Question

A mammoth bone is found to contain roughly 5% of the isotope C-14. Living elephant bones commonly contain about 40% C-14. Given that C-14 has a half-life of 5,730 years, how long ago did this mammoth die? Please show your work. (hint: no complex math is needed to solve this problem)

Answer 1

Answer:

The mammoth died 17.190 years ago.

Step-by-step explanation:

Given information:

• Half-life of C-14 is 5.700 years.

• Living elephant bones commonly contain about 40% C-14.

For every 5700 years, the elephant's bones losses a half-life of C-14. And supposing that the living mammoth bones had the same percentage of C-14.

If the mammoth when living has 40%, then:

• After 5.730 years, the mammoth bones will contain 20%.  
• After 11.460 years(2 half-lives), the mammoth bones will contain 10%.
• After 17.190  years(3 half-lives), the mammoth bones will contain 5%.

Answer 2

Answer:

$\large \boxed{\text{17 190 yr ago}}$

Step-by-step explanation

The amount of C-14 has declined from 40 % to 5 % of the original, that is, to ⅛ of the original amount.

The half-life of C-14 (5730 yr) is the time it takes for half of the isotope to decay.

We can make a table of the amount remaining after each successive half-life.

$\begin{array}{ccc}\textbf{Number of} && \textbf{Fraction} \\\textbf{half-lives} & \textbf{Years} & \textbf{remaining} \\0 &0 & 1 \\1 &5730 &\frac{1}{2 } \\\\2 &11460 &\frac{1}{4 } \\\\3 &17190 &\frac{1}{8} \\\\4 & 22920& \frac{1}{16} \\\end{array}\\\text{We see that the fraction of C-14 is reduced to \frac{1}{8} after three half-lives.}\\\text{The mammoth died \large \boxed{\textbf{17 190 yr ago}} }$