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Question: The half-life for radioactive decay of C-14 is 5730 years. An archaeological artefact containing woo...

The half-life for radioactive decay of C-14 is 5730 years. An archaeological artefact containing wood had only 80% of the C-14 found in a living tree. The age of the sample is?
a.) 1485 years
b.) 1845 years
c.) 530 years
d.) 4767 years

Explanation

Solution

Hint: The age of sample can be determined by t=2.303k\dfrac{2.303}{k}log[R]0[R]\dfrac{[R]_0}{[R]}, here substitute the values of k, [R]. The mentioned radioactive decay is a first order process, and the k represents the decay constant.

Complete step by step solution:
Firstly, let us know that we need to calculate the age of the sample. The age of sample can be found by t=2.303k\dfrac{2.303}{k}log[R]0[R]\dfrac{[R]_0}{[R]}, here k symbolises the decay constant and t is the time i.e. age and [R]0[R]\dfrac{[R]_0}{[R]} represent the ratio of total number of samples to the decayed samples.

Now, we are given with the half-life of C-14 i.e. 5730 years, so we will calculate the decay constant that is k=0.693t1/2\dfrac{0.693}{t}_{1/2}, t1/2_{1/2} is the representation of half-life.
Now substitute the value of t1/2_{1/2} then k=0.6935730\dfrac{0.693}{5730} years1^{-1}.
Now, we already know t= 2.303k\dfrac{2.303}{k}log[R]0[R]\dfrac{[R]_0}{[R]}, [R]0[R]\dfrac{[R]_0}{[R]} is 10080\dfrac{100}{80}, we are given 80% of decay, then t= 2.3030.6935730\dfrac{2.303}{\dfrac{0.693}{5730}}log 10080\dfrac{100}{80}= 1845 years (approximately).
Therefore, the age of the sample is 1845 years. The correct option is B.

Note: Don’t get confused between the symbols. Sometimes k, the decay constant is also represented by λ\lambda. The decay constant k and ratio of activity samples is different. The symbol k shows the dependence of half-life of the corresponding sample, whereas the ratio is how much activity is done by the sample during the decay.