Question

The half-time for radioactive decay of C-14 is 5700 years. An archaeological artifact contained wood that had only 75% of the C-14 found in living tree. Estimate the age of the sample. (log 2 = 0.3010, log 3 = 0.4771)​

Answer 1

Answer:

Firstly, let us know that we need to calculate the age of the sample. The age of sample can be found by t=2.303klog[R]0[R], here k symbolises the decay constant and t is the time i.e. age and [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, t1/2 is the representation of half-life.

Now substitute the value of t1/2 then k=0.6935730 years−1.

Now, we already know t= 2.303klog[R]0[R], [R]0[R] is 10080, we are given 80% of decay, then t= 2.3030.6935730log 10080= 1845 years (approximately).

Therefore, the age of the sample is 1845 years. The correct option is B.