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

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

\lambda=\frac{0.693}{5730  y}=1.21 \times 10^{-4}  y ^{-1}

Next, equation (25.11) relates to the actual number of atoms: N at t = 0 (the time when the _{}^{14}\text{C} equilibrium was destroyed) and N_t at time t (the present time). As discussed on page 1123, the activity just before the _{}^{14}\text{C} equilibrium was destroyed was 15 dis \text{min}^{-1} \text{ g}^{-1}; at the time of the measurement, it is 10 dis \text{min}^{-1} \text{ g}^{-1}. The corresponding numbers of atoms are equal to these activities divided by λ.

N_0=A_0 / \lambda=15 / \lambda \quad \text {  and  } \quad N_t=A_t / \lambda=10 / \lambda

Finally, we substitute into equation (25.12).

\ln \frac{N_t}{N_0}=\ln \frac{10 / \lambda}{15 / \lambda}=\ln \frac{10}{15}=-\left(1.21 \times 10^{-4}  y ^{-1}\right) t

-0.41=-\left(1.21 \times 10^{-4}  y^{-1}\right) t

t=\frac{0.41}{1.21  \times  10^{-4}  y ^{-1}}=3.4 \times 10^3  y

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}\text{C} and the other nonradioactive carbon isotopes ceases. If at the time equilibrium was destroyed the activity was 14 dis \text{min}^{-1} \text{ g}^{-1}, we would have determined the object to be 2.8 × 10³ years old, which is approximately a 17% error.