Chapter 12: Problem 49
What is the probability of a \({ }^{14} \mathrm{C}\) atom in atmospheric \(\mathrm{CO}_{2}\) decaying in your lungs during a single breath? The atmosphere is about \(0.03 \% \mathrm{CO}_{2}\). Assume you take in about 0.5 L of air in each breath and exhale it 3.5 s later.
Short Answer
Expert verified
The probability of a \( ^{14}C \) atom decaying in your lungs during a single breath is approximately 6.41 \times 10^{-5}.
Step by step solution
01
- Calculate the volume of CO2 in one breath
First, determine the volume of \( \text{CO}_2 \) in 0.5 L of air. Since the concentration of \( \text{CO}_2 \) in the atmosphere is 0.03%, you can find it by the formula: \[ V_{ \text{CO}_2 } = 0.5 \text{ L} \times 0.0003 = 0.00015 \text{ L} \]
02
- Determine the number of CO2 molecules in one breath
Using the ideal gas law and Avogadro's number, the number of \( \text{CO}_2 \) molecules in 0.00015 L of CO2 can be calculated. Avogadro's number is \( 6.022 \times 10^{23} \). The ideal gas law states \[ \text{PV} = \text{nRT} \] Assume standard conditions (T = 298 K and P = 1 atm) and note R = 0.0821 Lâ‹…atm/(molâ‹…K). From the ideal gas law, we get the number of moles: \[ n = \frac{ \text{PV} }{\text{RT}} \] \[ n = \frac{ 1 \times 0.00015 }{0.0821 \times 298} = 6.125 \times 10^{-6} \text{ moles} \] Then multiply by Avogadro's number to get the number of molecules: \[ N = 6.125 \times 10^{-6} \times 6.022 \times 10^{23} = 3.68 \times 10^{18} \text{ molecules} \]
03
- Determine the fraction of CO2 molecules that are \( ^{14}C \)
The fraction of \( ^{14}C \)-containing \( \text{CO}_2 \) in the atmosphere is about \( 1.3 \times 10^{-12} \). Therefore, the number of \( ^{14}C \)-containing \(\text{CO}_{2} \) molecules is: \[ N_{^{14}\text{C}} = 3.68 \times 10^{18} \times 1.3 \times 10^{-12} = 4.784 \times 10^6 \]
04
- Calculate the decay probability during one breath
The decay constant \( \text{(λ)} \) for \( ^{14}C \) is given by \[ \text{λ} = \frac{ \text{ln}(2) }{5730 \text{ years}} \] Convert the decay constant into seconds: \[ \text{λ} = \frac{ \text{ln}(2) }{5730 \times 365 \times 24 \times 3600} = 3.83 \times 10^{-12} \text{ s}^{-1} \] Determine the probability of decay during a 3.5-second breath: \[ P = 1 - \text{e}^{-λt} \] \[ P = 1 - \text{e}^{-3.83 \times 10^{-12} \times 3.5} \] For small λt, \[ P ≈ 3.83 \times 10^{-12} \times 3.5 ≈ 1.34 \times 10^{-11} \]
05
- Determine the probability of any \( ^{14}C \) atom decaying in one breath
The total probability of any \( ^{14}C \) atom decaying is the product of the number of \( ^{14}C \) atoms and the probability of one decaying: \[ P_{\text{total}} = 4.784 \times 10^6 \times 1.34 \times 10^{-11} = 6.41 \times 10^{-5} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
carbon-14 decay
Carbon-14 decay is a natural process that helps us date ancient materials. Carbon-14 (or \({}^{14}C\) ) is a radioactive isotope of carbon. It decays into nitrogen-14 over time by releasing a beta particle. This decay process is useful in radiocarbon dating.
When an organism dies, it stops absorbing carbon. The \({}^{14}C\) in its body begins to decay at a known rate which is characterized by its half-life. The half-life of \({}^{14}C\) is about 5730 years, meaning that half of the carbon-14 atoms will have decayed after this time period.
