Question

If a tree dies and the trunk remains undisturbed for 13,750 years, what percentage of the original 14c is still present? (the half-life of 14c is 5730 years.)

Answer 1

Answer:

18.94%.

Explanation:

• The decay of carbon-14 is a first order reaction.

• The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.

• The integration law of a first order reaction is:

kt = ln [A₀]/[A]

k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.

t is the time = 13,750 years.

[A₀] is the initial percentage of carbon-14 = 100.0 %.

[A] is the remaining percentage of carbon-14 = ??? %.

∵ kt = ln [Ao]/[A]

∴ (1.21 x 10⁻⁴ year⁻¹)(13,750 years) = ln (100.0%)/[A]

1.664 =  ln (100.0%)/[A]

Taking exponential for both sides:

5.279 = (100.0%)/[A]

∴ [A] = (100.0%)/5.279 = 18.94%.

Answer 2

If a tree dies and the trunk remains undisturbed for 13,750 years, 18.94% of the original C-14 will still be present.

What is radioactive decay?

Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation.

C-14 decays with a half-life (th) of 5730 years. We can calculate the rate constant (k) using the following expression.

k = ln2 / th = ln2 / 5730 y = 1.210 × 10⁻⁴ y⁻¹

The decay follows first-order kinetics. We can calculate the fraction of the original C-14 after 13,759 years using the following expression.

$[C]/[C]_0 = e^{-k.t} \\[C]/[C]_0 = e^{-(1.210.10^{-4}y^{-1} ).(13,750y)} = 0.1894 = 18.94 \%$

where,

• [C] is the amount of C after a time t.
• [C]₀ is the original amount of C.

If a tree dies and the trunk remains undisturbed for 13,750 years, 18.94% of the original C-14 will still be present.

Learn more about radioactive decay here: https://brainly.com/question/25013071