in the atomic weight (uranium 234 ( 234 U), uranium 235 ( 235 U) and uranium 238 ( 238 U)). The original unstable radioactive element which will decay is referred to as the parent isotope. The product of the decay is referred to as the daughter isotope (which may or may not be stable). The time required for one-half of the parent isotope to decay to daughter isotope is the half-life (T 1/2 ) . The half-life of elements is one way we can utilize the radioactive decay to determine the age of the rock. For example at the time of formation the rock will start with 1000 atoms of the parent isotope and 0 atoms of the daughter isotope. After one half-life we will now have 500 atoms of parent and 500 atoms of daughter isotopes. If one more half-life elapses one-half of the remaining parent isotopes will decay to daughter isotopes, leaving 250 parent isotopes atoms and 750 daughter isotopes atoms. After a third half- life the remaining parent will be 125 and daughter will be 875. Each parent-daughter isotope combination has an identified predictable number of years for the half-life. A table of commonly utilized parent daughter pairs with their half-life is Parent-Daughter Half-life (T 1/2 ) 238 U – 206 Pb 4.5 Billion years 235 U – 207 Pb 716 Million years 40 K – 40 Ar 1.3 Billion years 14 C – 14 N 5730 years Percent Parent Percent Daughter Halflives Elapsed Age 100 0 0 0.000(T 1/2 ) 84.1 15.9 0.25 0.25(T 1/2 ) 70.7 29.3 0.50 0.5(T 1/2 ) 50 50 1 1.0(T 1/2 ) 25 75 2 2.0(T 1/2 ) 12.5 87.5 3 3.0(T 1/2 ) 6.25 93.75 4 4.0(T 1/2 ) 3.125 96.875 5 5.0(T 1/2 )
Relative age Principles of stratigraphy Principle of original horizontality : as sediments are deposited, they will be deposited horizontally (or nearly horizontally). Principle of superposition : In a sequence of strata (layers of sedimentary rocks) the oldest rocks will be located at the bottom and the youngest will be at the top. Principle of inclusions : any rock, rock fragment or fossil included within another rock must be older than the rock it is included within. Principle of cross-cutting relations : Any feature (e.g. fault or dike), will be younger than the rock it cuts across. Principle of lateral continuity : sedimentary layers will spread out continuously in all directions until they abut against a boundary, taper off due to lack of material, or grade into another rock. Principle of faunal succession : Utilizes Index fossils to correlate the ages of distant rock units. Index fossils : Fossils of organisms which are easily identifiable, were widespread, and short Unconformities represent missing geologic time whether due to the representative rock being eroded or due to non-deposition. These occur in three types: Disconformity : missing geologic time between parallel layers. Angular unconformity : missing geologic time between non-parallel layers. Nonconformity : Missing time at the contact between non-layered igneous or metamorphic rock complexes. The Geologic Time Scale – originally developed through relative dating techniques and fossils. Age dates on boundaries are the result of numerical dating techniques. Eon Era Period Epoch Years 11 ka 2.5 ma Quaternary Holocene Pleistocene Neogene Pliocene Phanerozoic Cenozoic
5.3 ma 23.0 ma 33.9 ma 55.8 ma 66.5 ma 145.5 ma 201.6 ma 251.0 ma 299.0 ma 318.0 ma 359.0 ma 416.0 ma 444.0 ma 488.0 ma 542.0 ma 2.5 ga 3.85 ga 4.6 ga Miocene Paleogene Oligocene Eocene Paleocene Cretaceous Jurassic Triassic Permian Carboniferous Pennsylvanian Mississippian Devonian Silurian Ordovician Cambrian Proterozoic Archean Hadean Part I: Numerical Dating Procedure: a. Use a website for rolling “virtual dice” (for example: https://www.random.org/dice/ ); note that you may need to take multiple rolls to reach a total of 80 dice rolled. b. Roll 80 dice to begin the experiment c. Remove the indicated dice for your experiment after each roll (these represent the daughter product – for Experiment #1, this would be the even numbered dice ) d. Count the remaining dice in the experiment and record this in your data table in the “Number of parent” column. Add the number of daughter product you created in each roll to the total from the Geologic Time Tertiary Mesozoic Paleozoic Precambrian
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previous roll. An example has been done for you in the table below. NOTE: The sum of parent and daughter in each row should be 80. If not, you’ve made a mistake. Also note that the number of parent should never increase at any point in the experiment. e. Repeat the steps until all of your dice have “decayed” into daughter. You may add to your data table if necessary. The first trial of experiment #1 is done for you as an example. f. Repeat steps b through e two more times to complete three trials. g. Plot all three parent decay curves for each experiment on the blank charts. You may also make your plot in Excel if you prefer. Each point representing each roll must be visible on the chart, including the initial state. Experiment #1: Odd (Parent) vs. Even (Daughter) Trial 1 Trial 2 Trial 3 Initial Number parent Number daughter % parent remaining Number parent Number daughter % parent remaining Number parent Number daughter % parent remaining Roll # 80 0 100 80 0 100 80 0 100 1 45 35 56 35 45 44 43 37 54 2 21 59 26 14 66 18 24 56 30 3 9 71 11 8 72 10 13 67 16 4 5 75 6 5 75 6 9 71 11 5 3 77 4 1 79 1 5 75 6 6 1 79 1 0 80 0 3 77 4 7 0 80 0 1 79 1 8 0 80 0
