Question

Carbon-14 has a radioactive half-life of 5730 years, which means every 5730 years half the Carbon-14 in a substance goes away. If a bone started with 100 mg of Carbon-14, but only has 10 mg left when it is discovered, how old is the bone?

Answer 1

The following equation gives the quantitative relationship between the original number of nuclei present at time zero (N₀) and the number (N) at a later time t:

$N = N_0e^{−λt}$

λ is the decay constant

The relationship between the decay constant λ and the half-life t_1/2 is

$λ = \frac{ln(2)}{t_{1/2}} ≈ \frac{0.693}{t_{1/2}}$

$λ = \frac{0.693}{5730} = 1.209\times10^{-4}$

$N/N_0 = e^{−λt}$

$10/100 = e^{−1.209\times10^{-4}t}$

$ln(0.1) = −1.209\times10^{-4}t$

$-2.302 = −1.209\times10^{-4}t$

$t = 1.904\times10^{4} = 19040\; years$