Question

Suppose an archaeologist finds a camel tooth that contains 42% of original amount of C -

14. Find age of the tooth.​

Answer 1

8675

Explanation:

Solve 0.42No=Noe^-0.0001t for t.

Divide both sides by No.

0.42No = Noe^-0.0001t

.42=e^-0.0001t

Take the log of both sides

In 0.42 = In e^-0.0001t

Apply In e^x =x.

In 0.42 = -0.0001 to solve for t.

t=8,675

So, the camels tooth is about 8675 years old

Answer 2

`\large \boxed{\text{7150 yr}}`

Explanation:

Two important equations in radioactive decay are

`\begin{array}{rcl}t_{(1)/(2)} &= &(\ln2)/(k ) \text{ and}\\\\\ln (N_(0))/(N_(t)) &=& kt\\\\\end{array}`

We use them for carbon dating.

1. Calculate the decay constant

The half-life of ¹⁴C is 5730 yr.

`\begin{array}{rcl}t_{(1)/(2)}& = &(\ln2)/(k )\\\\k& = &\frac{\ln2}{t_{(1)/(2)}}\\\\ & = & \frac{\ln2}{\text{5730 yr}}\\\\ & = & 1.210 * 10^(-4)\text{ yr}^(-1)\\\end{array}`

2. Calculate the age of the tooth

`\begin{array}{rcl}\ln (N_(0) )/(N_(t)) & = & kt\\\\\ln ( N_(0))/(0.42 N_(0)) & = & 1.210 * 10^(-4)\text{ yr}^(-1) * t\\\\\ln 2.381 & = & 1.210 * 10^(-4)t \text{ yr}^(-1)\\0.8675 & = & 1.210 * 10^(-4)t \text{ yr}^(-1)\\t & = & \frac{0.8675}{1.210 * 10^(-4) \text{ yr}^(-1)}\\\\ & = & \textbf{7150 yr}\\\end{array}\\\text{The age of the sample is $\large \boxed{\textbf{7150 yr}}$}`