Question

carbon-14 is used to determine the time an organism was living. the amount of carbon-14 an organism has is constant with the atmosphere, but when an organism dies the carbon-14 decays at a half - life of 5,730 years. if an archeologist measured the amount of carbon-14 in an organism and it was 25% of the total amount of atmospheric c - 14, what is the age of the organism?
o 1432.5 years
o 5,730 years
o 11,460 years
o 17,190 years

Explanation:

Step1: Understand half - life concept

The half - life formula is $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, where $N$ is the final amount of the substance, $N_0$ is the initial amount, $t$ is the time elapsed, and $T_{1/2}$ is the half - life. Given that $N = 0.25N_0$ and $T_{1/2}=5730$ years.

Step2: Substitute values into formula

Substitute $N = 0.25N_0$ into $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, we get $0.25N_0=N_0(\frac{1}{2})^{\frac{t}{5730}}$. Divide both sides by $N_0$ (since $N_0
eq0$), we have $0.25 = (\frac{1}{2})^{\frac{t}{5730}}$.

Step3: Rewrite 0.25 as a power of $\frac{1}{2}$

Since $0.25=\frac{1}{4}=(\frac{1}{2})^2$, the equation becomes $(\frac{1}{2})^2 = (\frac{1}{2})^{\frac{t}{5730}}$.

Step4: Solve for $t$

If $a^m=a^n$, then $m = n$. So, $2=\frac{t}{5730}$. Multiply both sides by 5730, we get $t = 2\times5730=11460$ years.

Answer:

11,460 years