Lab12 Geologic Time Doezie.pdf (1)
docx
keyboard_arrow_up
School
Orange Coast College *
*We aren’t endorsed by this school
Course
A110
Subject
Geology
Date
Jan 9, 2024
Type
docx
Pages
11
Uploaded by SuperInternetBear20
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
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
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
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
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.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
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?