Problem 35
Question
A sample of \({ }^{210}\) Po initially weighed 2.000 grams. After 25 days, 0.125 gram of \({ }^{210}\) Po remained, the rest of the sample having decayed to the stable \({ }^{206} \mathrm{~Pb}\) isotope. Calculate the half-life of \(^{210}\) Po and the mass of \(^{206} \mathrm{~Pb}\) formed.
Step-by-Step Solution
Verified Answer
The half-life of \(^{210}Po\) is approximately 6.25 days, and 1.875 grams of \(^{206}Pb\) are formed.
1Step 1: Define the Decay Formula
The decay of a substance over time can be described by the formula \( N(t) = N_0 \times e^{-\lambda t} \), where \( N(t) \) is the remaining amount at time \( t \), \( N_0 \) is the initial amount, and \( \lambda \) is the decay constant.
2Step 2: Relate Decay Constant to Half-Life
Half-life \( T_{1/2} \) is related to the decay constant \( \lambda \) by the equation \( T_{1/2} = \frac{\ln(2)}{\lambda} \). Our goal is to find \( \lambda \) to calculate the half-life.
3Step 3: Express Given Information Mathematically
Initially, the sample is 2.000 grams, and after 25 days, it is 0.125 grams. So, \( N_0 = 2.000 \) and \( N(25) = 0.125 \). Substitute these values into the decay formula: \( 0.125 = 2.000 \times e^{-25 \lambda} \).
4Step 4: Solve for Decay Constant (λ)
Rearrange the equation: \( e^{-25 \lambda} = \frac{0.125}{2.000} = 0.0625 \). Take the natural logarithm of both sides to solve for \( \lambda \): \( \ln(e^{-25 \lambda}) = \ln(0.0625) \). This simplifies to \( -25 \lambda = \ln(0.0625) \), so \( \lambda = -\frac{\ln(0.0625)}{25} \approx 0.11096 \).
5Step 5: Calculate the Half-Life
Use the relationship \( T_{1/2} = \frac{\ln(2)}{\lambda} \) to calculate the half-life. Substitute \( \lambda \approx 0.11096 \) to find \( T_{1/2} = \frac{\ln(2)}{0.11096} \approx 6.25 \) days.
6Step 6: Determine Mass of 206Pb Formed
Since initially 2.000 grams of \(^{210}Po\) were present and 0.125 grams remain, the remaining material is the \(^{206} Pb\) formed. Thus, the mass of \(^{206}Pb\) is \( 2.000 - 0.125 = 1.875 \) grams.
Key Concepts
Decay ConstantRadioactive DecayLead Isotopes
Decay Constant
The decay constant, denoted as \( \lambda \), is a fundamental parameter in the study of radioactive decay. It represents the probability per unit time that a single nucleus will decay. The decay constant is crucial in determining how fast a radioactive substance will decay over a given period.
Understanding the decay constant allows us to calculate the half-life of a substance, which is the time it takes for half of the radioactive atoms in a sample to decay. If the decay constant is large, this means the substance decays quickly. Conversely, a smaller decay constant indicates a slower decay process.
In our decay formula, \( N(t) = N_0 \times e^{-\lambda t} \), the constant \( \lambda \) decides the rate of decay. When solving decay problems, we often first need to find \( \lambda \) before calculating the half-life or other important properties of the radioactive material. This makes \( \lambda \) indispensable in practical decay calculations.
Understanding the decay constant allows us to calculate the half-life of a substance, which is the time it takes for half of the radioactive atoms in a sample to decay. If the decay constant is large, this means the substance decays quickly. Conversely, a smaller decay constant indicates a slower decay process.
In our decay formula, \( N(t) = N_0 \times e^{-\lambda t} \), the constant \( \lambda \) decides the rate of decay. When solving decay problems, we often first need to find \( \lambda \) before calculating the half-life or other important properties of the radioactive material. This makes \( \lambda \) indispensable in practical decay calculations.
Radioactive Decay
Radioactive decay is the natural process by which an unstable atomic nucleus loses energy by emitting radiation. This process can result in the transformation of one element into another, as seen with the decay from the radioactive isotope polonium-210 ( ^{210} Po) to lead-206 ( ^{206} Pb).
The decay process follows a predictable pattern that can be quantified using mathematical models like the exponential decay formula. This pattern is an exponential decrease over time, where the amount of substance remaining after a given time \( t \) is calculated as \( N(t) = N_0 \times e^{-\lambda t} \). This formula helps predict how quickly a radioactive substance will diminish over time, based on the initial amount \( N_0 \) and the decay constant \( \lambda \).
In our context, understanding radioactive decay is essential for estimating not just how long the process takes but also the resultant amount and composition of the transformed material, like lead isotopes.
The decay process follows a predictable pattern that can be quantified using mathematical models like the exponential decay formula. This pattern is an exponential decrease over time, where the amount of substance remaining after a given time \( t \) is calculated as \( N(t) = N_0 \times e^{-\lambda t} \). This formula helps predict how quickly a radioactive substance will diminish over time, based on the initial amount \( N_0 \) and the decay constant \( \lambda \).
In our context, understanding radioactive decay is essential for estimating not just how long the process takes but also the resultant amount and composition of the transformed material, like lead isotopes.
Lead Isotopes
Lead isotopes, such as
^{206}
Pb, are the end products of radioactive decay chains and possess stable nuclei. Lead has four stable isotopes:
^{204}
Pb,
^{206}
Pb,
^{207}
Pb, and
^{208}
Pb. These isotopes are essential in various scientific fields, including geology and archaeology, where they help in dating rocks and artifacts.
When a radioactive substance like polonium-210 decays, it turns into a stable isotope of lead over time, as in our example where ^{210} Po decays to ^{206} Pb. This transformation is a critical component of understanding elemental stability and the net amount of decay products formed.
In practical scenarios, calculating how much of a lead isotope results from radioactive decay can provide insights into both the original decay material and its journey toward isotope stability, thus aiding in a multitude of research applications.
When a radioactive substance like polonium-210 decays, it turns into a stable isotope of lead over time, as in our example where ^{210} Po decays to ^{206} Pb. This transformation is a critical component of understanding elemental stability and the net amount of decay products formed.
In practical scenarios, calculating how much of a lead isotope results from radioactive decay can provide insights into both the original decay material and its journey toward isotope stability, thus aiding in a multitude of research applications.
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