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the principle of original horizontality. Hence, the correct order Determine the conversion ratio (CR) from actual units to
of events, from oldest to youngest, is C, tilting, B, D, A, E. model units:
9
21.8. Event A overlies B, so the principle of superposition dictates it 4.5 × 10 yr = 90 m model
9
is younger than B. Event E cuts across A, C, and D, so it must _ = _
4.5 × 10 yr
90 m model

9
9
be younger than these events. Event B cuts across C and D, so it 4.5 × 10 yr 4.5 × 10 yr
is younger than these events. Event C cuts across D, so C is _
90 m model

1 =

younger than D. 4.5 × 10 yr
9
Hence, the correct order of events, from oldest to −8 m model
_

youngest, is D, C, B, A, and E. CR = 2.0 × 10 yr
Based on its age, C was intruded during the Devonian Determine the duration of the Precambrian era, remembering
period. The basalt dike, E, was intruded during the Triassic, and that the age is negative because it is before the present.

the dinosaur bones in the sedimentary rocks are also of Triassic From Figure 21.17:

age. Therefore, the geologic periods missing from the record t beginning = −4.500 × 10 yr
9
due to the unconformity are t end = −5.51 × 10 yr
8
Middle Devonian through the end of the Permian. duration = ?
21.9. The original number of moles of uranium-235 would be the sum duration = t end − t beginning

8
9
of moles of parents and daughter isotopes. Hence, the original = −5.51 × 10 yr − (−4.500 × 10 yr)
9
number of moles of uranium-235 was 8(2 + 6). Th e fi rst half-life = 3.949 × 10 yr
would reduce the number of moles of uranium-235 to 4. Th e Determine the model dimension of the Precambrian era.
second half-life would reduce the remaining 4 moles to 2, giving 9
duration = 3.949 × 10 yr
the ratio of 2 moles of uranium-235 to 6 moles of lead-207.
−8 m model

Therefore 2 half-lives have passed. CR = 2.0 × 10 _
yr
number of half-lives = 2 d model = ?
8
time of half-life = 7.04 × 10 yr d model = duration (CR)
age = ? 9 −8 m model
_
(

age = (number of half-lives)(time of half-life) = 3.949 × 10 2.0 × 10 yr )
8
= (2)(7.04 × 10 yr) 9 −8 _
m model

= 3.949 × 10 (2.0 × 10 ) yr yr

9
= 1.41 × 10 yr
1
= 7.9 × 10 m model
21.10. Determine the number of half-lives. = 79 m
4
age = 2.2920 × 10 yr 21.13. Determine the conversion ratio (CR) based on duration of the
3
time of half-life = 5.730 × 10 yr 3
geologic time scale, which is 4.500 × 10 Myr:
number of half-lives = ?
d model

age = (number of half-lives)(time of half-life) d model = duration (CR) ∴ CR = _

duration
age
__
∴ (number of half-lives) =
28 m model
(time of half-life) CR = __

3
4
2.2920 × 10 yr 4.500 × 10 Myr
__
= 28

_
3
5.730 × 10 yr = _ m model

= 4 4.500×10 3 Myr
3
−3 m model

After the first half-life (5.730 × 10 yr), 50 g will remain. Aft er = 6.2 × 10 _

Myr
3
3
the second half-life (5.730 × 10 yr + 5.730 × 10 yr), 25 g will
3
3
remain. After the third half-life (5.730 × 10 yr + 5.730 × 10 yr + Determine the duration of the Cenozoic era, remembering that

3
5.730 × 10 yr), 12.5 g will remain. Aft er the fourth half-life the age is negative because it is before the present.
3
3
3
(5.730 × 10 yr + 5.730 × 10 yr + 5.730 × 10 yr + 5.730 × From Figure 21.17:
3
10 yr), 6.25 g will remain. t beginning = −65 Myr
duration = t end − t beginning
21.11. Th e first half-life would reduce the percentage of potassium-40 t end = 0 Myr = 0 Myr − (−65 Myr)

atoms to 50 percent. The second half-life would reduce the duration = ? = 65 Myr

percentage of potassium-40 atoms to 25 percent. Th e third Determine the model dimension of the Cenozoic era.
half-life would reduce the percentage of potassium-40 atoms to
duration = 65 Myr
12.5 percent. Therefore, 3 half-lives have elapsed. d model = duration (CR)

−3 m model

CR = 6.2 × 10 _
number of half-lives = 3 Myr d model = 65 Myr 6.2 × 10 _
(
−3 m model

9
time of half-life = 1.25 × 10 yr d model = ? Myr )
age = ? = 65(6.2 × 10 ) Myr _
−3
m model

age = (number of half-lives)(time of half-life) Myr
9
= (3)(1.25 × 10 yr) = 4.0 × 10 −1 m model
9
= 3.75 × 10 yr 21.14. Create a table that lists the geologic periods encompassing the

21.12. This problem requires a conversion ratio to be developed to age ranges of all of the life-forms present in the assemblage, and

convert actual units into model units. Then the duration of the shade the age range of each life-form. In this case, the three
Precambrian era must be calculated and multiplied by the
conversion ratio to obtain the model dimensions.
E-47 APPENDIX E Solutions for Group A Parallel Exercises 689