Question

An archaeologist analyses the wood in a prehistoric structure and finds that ${C^{14}}$ (Half-life = 5700 years) to ${C^{12}}$ is only one-fourth of that found in the cells buried plants. The age of the wood is about
A) 5700 years
B) 2850 years
C) 11400 years
D) 22800 years

Answer 1

Hint:The question is from the radioactive decay section of physics. We have to apply the carbon dating concepts to solve this problem.


Formula used:
$Age = \dfrac{{2.303 \cdot {t_{1/2}}}}{{0.693}}\log \dfrac{{{N_0}}}{N}$
Where ${N_0}$= Initial amount of substance, $N$= present amount of substance, ${t_{1/2}}$= half-life.


Complete answer:
The given values are,
$N$= present amount of substance = 0.25${N_0}$
${N_0}$= Initial amount of substance
${t_{1/2}}$= half-life = 5770 years
Then the age of the wood is about,
$Age = \dfrac{{2.303 \times 5770}}{{0.693}}\log \dfrac{1}{{0.25}}$
$Age = 11544.53 years$

Hence, the correct option is Option (C).

Additional Information:
Carbon dating is a method for determining the age of an object containing organic material by using the properties of radiocarbon(carbon-14). Willard Libby developed the technique in the late 1940s at the University of Chicago. It's also called Radiocarbon dating and carbon-14 dating.
The atomic number and mass of the decaying atoms determine the rate of radioactive decay. The ratio of radioactive isotopes to the estimated initial concentration of these isotopes at the moment of the organism's death can be used to calculate the approximate age of the decaying material. According to scientists, there hasn't been much of a shift in the ratio of carbon-12 to carbon-14 isotopes in the atmosphere, therefore their connection should mostly remain unchanged.


Note:The equation of radioactive decay is ${N_t} = {N_0}{e^{ - \lambda t}}$. Where, ${N_t}$ = number of nuclei present after some time t, ${N_0}$= initial number of nuclei present, \[\lambda \]= disintegration constant and t = time. While solving the question, pay attention to log values and don’t put natural log values instead of log10.