Question

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

Answer 1

Answer:

8675

Step-by-step 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

Answer:

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

Step-by-step explanation:

Two important equations in radioactive decay are

$\begin{array}{rcl}t_{\frac{1}{2}} &= &\dfrac{\ln2}{k } \text{ and}\\\\\ln \dfrac{N_{0}}{N_{t}} &=& kt\\\\\end{array}$

We use them for carbon dating.

1. Calculate the decay constant

The half-life of ¹⁴C is 5730 yr.

$\begin{array}{rcl}t_{\frac{1}{2}}& = &\dfrac{\ln2}{k }\\\\k& = &\dfrac{\ln2}{t_{\frac{1}{2}}}\\\\ & = & \dfrac{\ln2}{\text{5730 yr}}\\\\ & = & 1.210 \times 10^{-4}\text{ yr}^{-1}\\\end{array}$

2. Calculate the age of the tooth

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