91Ó°ÊÓ

To find the decay constant, we need to use the formula for the half-life of Carbon-14: \(\lambda = \dfrac{ln(2)}{T_{1/2}}\) The half-life of Carbon-14, \(T_{1/2}\), is approximately 5730 years. We first need to convert this to seconds to use in our formula: 1 year = 365.25 days \times 24 h/day \times 3600 s/h \(T_{1/2} = 5730\; \text{years}\; \cdot 365.25\; \frac{\text{days}}{\text{year}}\; \cdot 24\; \frac{\text{h}}{\text{day}}\; \cdot 3600\; \frac{\text{s}}{\text{h}} = 1.808 \times 10^{11} \:\text{s}\) Now we can plug this half-life value into the formula to find the decay constant: \(\lambda = \dfrac{ln(2)}{1.808 \times 10^{11}\; \text{s}} = 3.842 \times 10^{-12}\; \text{s}^{-1}\) So the decay constant for Carbon-14 is \(3.842 \times 10^{-12} \;\text{s}^{-1}\).

We are given that one mole of carbon has a mass of 12.011 grams and the relative abundance of Carbon-14 is \(1.3 \times 10^{-12}\). To find the number of Carbon-14 atoms in 1 gram of carbon, we first need to determine how many moles are present in 1 gram of carbon: \(\text{moles} = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{1.00\; \text{g}}{12.011\; \text{g/mol}} = 0.0833\; \text{mol}\) Now, we can find the number of Carbon-12 atoms in these moles: \(\text{atoms} = \text{moles} \cdot \text{Avogadro's number} = 0.0833\; \text{mol} \cdot 6.022 \times 10^{23}\; \frac{\text{atoms}}{\text{mol}} = 5.01 \times 10^{22}\; \text{atoms}\) Finally, we can find the number of Carbon-14 atoms by multiplying the number of Carbon-12 atoms by the relative abundance: \(\text{Carbon-14 atoms} = 5.01 \times 10^{22}\; \text{atoms} \cdot 1.3 \times 10^{-12} = 6.51 \times 10^{10}\; \text{atoms}\) There are \(6.51 \times 10^{10}\) Carbon-14 atoms in 1 gram of carbon.

Now we have the decay constant \(\lambda\) and the number of Carbon-14 atoms in 1 gram of carbon. We can use these values to calculate the activity using the formula: \(\text{Activity} = \lambda \cdot \text{number of atoms} = 3.842 \times 10^{-12} \;\text{s}^{-1} \cdot 6.51 \times 10^{10} \;\text{atoms} = 0.25\;\text{Bq}\) The Carbon-14 activity per gram of carbon in a living sample is 0.25 Bq, which confirms the statement given in the problem.