Question

An Archaeologist in Turkey discovers a spear head that contains 51% of its original amount of C-14. Find the age of the spear head to the nearest year.

Answer 1

Answer: 5566 years

Step-by-step explanation:

We know that the half-life of carbon-14 is 5730 years.

Let x be the original amount of C-14 , then the current amount of C-14 =$0.51x$

The radioactive half-life formula for C-14 is given by :-

$P=P_0(0.5)^{\dfrac{t}{t_{\frac{1}{2}}}}$, where P is the amount of C-14 at time t and  $P_0$ is the original amount.

$\Rightarrow\ 0.51x=x(0.5)^{\dfrac{t}{5730}}\\\\\Rightarrow\ 0.51=(0.5)^{\dfrac{t}{5730}}$

Taking log on both the sides , we get

$\log(0.51)=\dfrac{t}{5730}\log(0.5)\\\\\Rightarrow\ -0.292429823902=\dfrac{t}{5730}\times-0.301029995664\\\\\Rightarrowt=\dfrac{5730\times0.292429823902}{0.301029995664}=5566.29875791\approx5566\text{years}$

Hence, the age of the spear head is 5566 years.

Answer 2

Use given half-life for C-14 of 5,730 years.

The exponential decay function:
$y = e^{-kt}$
plug in half life, solve for 'e^(-k)'
$0.5 = e^{-5730k} \\ \\ 0.5 = (e^{-k})^{5730} \\ \\ e^{-k} = (0.5)^{1/5730}$

Sub into original decay function, plug in 0.51 for 'y' and solve for 't'

$y = e^{-kt} = (e^{-k})^t \\ \\ 0.51 = (0.5)^{t/5730} \\ \\ ln(0.51) = \frac{t}{5730} ln(0.5) \\ \\ t = 5730 \frac{ln 0.51}{1n 0.5}$