In our problem, we are calculating the probability of a \({}^{14}C\) atom decaying in your lungs during a single breath. This involves understanding the rate at which \({}^{14}C\) decay happens.
When an organism dies, it stops absorbing carbon. The \({}^{14}C\) in its body begins to decay at a known rate which is characterized by its half-life. The half-life of \({}^{14}C\) is about 5730 years, meaning that half of the carbon-14 atoms will have decayed after this time period.
In our problem, we are calculating the probability of a \({}^{14}C\) atom decaying in your lungs during a single breath. This involves understanding the rate at which \({}^{14}C\) decay happens.
radioactive isotopes
Radioactive isotopes are atoms that have an unstable nucleus. This instability causes them to lose energy by emitting radiation in a process called radioactive decay. Each radioactive isotope decays at a characteristic rate which is quantified by its half-life.
In our given scenario, \({}^{14}C\) is a radioactive isotope of carbon. It decays over time, allowing scientists to measure the amount of \({}^{14}C\) remaining in an object to determine its age. The decay constant (λ) of \({}^{14}C\) can be calculated using the relationship: \[ λ = \frac{ \text{ln}(2) }{ \text{half-life} } \] For \({}^{14}C\), this decay constant is very small, showing that \({}^{14}C\) atoms decay very slowly over time.
In the context of the exercise, we're interested in how often these decays might happen within a quick 3.5-second breath.
In our given scenario, \({}^{14}C\) is a radioactive isotope of carbon. It decays over time, allowing scientists to measure the amount of \({}^{14}C\) remaining in an object to determine its age. The decay constant (λ) of \({}^{14}C\) can be calculated using the relationship: \[ λ = \frac{ \text{ln}(2) }{ \text{half-life} } \] For \({}^{14}C\), this decay constant is very small, showing that \({}^{14}C\) atoms decay very slowly over time.
In the context of the exercise, we're interested in how often these decays might happen within a quick 3.5-second breath.
ideal gas law
The ideal gas law is a fundamental equation in chemistry and physics that describes how gases behave. It combines several gas laws into one formula: \[ \text{PV} = \text{nRT} \] Where:
- P is the pressure of the gas,
- V is the volume of the gas,
- n is the number of moles of the gas,
- R is the ideal gas constant (0.0821 Lâ‹…atm/(molâ‹…K)),
- T is the temperature in Kelvin.
probability calculation
Probability calculation is a branch of mathematics that helps predict how likely events are to occur. In our exercise, several probability calculations are needed to find out the chances of a \({}^{14}C\) atom decaying in a short period.
We start with finding the fraction of \( \text{CO}_2 \) molecules that contain \({}^{14}C \). Given the tiny fraction of \({}^{14}C\) in the atmosphere, this part involves multiplication by a very small number.
Next, we use the decay constant \(λ\) to compute the probability that a \({}^{14}C\) atom will decay during a specific time frame (in our problem, one breath taken over 3.5 seconds):
Finally, multiplying the number of \({}^{14}C\) atoms by the single atom decay probability gives the total probability of any \({}^{14}C\) atom decaying during that breath. This combination of small probabilities and large numbers shows how fundamental principles can predict everyday scenarios.
We start with finding the fraction of \( \text{CO}_2 \) molecules that contain \({}^{14}C \). Given the tiny fraction of \({}^{14}C\) in the atmosphere, this part involves multiplication by a very small number.
Next, we use the decay constant \(λ\) to compute the probability that a \({}^{14}C\) atom will decay during a specific time frame (in our problem, one breath taken over 3.5 seconds):
- The formula used is \[ P = 1 - e^{-λt} \] .
- For small values of λt, the probability can be approximated with \ P ≈ λt \
Finally, multiplying the number of \({}^{14}C\) atoms by the single atom decay probability gives the total probability of any \({}^{14}C\) atom decaying during that breath. This combination of small probabilities and large numbers shows how fundamental principles can predict everyday scenarios.