Question

Living matter has an activity of 260 Bq/kg due to carbon – 14. If a sample of wood from a
burial site has an activity of 155 Bq/kg, estimate the age of the site. (Assume the half – life
of carbon – 14 is 5730 years).

Answer 1

 The activity of the wood decreases as:

$$A_t=A_0(\frac{1}{2})^\frac{t}{h}$$ $$T_\frac{1}{2}=5730$$

$$A_0=260bq/kg$$

A=155dq-Activity at the time T

Therefore;

$$A=0_0(−\frac{T}{T}\frac{​1}{2}→T=T_\frac{1}{2}log_2[A_\frac{0}{A}]$$

$$5730\times log_2[\frac{260}{155}=4276years$$

Therefore the age of the site is 4276 years.