Question

The half life for the decay of carbon-14 is 5.73 times 10^3 years.
Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of wood from an archeological dig is measured to be 77.The activity in a similar-sized sample of fresh wood is measured to be 85.Calculate the age of the artifact. Round your answer to 2 significant digits.

Answer 1

Answer:

790 years

Explanation:

Given that;

0.693/t1/2 = 2.303/t log [A]o/[A]

So;

t1/2 =half life of carbon-14

t= age of the sample

[A]o= activity of the living sampoke

[A] = activity at time t

0.693/5.73 ×10^3 = 2.303/t log 85/77

1.21 × 10^-4 = 2.303/t log 1.1

1.21 × 10^-4 = 0.0953/t

t= 0.0953/1.21 × 10^-4

t= 790 years (to 2sf)