Question

An archeological specimen containing 9.42 g of carbon has an activity of 1.58 Bq. How old (in years) is the specimen? Do not enter unit.

Answer 1

The 14-C activity in the living organism was about 0.23 Bq/(gm C) (13.5 decays/min/(gm C)). Atmospheric testing of nuclear weapons dramatically increased the 14-C concentration in the atmosphere and biosphere, but because this is an archeological specimen, we'll assume this effect is not relevant. We will also neglect variations in the 14C abundance caused by variations in the production rate or exchange with "dead" carbon in the oceans or other old carbon reservoirs.

The basic decay equation is:

A(t) = Ao*exp(-lamda*t)

Solving for t:

t = -(1/lamda)*ln(A(t)/Ao)

where A(t) is the activity of a radioactive sample at time t (here this is 1.58Bq/(9.42gm) = 0.167 Bq/(gm C)
Ao is the initial activity (at time t = 0) assumed here to be 0.225 Bq/(gm C)
lamda is the decay constant. lamda is related to the halflife (T) by lamda = ln(2)/T

For 14-C, T = 5568 yr, so lamda = ln(2)/(5568yr) = 1.245*10^-4 yr^-1

So:

t = (5568 yr/ln(2))*ln(Ao/A(t))

Plugging in the values for A(t) and Ao;

t = (5568 yr/ln(2))*ln(0.23/0.167) = 2571.22 yr