Chapter 20: Problem 74
A sample of silver-112 gives a reading of 875 counts per hour on a radiation counter. After \(0.25 \mathrm{~h}\), the sample gives a reading of 400 counts per hour. What is the half-life (in hours) of silver-112?
Short Answer
Expert verified
The half-life of silver-112 is approximately 0.221 hours.
Step by step solution
01
Write Down the Exponential Decay Formula
The number of radioactive counts follows an exponential decay formula given by \( N(t) = N_0 e^{-kt} \), where \( N(t) \) is the number of counts at time \( t \), \( N_0 \) is the initial number of counts, and \( k \) is the decay constant. We need to find \( k \) first.
02
Insert the Given Values into the Decay Formula
From the problem statement, at \( t=0 \), \( N_0 = 875 \) counts per hour. At \( t=0.25 \text{ hours} \), \( N(0.25) = 400 \) counts per hour. Using the formula \( 400 = 875 e^{-0.25k} \), we can solve for \( k \).
03
Solve for the Decay Constant \( k \)
Rearrange the equation: \( e^{-0.25k} = \frac{400}{875} \). Taking the natural logarithm of both sides gives \( -0.25k = \ln\left(\frac{400}{875}\right) \). Calculate \( k \) by dividing by \(-0.25\): \( k = -4 \ln\left(\frac{400}{875}\right) \).
04
Calculate the Natural Logarithm
Calculate \( \frac{400}{875} \approx 0.4571 \) and then find \( \ln(0.4571) \approx -0.7828 \). Therefore, \( k = -4(-0.7828) = 3.1312 \).
05
Use the Decay Constant to Find the Half-Life
The half-life \( T \) is related to the decay constant \( k \) by the equation \( T = \frac{\ln(2)}{k} \). Substitute the calculated \( k \) into this equation: \( T = \frac{0.693}{3.1312} \).
06
Calculate the Half-Life
Perform the final calculation: \( T \approx \frac{0.693}{3.1312} \approx 0.221 \text{ hours} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Decay Formula
Radioactive decay follows a pattern called exponential decay. This mathematical concept describes how quantities decrease rapidly at first, then more slowly over time. The general formula for exponential decay is given by:\[ N(t) = N_0 \cdot e^{-kt} \]where:
- \(N(t)\) is the number of counts at time \(t\).
- \(N_0\) is the initial number of counts.
- \(k\) is the decay constant.
- \(e\) is the base of the natural logarithm, approximately equal to 2.718.
Half-Life Calculation
In radioactive decay, the half-life refers to the time it takes for half of the radioactive substance to decay. To compute the half-life, use the decay constant \(k\) derived from the exponential formula. The half-life \(T\) is calculated using this formula:\[ T = \frac{\ln(2)}{k} \]Key elements for understanding half-life calculation:
- \(\ln(2)\) represents the natural logarithm of 2, which is approximately 0.693.
- The half-life provides a measure of how quickly a radioactive material undergoes decay.
- Half-life is often practical in fields like archeological dating and nuclear medicine.
Decay Constant
The decay constant \(k\) is a crucial parameter in the exponential decay formula. It signifies the rate at which a radioactive substance decays. Here's how we can find \(k\):- Start with the rearranged formula: \( e^{-kt} = \frac{N(t)}{N_0} \)- Use the natural logarithm to solve: \(-kt = \ln\left(\frac{N(t)}{N_0}\right)\)- Solve for \(k\) by isolating it: \(k = -\frac{\ln\left(\frac{N(t)}{N_0}\right)}{t}\)This formula demonstrates how the decay constant is computed using the initial and observed quantities over a set period. Understanding \(k\) is vital for predicting decay over time and forms the basis of more complex tasks like calculating half-life and determining the age of samples in various scientific fields. Calculating \(k\) lets us quantify decay processes accurately.