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Dinosaur fossils are too old to be reliably dated using carbon-14, which has a half-life of about 5730 years. Suppose we had a 68 million year old dinosaur fossil. How much of the living dinosaur's 14C would be remaining today? (Round your answer to five decimal places.)

Respuesta :

Edufirst

Answer:

  • 0.00000

Explanation:

The half-life of a radioisotope, in this case carbon-14, is the time that a sample requires to reduce its amount to half, and it is a constant for every radioisotope (it does not change with the amount of sample).

Then, the formula for the remaining amount of a radioisotope is:

  • A / A₀ = (1/2)ⁿ

Where:

  • A is the final amount of the element,
  • A₀ is the initial amount of the element,
  • A/A₀ is ratio of remaining amount to the original amount, and
  • n is the number of half-lives elapsed

The number of half-lives for carbon-14 elapsed for the dinosaur fossil is:

  • n = 68 million years / 5730 years ≈  11,867

Then, A / A₀ = (1/2)ⁿ = (1/2)¹¹⁸⁶⁷ ≈ 0.00000 .

The number is too small, and when you round to five decimal places the result is zero. That is why carbon-14 cannot be used to date dinosaur fossils, given that they are too old.

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