Question

The bones of a prehistoric man found in the desert of New Mexico contain approximately 5% of the original amount of carbon 14. If the half-life of carbon 14 is 5700 years, approximately how long ago did the man die?

Short answer

The man died approximately $24,636$years ago.

Step-by-step solution

Step 1. Given information  

The bones of a prehistoric man found in the desert of New Mexico contain approximately 5% of the original amount of carbon 14. If the half-life of carbon 14 is 5700 years.

Step 2. Required to find  

Approximately how long ago did the man die?

Step 3. Finding approximately how long ago the man died

Let, $A\left(t\right)=A_0{e}^{kt}$

Given, $t=5700,A\left(5700\right)=\frac{A_0}{2}$

$\frac{A_0}{2}=A_0{e}^{k\left(5700\right)}$

$\Rightarrow \frac{1}{2}=e^{5700k}$

Taking log on both sides,

$\ln \left(\frac{1}{2}\right)=\ln e^{5700k}$

$\Rightarrow \ln \left(\frac{1}{2}\right)=5700k$

$\Rightarrow \frac{\ln \left(\frac{1}{2}\right)}{5700}=k$

$\Rightarrow k=\frac{-0.6931}{5700}$

$\Rightarrow k=-1.216\times {10}^{-4}$

The equation becomes,

$A\left(t\right)=A_0{e}^{-1.216\times {10}^{-4}t}$

Put $A\left(t\right)=0.05A_0$

$0.05A_0=A_0{e}^{-1.216\times {10}^{-4}t}$

role="math" localid="1646624433323" $\Rightarrow 0.05=e^{-1.216\times {10}^{-4}t}$

Taking log on both sides,

$\ln 0.05=\ln e^{-1.216\times {10}^{-4}t}$

$\Rightarrow \ln 0.05=-1.216\times {10}^{-4}t$

$\Rightarrow t=\frac{\ln 0.05}{-1.216\times {10}^{-4}}$

$\Rightarrow t=\frac{-2.9957}{-1.216\times {10}^{-4}}$

$\Rightarrow t=24636$