Question

Carbon-14 has a half-life of 5720 years and this is a first order reaction. If a piece of wood has converted 75% of the carbon-14, then how old is it?
A) 11440 years
B) 2375 years
C) 4750 years
D) 4290 years
E) 1430 years

Answer 1

The amount of the substance (At) at any time t is calculated through the equation,
                                  At = (A0)(e^-kt)
where A0 is the initial amount and k is constant. 

For the half-life,
                              ln (At / A0) = ln 0.5 = -k(5270 years)
The value of k is 1.3153x10^-4 / year.
Substituting this value to the same equation, where At = 0.25A0
                                    ln 0.25 = -(1.3153 x 10^-4)(t)
The value of t from this equation is 10540 years. The closest value to this is letter A. 
                                

Answer 2

The value of the time $t$ is $\boxed{\bf t=2,375\text{\bf\ years}}$.

The correct option is $\boxed{\bf option\ B}$

Further explanation:

Carbon-14 is a weakly radioactive isotope of carbon.

Radiocarbon dating is the archaeological process to determine the age of an object which should be of organic material with the properties of radiocarbon.

The radiocarbon decay is the amount of carbon which is converted into wood.

The amount $A(t)$ of the substance at any time $t$ is calculated as follows:

$\boxed{A(t)=A_{0}e^{-kt}}$     ......(1)

Here, $A_{0}$ is the initial amount of the substance and $k$ is the constant.

The concentration in the initial amount is always $100\%$.

Given:

The half-life of the Carbon-14 is $5270\text{ years}$. The concentration of the carbon in the piece of wood is $75\%$.

Calculation:

Step 1:

First we calculate the rate constant of the reaction.

The half-life of Carbon-14 is $5270\text{ years}$.

Substitute $5270$ for $t$ and $\frac{A_{0}}{2}$ for $A(t)$ in the equation (1) to obtain the value of $k$.

$\begin{aligned}\dfrac{A_{0}}{2}&=A_{0}e^{-k\cdot 5270}\\ \dfrac{1}{2}&=e^{-5270k}\\ln(0.5)&=-5270k\\-0.693&=-5270k\end{aligned}$

Further solve the above equation to obtain the value of $k$.

$\begin{aligned}0.693&=5270k\\k&=\dfrac{5270}{0.693}\\k&=1.21154\times 10^{-4}\end{aligned}$

Therefore, the value of $k$ is $k=1.21154\times 10^{-4}$.

Step 2:

Now we calculate the age of wood.

The concentration of initial carbon $(A_{0})$ is $100\%$.

The amount of the carbon remaining in the wood is $75\%$ which is the final concentration.  

The value of $t$ is obtained from the equation (1) as follows:

$\begin{aligned}A(t)&=A_{0}e^{-kt}\\ \dfrac{A(t)}{A_{0}}&=e^{-kt}\\ \dfrac{A(t)}{A_{0}}&=\dfrac{1}{e^{kt}}\\e^{kt}&=\dfrac{A_{0}}{A(t)}\\t&=\dfrac{1}{k}ln\left(\dfrac{A_{0}}{A(t)}\right)\end{aligned}$

Therefore, the expression for $t$ is $t=\frac{1}{k}ln(\frac{A_{0}}{A(t)})$.

Now, substitute $1.21154\times 10^{-4}$ for $k$, $100$ for $A_{0}$ and $75$ for $A(t)$ in the expression of $t$.

$\begin{aligned}t&=\dfrac{1}{k}ln\left(\dfrac{A_{0}}{A(t)}\right)\\&=\dfrac{1}{1.21154\times 10^{-4}}\times ln\left(\dfrac{100}{75}\right)\\&=2374.5\\ &\approx 2375\end{aligned}$

Therefore, the value of $t$ is $2375\text{ years}$.

Thus, the age of sample of Carbon-14 is $\boxed{\bf 2,375\text{\bf\ years}}$.

This implies that the correct option is $\boxed{\bf option B}$.

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Answer details:

Grade: Junior school

Subject: Mathematics

Chapter: Exponential  function

Keywords: Half-life, carbon-14, isotope, reaction, wood, concentration, amount, initial amount, final point, time, years, age of wood, exponents, scientific form, radiocarbon. Radiocarbon decay, carbon dating.