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tatyana61 [14]
3 years ago
13

In 1985, during the construction of a skyscraper in Austin, Texas, the remains of a mastodon were unearthed. If only 1/64 of the

original carbon-14 remained, and carbon-14 has a half life of 5730 years, approximately how old were the mastodon bones?
Physics
1 answer:
Art [367]3 years ago
8 0

Answer:

  • <u>34,380 years</u>

Explanation:

The remaining amount of a radioactive isotope is found as the product of the original amount by (1/2) raised to the number of half-lives elapsed. The formla is:

  • M = M₀ × (1/2)ⁿ

Where M is the remaining amoun, M₀ i s the initial amount, and n is the number of half-lives.

Here, 1/64 of the original carbon-14 remained, meaning that M/M₀ = 1/64.

Then, you can substitute in the equation and solve:

  • (1/64) = (1/2)ⁿ
  • (1/2⁶) = (1/2ⁿ)
  • 2⁶ = 2ⁿ
  • 6 = n

Then, 6 half-lives elapsed since the mastodon died and the remains were dated.

Then, you must multiply 6 by the <em>half-life </em>time:

  • 6 × 5730 years = 34,380 years ← answer
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