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%.