Question

An archaeological specimen containing $^{14}C$ 40 counts in 5 min per gram of carbon. A specimen of freshly cut wood gives 20.3 counts per gram of carbon per minute. The counter used recorded a background count of 5 counts per minute in absence of any $^{14}C$ containing samples. What is the age of the sample? [${{T}_{50}}$ of $^{14}C$ = 5668 ]
[A] 13328 years
[B] 17321 years
[C] 14328 years
[D] None of these

Answer 1

To solve this you can use the principle of radiocarbon dating. Use the cpm given to find the activity of both the samples and do not forget to subtract the error. You can use the formula-
$t=\dfrac{{{t}_{\text {1/2}}}\times 2.303}{0.693}\log \dfrac{activity\text{ of }{}^{14}C\text{ }of\,living\text{ organism}}{activity\text{ of }{}^{14}C\text{ of the dead organism}}$

Complete step by step solution:
We know that the half-life period of a radioactive element is the time required for the substance to reduce to half of its initial value.
Here, the half-life period of carbon is given to us as 5668.
Now, it is given to us that the archaeological specimen gives 40 counts per 5 minute for 1 gram of carbon. Therefore, for 1 minute it will be $\dfrac{40}{5}$ i.e. 8 counts per minute.
And a freshly cut sample of wood gives 20.3 counts per gram of carbon per minute.
However, the counter recorded a backlog of 5 counts per minute.
Therefore, activity of wood sample = $20.3 – 5$ $= 15.3$
And activity of the archaeological sample = $8 – 5$ $= 3$.
The sample of wood is non-radioactive and the archaeological specimen is radioactive. Therefore, we can use the formula of radio-carbon dating here.
We can calculate the age of the sample in terms of activity and half-life of the radioactive carbon by using the formula-
$t=\dfrac{{{t}_{\text {1/2}}}\times 2.303}{0.693}\log \dfrac{activity\text{ of wood sample}}{activity\text{ of }{}^{14}C\text{ of the archaeological sample}}$
Therefore, putting the values in the above equation we will get-
$t=\dfrac{5668\times 2.303}{0.693}\log \dfrac{15.3}{3}=13327.8\text{ years}$
From the above calculation we can see that the age of the sample is 13327.8 i.e. almost equal to 13328 years.

Therefore, the correct answer is option [A] 13328 years.

Note: In the above question, we have used the principle of radiocarbon dating to find out the age of the sample. The Carbon-14 method is a radioactive method that we can use to determine the age of any object that contains organic materials. This process is carried out by a radioactive isotope of carbon carbon-14. Therefore, this method is called the carbon-14 method or radiocarbon dating. However, we cannot use this method to determine the age of very old rocks and other species due to the short half-life span of carbon. For species older than 50,000 years, we will need an element with a higher half-life span.