A paleontologist inspects an undisturbed sedimentary sequence containing two distinct volc… — JAMB UTME

A paleontologist inspects an undisturbed sedimentary sequence containing two distinct volcanic ash layers: Layer $X$ (the lower bed) and Layer $Y$ (the upper bed), which encapsulate an intermediate fossiliferous sedimentary stratum. Mass spectrometry reveals that potassium-bearing minerals in Layer $X$ have a $^{40}\text{K}$ to $^{40}\text{Ar}$ ratio of $1:3$ , whereas minerals in Layer $Y$ have a $^{40}\text{K}$ to $^{40}\text{Ar}$ ratio of $1:1$ . Given that the half-life of $^{40}\text{K}$ is $1.3 \times 10^9\text{ years}$ , which of the following deductions regarding the age and geological significance of the fossil in the intermediate layer is correct?

The fossilized organism existed between $1.3 \times 10^9$ and $2.6 \times 10^9$ years ago, bounded by absolute radiopaque ages of the surrounding strata based on the law of superposition.Answer
B
The fossilized organism existed between $0.65 \times 10^9$ and $1.3 \times 10^9$ years ago because radio-isotopic decay rates accelerate in deeper geological layers due to environmental pressures.
C
The fossilized organism is older than $2.6 \times 10^9$ years because upper volcanic ash beds indicate primary ecological succession over older underlying fossil beds.
D
The fossilized organism must be classified as an analogous transitional form because identical radiometric ratios reflect functional rather than chronological relationships.

Answer

The fossilized organism existed between

1.3 \times 10^9

and

2.6 \times 10^9

years ago, bounded by absolute radiopaque ages of the surrounding strata based on the law of superposition.

Layer X contains a

1:3

ratio of

^{40}\text{K}

^{40}\text{Ar}

, meaning

25\%

of the parent isotope remains, corresponding to two half-lives (

2.6 \times 10^9\text{ years}

). Layer Y contains a

1:1

ratio, meaning

50\%

of the parent isotope remains, corresponding to one half-life (

1.3 \times 10^9\text{ years}

). Under the law of superposition, sedimentary strata between two dated volcanic beds fall chronologically between those bracketed age limits.

Step-by-Step Solution

Determine the age of lower Layer X using half-life calculation

Ratio

^{40}\text{K}:^{40}\text{Ar} = 1:3

indicates that

1/4

(

25\%

) of the original

^{40}\text{K}

remains. This corresponds to

2

half-lives:

2 \times 1.3 \times 10^9 = 2.6 \times 10^9\text{ years}

When

^{40}\text{K}

decays into

^{40}\text{Ar}

, the total initial parent quantity equals parent plus daughter products (

1 + 3 = 4

). Remaining fraction is

1/4 = (1/2)^2

Determine the age of upper Layer Y using half-life calculation

Ratio

^{40}\text{K}:^{40}\text{Ar} = 1:1

indicates that

1/2

(

50\%

) of the original

^{40}\text{K}

remains. This corresponds to

1

half-life:

1 \times 1.3 \times 10^9 = 1.3 \times 10^9\text{ years}

Total initial parent quantity is

1 + 1 = 2

. Remaining fraction is

1/2 = (1/2)^1

Apply the Law of Superposition to place the fossil in time

Since Layer X is below the fossil stratum and Layer Y is above it, the fossil is older than Layer Y (

1.3 \times 10^9

years) and younger than Layer X (

2.6 \times 10^9

years).

In undisturbed sedimentary rock sequences, deeper rock layers are older than superior rock layers.

Key Concept

Integration of Radiometric Dating and Stratigraphic Superposition in Paleontology

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic