Question

Archeologists can determine the age of artifacts made of wood or bone by measuring the concentration of radioactive isotope 14C present in the object. The amount of isotope decreases in a first-order process. If 15.5% of the original amount of 14C is present in a wooden tool at the time of analysis, what is the age of the tool

Answer 1

Answer:

1377.8years

Explanation:

Using

0.693/t1/2 = 2.303/t log Ao/A

Where t1/2= half life

t= time or age of the bone

Ao= amount of radioactive material at time t=0

A= amount of radioactive material at time t

0.693/5670 = 2.303/t log Ao/Ao-0.155Ao

0.693/5670= 2.303/t log Ao/0.845Ao

0.693/5670= 2.303/t log 1/0.845

0.693/5670= 0.1684/t

t= 1377.8years

Answer 2

Answer:

Age of the tool = 15411.75 years

Explanation:

Assume half life of 14C = 5730 years

Calculate rate constant using formula:

$t_{1/2} = \frac{ln 2}{k}$ thus $k = \frac{ln 2}{t_{1/2}} = 1.21E-4$

Assuming initial concentration, Co of 14C = 100%

Final concentration, C of 14C = 15.5%

Use formula:

$ln \frac{C}{Co} = kt$

thus

$ln \frac{100}{15.5} = (1.21E-04) ( t )$

t = 15411.75 years