# Question d1146

Jun 8, 2017

$\text{1.0 pg}$

#### Explanation:

You can't answer this question without knowing the half-life of carbon-14, so look that up before doing anything else.

You'll find it listed as

${t}_{\text{1.2" = "5730 years}}$

https://en.wikipedia.org/wiki/Carbon-14

Now, a radioactive isotope's nuclear half-life tells you the time needed for half of an initial sample to undergo radioactive decay.

In other words, the amount of this radioactive isotope is halved with the passing of every half-life.

So, let's say that you start with an unknown amount of carbon-14, let's say ${A}_{0}$. After a period of time equal to one half-life passes, you will be left with

A_ (1 xx t _"1/2") = A_0 * 1/2 = A_0/2 -> " after 5730 years"

After another period of time equal to one half-life passes, you will be left with

A_ (2 xx t_ "1/2") = A_0/2 * 1/2 = A_0/4 ->" after 2" xx "5730 years"

This means that after

$2 \times \text{5730 years" = "11460 years}$

pass, the sample of carbon-14 will be down to $\frac{1}{4} \text{th}$ of its initial value.

Since this is how much time passed since the plant died, you can say that $\text{0.25 pg}$ represents $\frac{1}{4} \text{th}$ of the initial mass of carbon-14 present in the plant.

Therefore, you will have

"initial mass" = 4 xx "0.25 pg" = color(darkgreen)(ul(color(black)(1.0 color(white)(.)"pg")))#

The answer is rounded to two sig figs, the number of sig figs you have for the mass of carbon-14 that remains after $11460$ years.