Question 43.4: The Activity of Carbon At time t = 0, a radioactive sample c...
The Activity of Carbon
At time t = 0, a radioactive sample contains 3.50 μg of pure { }_6^{11} C, which has a half-life of 20.4 min.
(A) Determine the number N_0 of nuclei in the sample at t = 0.
(B) What is the activity of the sample initially and after 8.00 h?
(A) Conceptualize The half-life is relatively short, so the number of undecayed nuclei drops rapidly. The molar mass of { }_6^{11} C is approximately 11.0 g/mol.
Categorize We evaluate results using equations developed in this section, so we categorize this example as a substitution problem.
Find the number of moles in 3.50 μg of pure { }_6^{11} C:
n=\frac{3.50 \times 10^{-6} g}{11.0 g/ \text{ mol}}=3.18 \times 10^{-7} \text{ mol}Find the number of undecayed nuclei in this amount of pure { }_6^{11} C :
N_0=\left(3.18 \times 10^{-7} \text{ mol}\right)\left(6.02 \times 10^{23} \text{ nuclei /mol}\right)=1.92 \times 10^{17} \text{ nuclei}(B) Find the initial activity of the sample using Equations 43.7 and 43.8:
R=\left|\frac{d N}{d t}\right|=\lambda N=\lambda N_0 e^{-\lambda t}=R_0 e^{-\lambda t} (43.7)
T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda} (43.8)
\begin{aligned}R_0 & =\lambda N_0=\frac{0.693}{T_{1 / 2}} N_0=\frac{0.693}{20.4 \min}\left(\frac{1 \ min}{60 s}\right)\left(1.92 \times 10^{17}\right) \\& =\left(5.66 \times 10^{-4} s^{-1}\right)\left(1.92 \times 10^{17}\right)=1.09 \times 10^{14} Bq\end{aligned}Use Equation 43.7 to find the activity at t = 8.00 h = 2.88 \times 10^4 s:
R=R_0 e^{-\lambda t}=\left(1.09 \times 10^{14} Bq\right) e^{-\left(5.66 \times 10^{-4} s^{-1}\right)\left(2.88 \times 10^4 s\right)}=8.96 \times 10^6 Bq