Question

A hiker in Africa discovers a skull that contains 32% of its original amount of C-14. find the age of the skull.

Answer 1

Answer:

9419.3 years.

Step-by-step explanation:

Let the initial amount of C-14 be 100 units.

We have been given that a hiker in Africa discovers a skull that contains 32% of its original amount of C-14. We are asked to find the age of the skull.

We will use half life formula to solve our given problem.

$A=a\cdot(\frac{1}{2})^{\frac{t}{h}}$, where,

A = Amount left after t years,

a = Initial amount,

t = time,

h = Half life.

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

32% of 100 units would be 32.

$32=100\cdot(\frac{1}{2})^{\frac{t}{5730}}$

$\frac{32}{100}=\frac{100\cdot(\frac{1}{2})^{\frac{t}{5730}}}{100}$

$0.32=(0.5)^{\frac{t}{5730}}$

Now, we will take natural log of both sides.

$\text{ln}(0.32)=\text{ln}((0.5)^{\frac{t}{5730}})$

Using log property $\text{ln}(a^b)=b\cdot\text{ln}(a)$, we will get:

$\text{ln}(0.32)=\frac{t}{5730}\cdot \text{ln}(0.5)$

$\frac{\text{ln}(0.32)}{\cdot \text{ln}(0.5)}=\frac{t\cdot \text{ln}(0.5)}{5730\cdot \text{ln}(0.5)}$

$1.64385618977=\frac{t}{5730}$

$\frac{t}{5730}=1.64385618977$

$\frac{t}{5730}*5730=1.64385618977*5730$

$t=9419.295967$

$t\approx 9419.3$

Therefore, the age of skull is approximately 9419.3 years.