From the half-life for 14C decay, 5715 year, determine the age of the artifact.

A wooden radioactive artifact present in a Chinese Temple comprising of ${}^{14}C$ activity was decaying at the rate of $38.0$ counts per minute, whereas for a standard of zero age for ${}^{14}C$, the standard rate of decaying activity is 58.2 counts per minute.

This article aims to find the age of the artifact on the basis of its decaying activity of ${}^{14}C$.

The main concept behind this article is Radioactive Decay of ${}^{14}C$, which is a radioactive isotope of Carbon $C$ and Half-Life.

Radioactive Decay is defined as an activity involving energy loss of an unstable atomic nucleus in the form of radiation. A material comprising unstable atomic nuclei is called a radioactive material.

The half-life of radioactive material $t_{\frac{1}{2}}$ is defined as the time required to reduce the concentration of given radioactive material to one-half based on radioactive decay. It is calculated as follows:

$$t_{\frac{1}{2}}=\frac{ln2}{k}=\frac{0.693}{k}$$

Where:

$t_{\frac{1}{2}}=$ Half-Life of Radioactive Material

$k=$ Decay Constant

The age $t$ of the radioactive sample is found in terms of its decaying rate $N$ in comparison to its standard decaying rate at zero age $N_o$ as per the following expression:

$$N=N_o{e}^{{\displaystyle \frac{-t}{k}}}$$

$$e^{{\displaystyle \frac{-t}{k}}}=\frac{N}{N_o}$$

Taking $Log$ on both sides:

$$Log\left(e^{{\displaystyle \frac{-t}{k}}}\right)=Log\left(\frac{N}{N_o}\right)$$

$$\frac{-t}{k}=Log\left(\frac{N}{N_o}\right)$$

Hence:

$$t=\frac{Log\left({\displaystyle \frac{N}{N_o}}\right)}{-k}$$

Expert Answer

The half-life of ${}^{14}C$ Decay $=5715Years$

Decaying rate $N=38countspermin$

Standard Decaying rate $N_o=58.2countspermin$

First, we will find the decay constant of ${}^{14}C$ Radioactive Material as per the following expression for Half-Life of radioactive material $t_{\frac{1}{2}}$:

$$t_{\frac{1}{2}}=\frac{ln2}{k}=\frac{0.693}{k}$$

$$k=\frac{0.693}{t_{\frac{1}{2}}}$$

Substituting the given values in the above equation:

$$k=\frac{0.693}{5715Yr}$$

$$k=1.21\times {10}^{-4}{\mathrm{Y}\mathrm{r}}^{-1}$$

The age $t$ of the artifact is determined by the following expression:

$$t=\frac{Log\left({\displaystyle \frac{N}{N_o}}\right)}{-k}$$

Substituting the given values in the above equation:

$$t=\frac{Log\left({\displaystyle \frac{38countspermin}{58.2countspermin}}\right)}{-1.21\times {10}^{-4}{\mathrm{Y}\mathrm{r}}^{-1}}$$

$$t=3523.13Yr$$

Numerical Result

The age $t$ of the ${}^{14}C$ artifact is $3523.13$ Years.

$$t=3523.13Yr$$

Example

Radioactive Isotope of Carbon ${}^{14}C$ has a half-life of $6100$ years for radioactive decay. Find the age of an archaeological wooden sample with only $80$ of the ${}^{14}C$ available in a living tree. Estimate the age of the sample.

Solution

The half-life of ${}^{14}C$ Decay $=6100Years$

Decaying rate $N=80$

Standard Decaying rate $N_o=100$

First, we will find the decay constant of ${}^{14}C$ Radioactive Material as per the following expression for Half-Life of radioactive material $t_{\frac{1}{2}}$:

$$t_{\frac{1}{2}}=\frac{ln2}{k}=\frac{0.693}{k}$$

$$k=\frac{0.693}{t_{\frac{1}{2}}}$$

Substituting the given values in the above equation:

$$k=\frac{0.693}{5730Yr}$$

$$k=1.136\times {10}^{-4}{\mathrm{Y}\mathrm{r}}^{-1}$$

The age $t$ of the wooden sample is determined by the following expression:

$$t=\frac{Log\left({\displaystyle \frac{N}{N_o}}\right)}{-k}$$

Substituting the given values in the above equation:

$$t=\frac{Log\left({\displaystyle \frac{80}{100}}\right)}{-1.136\times {10}^{-4}{\mathrm{Y}\mathrm{r}}^{-1}}$$

$$t=1964.29Yr$$

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