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?

Answer 1

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