Question

The fraction of a radioactive isotope remaining at time t is${\left(\frac{1}{2}\right)}^{t/t_{1/2}}$ , where$t_{1/2}$ is the half-life. If the half-life of carbon-14 is 5730yr , what fraction of carbon- 14 in a piece of charcoal remains after

(a) 10.0yr;

(b) $10.0\times {10}^{3}yr$;

(c) $10.0\times {10}^{4}yr$?

(d) Why is radiocarbon dating more reliable for the fraction remaining in part (b) than that in part (a) or in part (c)?

Short answer

1. Fraction of Carbon-14 in a piece of charcoal after 10.0yr is 0.999.
2. Fraction of Carbon-14 in a piece of charcoal after$10.0\times {10}^{3}yr$ is 0.298.
3. Fraction of Carbon-14 in a piece of charcoal after $10.0\times {10}^{4}yr$ is $5.50\times {10}^{-6}$.
4. Part (b) radiocarbon dating is more reliable because sufficient amount $C^{14}$ isotope is remaining in the charcoal. While in part (a) the fraction of charcoal decayed is very less and in part (c) the fraction of the charcoal remaining is very less.

Step-by-step solution

Radiocarbon dating

Radio carbon dating is a process that helps in estimation of the age of an object depending on the carbon content present in it.

Find the fraction of carbon- 14 in a piece of charcoal remains after  

(a)

Consider the given information:

Half-life, $t_{1/2}=5730yr$

Time, t = 10.yr

After 10.0 yr , the fraction of carbon-14 in a piece of charcoal is calculated as shown below.

$$\begin{gathered} Fraction={\left(\frac{1}{2}\right)}^{t/t_{1/2}} \\ ={\left(\frac{1}{2}\right)}^{\frac{10}{5730}} \\ =0.999 \end{gathered}$$

Therefore, the fraction of Carbon-14 in a piece of charcoal after 10.0 is 0.999.

Find the fraction of carbon- 14 in a piece of charcoal remains after  10.0×103yr

(b)

Consider the given information:

Half-life, $t_{1/2}=5730yr$

Time, $t=10.0\times {10}^{3}yr =1\times {10}^{4}yr$

After $10.0\times {10}^{3}yr$, the fraction of carbon-14 in a piece of charcoal is calculated as shown below.

$$\begin{gathered} Fraction={\left(\frac{1}{2}\right)}^{t/t_{1/2}} \\ ={\left(\frac{1}{2}\right)}^{\frac{1\times {10}^4}{5730}} \\ =0.298 \end{gathered}$$

Therefore, the fraction of Carbon-14 in a piece of charcoal after $10.0\times {10}^{3}yr$ is $0.298$.

Find the fraction of carbon- 14 in a piece of charcoal remains after  10.0×104yr

(c)

Consider the given information:

Half-life, $t_{1/2}=5730yr$

Time, $t=10.0\times {10}^{4}yr$

After $10.0\times {10}^{4}yr$, the the fraction of carbon-14 in a piece of charcoal is calculated as shown below.

$$\begin{gathered} Fraction={\left(\frac{1}{2}\right)}^{t/1/2} \\ ={\left(\frac{1}{2}\right)}^{\frac{10.0\times {10}^4}{5730}} \\ =5.50\times {10}^{-6} \end{gathered}$$

Therefore, the fraction of Carbon-14 in a piece of charcoal after $10.0\times {10}^{4}yr$ is $5.50\times {10}^{-6}$.

Explain the reason for radiocarbon dating more reliable for the fraction remaining in part (b) than that in part (a) or in part (c)

(d) From the above calculations it is evident that the fraction of carbon that has decayed in part (a) is very less as the time period is only 10 years whereas the amount of carbon that is left in the charcoal in part (c) is very less as the time period of the decay is $10.0\times {10}^4 years$. So, the estimation of the time period in both these cases can be erroneous. But in part (b) the quantity of carbon decayed is satisfactory for dating. That is why radiocarbon dating is more reliable in part (b).