Question

A 12.43 -g fragment of charcoal is to be carbon dated. Measurements show that it has an activity of 105 decays/min. How many years ago did the tree from which the charcoal was produced die? (Hint: The half-life of \({ }_{6}^{14} \mathrm{C}\) is \(5730 \mathrm{yr}\), and the \({ }_{6}^{14} \mathrm{C} /{ }_{6}^{12} \mathrm{C}\) ratio in living organic matter is \(1.20 \cdot 10^{-12}\).)

Short answer

Question: Given that a charcoal fragment has a mass of 12.43 g, an activity of 105 decays/min, and the half-life of Carbon-14 is 5730 years, estimate the time elapsed since the tree from which the charcoal was produced died. Answer: The tree from which the charcoal was produced approximately died 5.84 years ago.

Step-by-step solution

Identify the given information

We are given: - Mass of the charcoal fragment: 12.43 g - Activity of the charcoal fragment: 105 decays/min - Half-life of Carbon-14: 5730 years - Ratio of Carbon-14 to Carbon-12 in living organisms: \(1.20 \times 10^{-12}\)

Calculate the number of moles of carbon in the sample.

First, we must determine the number of moles of carbon in the sample. The molar mass of carbon is 12.01 g/mol, so we can calculate the moles of carbon in the sample using the following formula: Moles of carbon = (mass of the sample) / (molar mass of carbon) Moles of carbon = \(12.43\:\text{g} \div 12.01\:\text{g/mol} = 1.035\:\text{mol}\)

Calculate the initial activity of the sample.

Now, let's calculate the initial activity of the sample when it was part of a living organism. To do this, we will use the given C-14 to C-12 ratio: Initial ratio of C-14 to C-12 = \(1.20 \times 10^{-12}\) Since Carbon-12 is the most abundant isotope and its mass is nearly equal to the molar mass of carbon, we can assume the number of moles of C-12 in the sample is equal to the total moles of carbon: Moles of C-12 = 1.035 mol Now, we can use the ratio to find the initial moles of C-14: Moles of C-14 = (Moles of C-12) × (ratio of C-14 to C-12) Moles of C-14 = \(1.035\:\text{mol} \times 1.20 \times 10^{-12} = 1.242 \times 10^{-12}\:\text{mol}\) The activity of Carbon-14 is the number of decays per minute, which can be calculated using the following formula: Activity = (Number of atoms in the sample) × (Decay constant) The decay constant, λ, can be calculated from the half-life using the following equation: λ = ln(2) / (half-life) λ = \(ln(2) \div 5730\: \text{yr} = 1.210\times 10^{-4}\: \text{yr}^{-1}\) Now, we can find the decay constant per minute: λ (1 minute) = \(1.210\times 10^{-4} \: \text{yr}^{-1} \times \frac{1\: \text{yr}}{525600\: \text{min}} = 2.3024 \times 10^{-10} \: \text{min}^{-1}\) Knowing the decay constant per minute, we can calculate the initial activity: Initial activity = (Moles of C-14) × (Avogadro's number) × (Decay constant per minute) Initial activity = \((1.242 \times 10^{-12}\:\text{mol}) \times (6.022\times 10^{23}\: \text{atoms/mol}) \times (2.3024 \times 10^{-10}\:\text{min}^{-1}) = 172.52\:\text{decays/min}\)

Use the decay formula to find the time elapsed

Now, we can use the decay formula to calculate the time elapsed since the tree died: Final activity = Initial activity × \(e^{-\lambda t}\) Where: - Final activity = 105 decays/min - Initial activity = 172.52 decays/min - λ = decay constant per minute - t = time elapsed We can rearrange the formula for t: \(t = \frac{1}{\lambda} \ln(\frac{\text{Initial activity}}{\text{Final activity}})\) Plugging in the values, we get: \(t = \frac{1}{2.3024 \times 10^{-10}\: \text{min}^{-1}} \ln (\frac{172.52 \: \text{decays/min}}{105 \: \text{decays/min}}) = 3067680\:\text{min}\) Now, we convert minutes to years: Time elapsed (years) = \(\frac{3067680\:\text{min}}{525600\:\text{min/yr}} = 5.84\:\text{yr}\) Therefore, the tree from which the charcoal was produced died approximately 5.84 years ago.