Question

A fossilized leaf contains 70% of its normal amount of carbon 14. How old is the fossil? Use 5700 years as the half-life of carbon 14.

Answer 1

To calculate the age of a fossil with 70% of its original carbon-14 remaining, the half-life of carbon-14, which is 5700 years, is used. After solving the half-life decay formula, it is determined that the leaf is approximately 2438 years old.

Step-by-step explanation:

If a fossilized leaf contains 70% of its normal amount of carbon 14 (C-14), you can calculate how old the fossil is using the half-life of C-14, which is 5700 years. To find out how many half-lives have passed for the leaf to have 70% of its C-14 remaining, you can use the half-life formula:

N_t = N_0(1/2)^(t/T)

Where N_t is the remaining amount of the isotope, N_0 is the original amount of the isotope, t is the time that has passed, and T is the half-life of the isotope.

If we take the initial amount of C-14 (100%) and the remaining amount of C-14 (70%), we can determine that no full half-life has occurred yet because the percentage of C-14 is more than 50%.

Now we need to solve for t, so we rearrange the formula to get:

t = T (log(N_t/N_0) / log(1/2))

t = 5700 * (log(0.7) / log(0.5))

Using a calculator, we find: t ≈ 2438 years approximately.

Therefore, the fossilized leaf is approximately 2438 years old.

Answer 2

Using the decay rate of carbon-14 with a half-life of 5,700 years, a fossilized leaf with 70% of its original carbon-14 is about 1,140 years old.

Step-by-step explanation:

The fossilized leaf is approximately 1,140 years old.

When a living organism dies, it stops absorbing carbon-14 (¹4C), and the existing ¹4C in its remains begins to decay at a known rate, the half-life. In the case of carbon-14, the half-life is 5,700 years. To find out how many half-lives have passed for a sample that has 70% of its original ¹4C remaining, we can use the formula: Age = (Half-life)*log(N/N0)/log(0.5), where N is the remaining amount of ¹4C, and N0 is the original amount of ¹4C.

Since the fossil has 70% of the ¹4C, N/N0 is 0.7, and the log(0.7)/log(0.5) is approximately 0.2. Multiplying 0.2 by the half-life of 5,700 years gives an age of 1,140 years. Therefore, the fossil is approximately 1,140 years old because two tenths of a half-life has passed.