Answer:
18.94%.
Explanation:
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%.
If a tree dies and the trunk remains undisturbed for 13,750 years, 18.94% of the original C-14 will still be present.
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.
[tex][C]/[C]_0 = e^{-k.t} \\[C]/[C]_0 = e^{-(1.210.10^{-4}y^{-1} ).(13,750y)} = 0.1894 = 18.94 \%[/tex]
where,
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
