Question

4 bellringer: radioactive decay carbon - 14 has a half - life of 5730 years. after 34,380 years, how many grams of carbon - 14 would remain from what was originally a 50.00 g sample? give your answer with 4 sig figs and units (with a space between the number and unit)

Explanation:

Step1: Calculate number of half - lives

The formula to find the number of half - lives $n$ is $n=\frac{t}{T_{1/2}}$, where $t$ is the time elapsed and $T_{1/2}$ is the half - life. Given $t = 34380$ years and $T_{1/2}=5730$ years. So, $n=\frac{34380}{5730}=6$.

Step2: Use decay formula

The radioactive decay formula is $N = N_0(\frac{1}{2})^n$, where $N_0$ is the initial amount and $N$ is the final amount. Here, $N_0 = 50.00$ g and $n = 6$. Then $N=50.00\times(\frac{1}{2})^6$.
$N = 50.00\times\frac{1}{64}=0.78125$ g.
Rounding to 4 significant figures, $N = 0.7813$ g.

Answer:

0.7813 g