Question

A piece of fossilized wood has a carbon-14 radioactivity that is 1/4 that of new wood. the half-life of carbon-14 is 5730 years. how old is the cloth?

Answer 1

1.146 x 10⁴ year.

Step-by-step 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 = ??? years.

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

[A] is the remaining percentage of carbon-14 = 1/4[A₀] = 25.0 %.

∵ kt = ln [Ao]/[A]

∴ (1.21 x 10⁻⁴ year⁻¹)(t) = ln (100.0%)/[25.0 %]

(1.21 x 10⁻⁴ year⁻¹)(t) = 1.386.

∴ t = 1.386/ (1.21 x 10⁻⁴ year⁻¹) = 1.146 x 10⁴ year.