Question

13. An important tool in archeological research is radiocarbon dating, developed by the American chemist Willard E Libby.3 This is a means of determining the age of certain wood and plant remains, and hence of animal or human bones or artifacts found buried at the same levels. Radiocarbon dating is based on the fact that some wood or plant remains contain residual amounts of carbon-14, a radioactive isotope of carbon. This isotope is accumulated during the lifetime of the plant and begins to decay at its death. Since the half-life of carbon-14 is long (approximately 5730 years4), measurable amounts of carbon-14 remain after many thousands of years. If even a tiny fraction of the original amount of carbon-14 is still present, then by appropriate laboratory measurements the proportion of the original amount of carbon-14 that remains can be accurately determined. In other words, if Q(t) is the amount of carbon-14 at time t and Qo is the original amount, then the ratio Q(t)/Q0 can be determined, as long as this quantity is not too small. Present measurement techniques permit the use of this method for time periods of 50,000 years or more. (a) Assuming that Q satisfies the differential equation Q' = ?rQ, determine the decay constant r for carbon-14. (b) Find an expression for Q(t) at any time t, if Q(0) = Qo. (c) Suppose that certain remains are discovered in which the current residual amount of carbon-14 is 20% of the original amount. Determine the age of these remains. .

Answer 1

a) r=-0.00012097

b)
`\large\ \boxed{Q(t)=Q_0e^(-0.00012097*t)}`

c) 13,304.65 years

Explanation:

a)

Q(t) satisfies the differential equation Q'(t) = rQ(t), hence

`\large (Q'(t))/(Q(t))=r`

Integrating on both sides

`\int (Q'(t))/(Q(t))dt=\int rdt\rightarrow log(Q(t))=rt+C`

where C is a constant. Taking the exponential on both sides

`\large e^(log(Q(t)))=e^(rt+C)=e^Ce^(rt)\rightarrow Q(t)=C_0e^(rt)`

In this case,
`\large C_0` is the initial value Q(0), the amount of the initial value of carbon-14 and we have

`\large Q(t)=Q_0e^(rt)`

As we know that the half-life of carbon-14 is approximately 5,730 years we have

`\large Q(5,730)=(Q_0)/(2)\rightarrow Q_0e^(5,730r)=(Q_0)/(2)\rightarrow e^(5,730r)=(1)/(2)`

Taking logarithm on both sides

`\large log(e^(5,730r))=log((1)/(2))\rightarrow 5,730r=-0.693147\rightarrow r=(-0.693147)/(5,730)`

and

`\large \boxed{r=-0.00012097}`

b)

`\large \boxed{Q(t)=Q_0e^(-0.00012097*t)}`

c)

We want to find a value of t for which

Q(t) = 20% of
`\large Q_0` =
`\large\ 0.2Q_0`

`\large Q(t)=0.2Q_0\rightarrow Q_0e^(-0.00012097*t)=0.2Q_0\rightarrow e^(-0.00012097*t)=0.2\rightarrow\\-0.00012097*t=log(0.2)\rightarrow t=(log(0.2))/(-0.00012097)`

and

`\large \boxed{t\approx 13,304.65\;years}`