Chapter 15: Problem 15
Iodine-123 is important for medical imaging studies and follows first-order decay kinetics. A \(15-\mu \mathrm{g}\) sample of I-123 has decayed to \(7.5 \mu \mathrm{g}\) after 13 hours. After how much time will it decay to only \(1.5 \mu \mathrm{g}\) ?
Short Answer
Expert verified
It will take about 43.3 hours for the sample to decay to 1.5 µg.
Step by step solution
01
Understand First-Order Kinetics
First-order decay kinetics means the rate of decay depends linearly on the concentration of the undecayed substance. The formula is given by \( N(t) = N_0 e^{-kt} \), where \( N(t) \) is the remaining quantity, \( N_0 \) is the initial quantity, \( k \) is the decay constant, and \( t \) is time.
02
Find the Decay Constant
Use the first-order decay formula with the given data to find the decay constant, \( k \). You know \( N_0 = 15 \, \mu\text{g} \) and \( N(t) = 7.5 \, \mu\text{g} \) when \( t = 13 \, \text{hours} \): \[7.5 = 15 e^{-13k}\]Rearrange and solve for \( k \):\[e^{-13k} = \frac{7.5}{15} = 0.5\]Take the natural logarithm of both sides:\[-13k = \ln(0.5)\]Solve for \( k \):\[k = -\frac{\ln(0.5)}{13}\]
03
Calculate Time for Further Decay
Now, use the decay constant \( k \) to find the time \( t \) when the quantity decays to \( 1.5 \, \mu\text{g} \):\[1.5 = 15 e^{-kt}\]Rearrange it and solve for \( t \):\[e^{-kt} = \frac{1.5}{15} = 0.1\]Take the natural logarithm of both sides:\[-kt = \ln(0.1)\]Use the previously calculated \( k \):\[t = -\frac{\ln(0.1)}{k}\]Substitute \( k = -\frac{\ln(0.5)}{13} \) into the equation for \( t \):\[t = -\frac{\ln(0.1)}{-\frac{\ln(0.5)}{13}}\]Simplify to find \( t \).
04
Solve for Time t
Using the values of natural logarithms:\[t = \frac{13 \times \ln(0.1)}{\ln(0.5)}\]Calculate this to find:\[t \approx 43.3 \, \text{hours}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Decay Constant
In the world of first-order kinetics, the decay constant, often represented by the letter \( k \), is a crucial parameter. It indicates how quickly a substance undergoes radioactive decay. The decay constant has the unit of reciprocal time, such as \( ext{hour}^{-1} \) or \( ext{second}^{-1} \), depending on the time unit used in the calculations. To find the decay constant, we use the first-order kinetics equation:- \( N(t) = N_0 e^{-kt} \), where: - \( N(t) \) is the quantity remaining after time \( t \), - \( N_0 \) is the initial amount, - \( e \) is the base of the natural logarithm, - \( k \) is the decay constant, and - \( t \) is time.By substituting known values into this equation, you can solve for \( k \). For example, when a 15 µg sample decays to 7.5 µg in 13 hours, you can rearrange the equation to find that \( k = -\frac{\ln(0.5)}{13} \). This value reflects the rate at which the sample of Iodine-123 decays over time. Understanding \( k \) helps predict how much of the substance will remain after any given time.
Half-Life
Half-life, often denoted by \( t_{1/2} \), is the time required for a radioactive substance to reduce to half its initial quantity. It's a practical way to understand the speed of a reaction, particularly in first-order kinetics like radioactive decay.For first-order reactions, the half-life is related to the decay constant by the equation:- \( t_{1/2} = \frac{\ln(2)}{k} \)This equation shows:- A larger decay constant \( k \) means a shorter half-life, indicating faster decay.- A smaller \( k \) leads to a longer half-life, suggesting slower decay.By understanding and calculating the half-life, you can quickly estimate how long a sample will take to reduce to specific amounts. This property is particularly useful in fields like medicine, where materials with specific half-lives are used for diagnostic purposes, ensuring they last just long enough to be effective without remaining in the body for extended periods.
Natural Logarithm
The natural logarithm is a mathematical function denoted as \( \ln \), which plays a crucial role when dealing with exponential functions such as those found in first-order reactions. The natural logarithm arises naturally when solving equations of the form \( e^x = a \) for \( x \). It is the inverse of the exponential function with the base \( e \).In the context of decay calculations, the natural logarithm helps solve for time \( t \) or the decay constant \( k \). For example, if you rearrange the decay equation:- \( e^{-kt} = \frac{1.5}{15} \), You'd take the natural logarithm of both sides:- \( -kt = \ln(0.1) \).This allows you to straightforwardly isolate variables like \( k \) or \( t \). The properties of logarithms such as \( \ln(ab) = \ln(a) + \ln(b) \) and \( \ln(a^b) = b \ln(a) \) make calculations more manageable. Understanding these can help simplify complex exponential problems and make them more approachable.