Question 25.4: Applying the Integrated Rate Law for Radioactive Decay: Radi......

Analyze
The solution requires three equations: (25.11), (25.12), and (25.13).

rate of decay α N   and    rate of decay = A = λN       (25.11)

\ln (\frac{N_{t}}{N_{0}}) = −λt        (25.12)

t_{1/2} = \frac{\ln (2)}{λ}         (25.13)

Solve
Equation (25.13) is used to determine the decay constant.

λ = \frac{0.693}{5730  a} = 1.21 × 10^{−4}  a^{−1}

Next, equation (25.11) relates to the actual number of atoms: N at t = 0 (the time when the ^{14}C equilibrium was destroyed) and N_{t} at time t (the present time). As discussed on page 1182, the activity just before the ^{14}C equilibrium was destroyed was 15 dis min^{−1}  g^{−1}; at the time of the measurement, it is 10 dis min^{−1}  g^{−1}. The corresponding numbers of atoms are equal to these activities divided by λ.

N_{0} = A_{0}/λ = 15/λ     and      N_{t} = A_{t}/λ = 10/λ

Finally, we substitute into equation (25.12).

\ln \frac{N_{t}}{N_{0}} = \ln \frac{10/λ}{15/λ} = \ln \frac{10}{15} = −(1.21 × 10^{−4}  a^{−1})t

−0.41 = −(1.21 × 10^{−4}  a^{−1})t

t = \frac{0.41}{1.21  ×  10^{−4}  a^{−1}} = 3.4 × 10³ a

Assess
In the previous example, we observed that the activity depends on the amount of material. Therefore, the results of radiocarbon dating depend on knowing the activity at the time the equilibrium between ^{14}C and the other nonradioactive carbon isotopes ceases. If at the time equilibrium was destroyed the activity was 14 dis min^{−1}  g^{−1}, 14 dis we would have determined the object to be 2.8 × 10³ years old, which is approximately a 17% error.