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:

\[N=N_o\ e^\dfrac{-t}{k}\]

\[e^\dfrac{-t}{k}=\frac{N}{N_o}\]

Taking $Log$ on both sides:

\[Log\left(e^\dfrac{-t}{k}\right)=\ Log\ \left(\frac{N}{N_o}\right)\]

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

Hence:

\[t\ =\ \frac{Log\ \left(\dfrac{N}{N_o}\right)}{-k}\]

## Expert Answer

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

**Decaying rate** $N\ =\ 38\ counts\ per\ min$

**Standard Decaying rate** $N_o\ =\ 58.2\ counts\ per\ min$

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:

\[k\ =\ \frac{0.693}{5715\ Yr}\]

\[k\ =\ 1.21\ \times\ {10}^{-4}\ {\rm Yr}^{-1}\]

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

\[t\ =\ \frac{Log\ \left(\dfrac{N}{N_o}\right)}{-k}\]

Substituting the given values in the above equation:

\[t\ =\ \frac{Log\ \left(\dfrac{38\ counts\ per\min}{58.2\ counts\ per\ min}\right)}{-1.21\ \times\ {10}^{-4}\ {\rm Yr}^{-1}}\]

\[t\ =\ 3523.13\ Yr\]

## Numerical Result

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

\[t\ =\ 3523.13\ Yr\]

## 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** $=\ 6100\ Years$

**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:

\[k\ =\ \frac{0.693}{5730\ Yr}\]

\[k\ =\ 1.136\ \times\ {10}^{-4}\ {\rm Yr}^{-1}\]

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

\[t\ =\ \frac{Log\ \left(\dfrac{N}{N_o}\right)}{-k}\]

Substituting the given values in the above equation:

\[t\ =\ \frac{Log\ \left(\dfrac{80\ %}{100\ %}\right)}{-1.136\ \times\ {10}^{-4}\ {\rm Yr}^{-1}}\]

\[t\ =\ 1964.29\ Yr\]