Ask Question
16 August, 19:06

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?

+1
Answers (1)
  1. 16 August, 19:12
    0
    34,380 years

    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 half-life time:

    6 * 5730 years = 34,380 years ← answer
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers