Question

The half-life for the radioactive decay of $^{14}C$ is $5730$ years. An archaeological artifact containing wood had only $80\% $ of the $^{14}C$ found in a living tree. Estimate the age of the sample.

Answer 1

Hint: Radioactive decay is related to the half-life. The time required for an isotope to be reduced to half of its original mass through radioactive decay is the half-life of that substance. Through radioactive decay, unstable isotopes undergo decay by emitting radiation. By calculating half-life we can predict the presence and absence of a particular isotope.

Complete step by step answer:
We know that half-life is the time required for an isotope to decay to half of its initial mass.
Half-life is denoted as ${t_{1/2}}$.
In the question, the isotope given is $^{14}C$
The half-life of $^{14}C$, ${t_{1/2}} = 5730years$.
The formula to find out the age of the sample is given below:
$t = \dfrac{{2.303}}{k}\log \dfrac{{{{[R]}_0}}}{{[R]}}$
Here k is the decay constant. ${[R]_0}$ and $[R]$ are initial and final concentration respectively.
In the question, it is given that only $80\% $ of the $^{14}C$ is found in a living tree. That is the final concentration $[R]$ is $80\% $.
We should take the initial concentration ${[R]_0}$ to be $100\% $.
The formula to find out decay constant, k is given below:
$k = \dfrac{{0.693}}{{{t_{1/2}}}}$.
Using the above formula, first, we have to find out decay constant and then substitute in the equation $t = \dfrac{{2.303}}{k}\log \dfrac{{{{[R]}_0}}}{{[R]}}$.
Decay constant, $k = \dfrac{{0.693}}{{{t_{1/2}}}} = \dfrac{{0.693}}{{5730}}year{s^{ - 1}}$
Therefore $t = \dfrac{{2.303}}{k}\log \dfrac{{{{[R]}_0}}}{{[R]}} = \dfrac{{2.303}}{{\dfrac{{0.693}}{{5730}}}}\log \dfrac{{100}}{{80}} = 1845.3722years$.
Thus the answer is $1845years$ approximate.
Therefore the age of the sample is $1845years$.

Note:
Sometimes the decay constant is also denoted by a symbol $\lambda $ instead of the symbol $k$.
Similarly, the initial concentration and final concentration are also represented as ${N_0}$ and $N$ respectively.
Thus the formula for time can also be written as $t = \dfrac{{2.303}}{\lambda }\log \dfrac{{{N_0}}}{N}$.
Radioactive decay is first-order reactions. Therefore decay constant can be calculated as follows:
$\lambda = \dfrac{{2.303}}{t}\log \dfrac{{{N_o}}}{N}$