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 ca

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 carbon-14 remained, and carbon-14 has a half life of 5730 years, approximately how old were the mastodon bones?

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Nho 4 years 2021-08-12T09:57:25+00:00 1 Answers 11 views 0

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  1. Thunguyet
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    2021-08-12T09:58:31+00:00
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    Answer:

    • 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

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