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A fossilized leaf contains 33% of its normal amount of carbon 14. How old is the fossil (to the nearest year)? Use

5600 years as the half-life of carbon 14.

User Blenderer
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1 Answer

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Answer: The the fossil is approximately 8,267 years old.

Step-by-step explanation: To solve this problem, we can use the formula for exponential decay:

N(t) = N₀ * (1/2)^(t/T)

where:

N(t) = the amount of carbon-14 remaining after time t

N₀ = the initial amount of carbon-14

T = the half-life of carbon-14

Let's assume that the fossilized leaf originally contained 100% of its normal amount of carbon-14. Since it now contains only 33% of its normal amount, we can say that:

N(t) = 0.33N₀

Substituting this into the formula above, we get:

0.33N₀ = N₀ * (1/2)^(t/5600)

Dividing both sides by N₀ and taking the logarithm base 2 of both sides, we get:

t/5600 = log₂(0.33)

t = 5600 * log₂(0.33)

t ≈ -14187 years

This result is negative, which doesn't make sense in the context of the problem. It means that the fossil must be older than our initial assumption of 100% carbon-14. Let's assume instead that the fossil originally contained 50% of its normal amount of carbon-14 (which is more realistic). Then, we can say that:

N(t) = 0.5N₀

Substituting this into the formula above, we get:

0.5N₀ = N₀ * (1/2)^(t/5600)

Dividing both sides by N₀ and taking the logarithm base 2 of both sides, we get:

t/5600 = log₂(0.5)

t = 5600 * log₂(0.5)

t ≈ 8267 years

Therefore, the fossil is approximately 8267 years old.

User Damon Maria
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