Radioactivity and half life

Question

If a tree dies and the trunk remains undisturbed for 1.685 × 10⁴ years, what percentage of the original ¹⁴C is still present? (The half-life of ¹⁴C is 5730 years.)

Answer 1

Half life problems are exponential problems.

If you start with an amount A₀ then after 1 half-life there is (obviously) 1/2 the material left or 1/2A₀ left. After 2 half-lives then there is 1/2 of that amount left or (1/2)•(1/2)•(A₀), so for 3 half-lives, there is (1/2)³•A0 left, etc.

So generically, the amount left equals A₀•(1/2)^{(# of half lives)}

To calculate the number of half lives, we take the number of years gone by divided by 5730 years.

For 1.685 x 10⁴ years, the number of half lives is (1.685 x 10⁴)/5730 = 2.94066 and (1/2)^2.94066 = 0.13025 so A(1.685 x 10⁴) = 0.13025A₀ or 13% of the original amount.