Experiment #2: 1-5 Parent vs 6 Daughter Follow the same procedure as before, only in this experiment ONLY dice that come up as 6’s will be considered “decayed” to daughter product. Complete the table and plot on the following page. Trial 1 Trial 2 Trial 3 Initial Number parent Number daughter % parent remaining Number parent Number daughter % parent remaining Number parent Number daughter % parent remaining Roll # 80 0 100 80 0 100 80 0 100 1 67 13 84 66 14 83 66 14 83 2 51 29 64 56 24 70 54 26 68 3 42 38 53 49 31 61 44 36 55 4 38 42 48 40 40 50 36 44 45 5 33 47 41 35 45 44 27 53 34 6 27 53 34 30 50 38 23 57 29 7 22 58 28 23 57 29 20 60 25 Trial 1 - Trial 2 - Trial 3 - 0 10 20 30 40 50 60 70 80 90 100 0 1 2 3 4 5 6 7 8 Percent Parent Number of rolls Odd vs even
8 18 62 23 22 58 28 16 64 20 9 16 64 20 19 61 24 12 68 15 10 13 67 16 14 66 18 9 71 11 11 9 71 11 13 67 16 8 72 10 12 8 72 10 7 73 9 6 74 7 13 6 74 7 6 74 7 5 75 6 14 5 75 6 4 76 5 3 77 4 15 5 75 6 4 76 5 2 78 3 16 3 77 3 3 77 4 2 78 3 17 2 78 2 3 77 4 2 78 3 18 2 78 2 3 77 4 0 80 0 19 2 78 2 3 77 4 20 1 79 1 0 80 0 21 1 79 1 22 0 80 0 23 24
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A half life is the time required for one-half of the parent isotope to decay to daughter isotope 2. Why were dice used for these experiments? Dice were used to simulate radioactive decay because the roll is a random event just like the decay of an atom 3. In the odd:even experiment (Experiment #1) about how many dice rolls was one half-life? Examine the plot you created. 1 roll 4. In the 1-5:6 experiment (Experiment #2) about how many dice rolls was one half life? Examine the plot you created. 3-4 rolls 5. Imagine a third experiment: if the daughter isotopes were represented by 5’s and 6’s would a half-life a. Take more rolls than the two you just did? b. Take fewer rolls then the two you just did? c. Be between the two you just did?
6. If you used 10-sided dice and removed the evens. Would it a. Take more rolls than the odd:even trial? b. Take fewer rolls than the odd:even trial? c. Be about the same as the odd:even trial? 7. If you used 10-sided dice and only removed the 6’s. Would it a. Take more rolls than the 1-5:6 trial? b. Take fewer rolls than the 1-5:6 trial? c. Be about the same as the 1-5:6 trial? Parent-Daughter Half-life (T 238 U – 206 Pb 4.5 Billion years 235 U – 207 Pb 716 Million years 40 K – 40 Ar 1.3 Billion years 14 C – 14 N Percent Parent Percent Daughter Half-lives Elapsed Age 100 0 0 0.000(T 1/2 ) 84.1 15.9 0.25 0.25(T 1/2 ) 70.7 29.3 0.50 0.5(T 1/2 ) 50 50 1 1.0(T 1/2 ) 25 75 2 2.0(T 1/2 ) 12.5 87.5 3 3.0(T 1/2 ) 6.25 93.75 4 4.0(T 1/2 ) 3.125 96.875 5 5.0(T 1/2 ) Applying Numerical Dating A basaltic lava flow exposed in a road cut is found to contain crystals of the mineral zircon which can be analyzed for uranium-235 and lead-207. The results show there is 84% 235 U and 16% 207 Pb in the zircon. Refer to the charts above to help answer the questions. 8. How many half-lives have passed? .25 half lives 9. What is the age of the basalt? 4.5 billion years 10. During which geologic period did the basalt form? proterozoic 11. What is the relative age of the rock above the lava? 12. What is the relative age below the lava?
A piece of charcoal from a burnt tree was retrieved from a lava flow in the Mojave Desert. Trees exchange carbon with the atmosphere while they are alive, including the unstable Carbon-14 isotope. Upon analysis at the lab it was found that only 25% of the original Carbon-14 parent isotope is present in this sample. Use the charts on the previous page to help you answer the following questions. 13. How many half-lives have passed? 2 half lives 14. How long ago was the tree burnt by the lava flow? 5730 years 15. During which geologic period did the lava flow erupt? Epoch ? 16. Would it make sense to apply Carbon-14 dating techniques to the lava flow in the example on the previous page? Why or why not? Relative Dating Problems: Examine the cross-section below and use the principles of relative dating to place the rock units below in their order of occurrence from earliest (#1) to latest (#9). Note which units are sedimentary, igneous, or metamorphic.
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1. _________breca________________ (happened first) 2. ________ sandstone______________ 3. _________conglomerate________________ 4. _________rippled sandstone_______________ 5. _________Slitstone________________ 6. _________shale________________ 7. _________andesite dike________________ 8. _________limestone________________ 9. _________basalt dike ______________ (happened last) Use the drawing below to answer the following questions. The key to the lithologic patterns for the diagram is at the end of this section.
Principle of cross cutting relationships 6. Which principle indicates that object A is older than intrusion B ? Principle of inclusions 1. Which stratigraphic principle shows that a deformational event occurred after the deposition of layers 1 through 4? Principle of original horizontality 2. Which principle indicates that layer 2 is older thanlayer 3? Principle of superposition 3. Which principle indicates that intrusion A intruded before a major erosional event? Principle of cross cutting relationships 4. Which two principles show that layers 5 and 6 were deposited after the erosion of the tilted layers? Principle of superposition and princip le of original horizonta lity 5. Which principle indicates that fault b occurred after fault a?