Question

An archeological specimen containing 7.5 g of carbon has an activity of 1.1 Bq. How old (in years) is the specimen? Carbon-14 has a half-life of 5730 years, and an activity of 0.23 Bq per gram of carbon in a living organism.

Answer 1

A(t) = Ao*exp(-λ*t)
Solving for t:
t = - (1/λ)*ln(A(t)/Ao)

where
A(t) is the activity of a radioactive sample at time t = 1.1/7.5 = 0.147 Bq/(gm C)
Ao is the initial activity (at time t = 0) = 0.23 Bq/(gm C)
λ is the decay constant. λ is related to the halflife (T) by λ = ln(2)/T

For 14-C, T = 5730 yr, so λ = ln(2)/(5730) = 1.21*10^-4 yr^-1

Substituting Values,

t = - 1/(1.21*10^-4)*ln(0.147/0.23)
t = 0.4476 *   1/(1.21*10^-4)
t = 3699.2 years