Problem 99
Question
The formula \(\log _{10}(1-k)=\frac{-.3}{H}\) models the relationship between the half-life \(H\) of a radioactive material and its rate of decay \(k\). Find the rate of decay of the iodine isotope \(I-131\) if its half-life is 8 days. Round to four decimal places.
Step-by-Step Solution
Verified Answer
The rate of decay \( k \) is approximately 0.0856.
1Step 1: Set up the equation
Given the equation \( \log_{10} (1-k) = \frac{-0.3}{H} \), we need to find the rate of decay \( k \) when the half-life \( H \) is 8 days. Substitute 8 for \( H \) in the equation: \( \log_{10} (1-k) = \frac{-0.3}{8} \).
2Step 2: Simplify the right side of the equation
Calculate \( \frac{-0.3}{8} \) which simplifies to \( -0.0375 \). So the equation becomes \( \log_{10} (1-k) = -0.0375 \).
3Step 3: Solve for (1-k) using the inverse log function
Apply the inverse of the log function to both sides to solve for \( 1-k \). This means \( 1-k = 10^{-0.0375} \).
4Step 4: Calculate 10 to the power of -0.0375
Use a calculator to find \( 10^{-0.0375} \). This calculates to approximately \( 0.9144 \). So, \( 1-k = 0.9144 \).
5Step 5: Solve for k
To find \( k \), subtract \( 0.9144 \) from 1: \( k = 1 - 0.9144 \).
6Step 6: Calculate k
Perform the subtraction: \( k = 0.0856 \). Therefore, the rate of decay \( k \) is approximately 0.0856 when rounded to four decimal places.
Key Concepts
Half-lifeRadioactive IsotopeLogarithmic Functions
Half-life
The concept of half-life is crucial when studying radioactive substances. Half-life refers to the time required for half of a given sample of radioactive nuclei to decay. It is a constant value specific to each radioactive isotope, such as iodine isotope \(I-131\), and remains the same regardless of the initial quantity of the substance. This means that no matter how much you start with, half of it will have decayed after each half-life period.
For instance, if you start with a sample containing 100 grams of \(I-131\), after its half-life of 8 days, only 50 grams of the isotope would remain. After another 8 days, just 25 grams would be left. This exponential decay pattern continues indefinitely. The concept of half-life is key to predicting how radioactive materials will behave over time and is widely used in various fields such as chemistry, radiology, and environmental science.
For instance, if you start with a sample containing 100 grams of \(I-131\), after its half-life of 8 days, only 50 grams of the isotope would remain. After another 8 days, just 25 grams would be left. This exponential decay pattern continues indefinitely. The concept of half-life is key to predicting how radioactive materials will behave over time and is widely used in various fields such as chemistry, radiology, and environmental science.
Radioactive Isotope
A radioactive isotope, or radioisotope, is an unstable atom that undergoes radioactive decay, releasing radiation in the form of particles or electromagnetic waves. This process transforms the original atom into a different element or a different isotope of the same element.
Radioactive isotopes are naturally occurring but can also be artificially created. They have a wide range of applications in medicine, industry, and research. For example:
Understanding radioactive isotopes is essential for safely handling and making the most of their benefits, while also mitigating risks associated with their radioactivity.
- In medicine, radioisotopes are used in imaging and cancer treatments, such as iodine-131 used for treating thyroid cancer.
- In industry, they help in detecting leaks and assessing the thickness of materials.
- In research, they are crucial for techniques like carbon dating.
Understanding radioactive isotopes is essential for safely handling and making the most of their benefits, while also mitigating risks associated with their radioactivity.
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions, and they play a vital role in equations involving exponential growth or decay. In the context of radioactive decay, the logarithmic function helps describe the relationship between the rate of decay and time, particularly in calculating the half-life of a substance.
The logarithmic equation \( \log_{10}(1-k) = \frac{-0.3}{H} \) from the exercise represents such a relationship, where \(k\) stands for the rate of decay and \(H\) the half-life of the radioactive material.
This form of equation allows us to solve for \(k\), demonstrating how logarithms can unravel exponential relationships and giving us practical insights into how substances decay. They also allow for complex calculations to be broken down into more manageable steps, especially when dealing with very large or very small numbers, which are common in nuclear physics.
The logarithmic equation \( \log_{10}(1-k) = \frac{-0.3}{H} \) from the exercise represents such a relationship, where \(k\) stands for the rate of decay and \(H\) the half-life of the radioactive material.
This form of equation allows us to solve for \(k\), demonstrating how logarithms can unravel exponential relationships and giving us practical insights into how substances decay. They also allow for complex calculations to be broken down into more manageable steps, especially when dealing with very large or very small numbers, which are common in nuclear physics.
Other exercises in this chapter
Problem 97
Explain why the graph of the function \(y=\log _{b} x\) contains the point (1,0) no matter what \(b\) is.
View solution Problem 98
\(\log _{3} 10\) is between which two integers? Explain your answer.
View solution Problem 100
The formula \(\mathrm{pH}=-\log _{10}\left(\mathrm{H}^{+}\right)\) gives the \(\mathrm{pH}\) for a liquid, where \(\mathrm{H}^{+}\) stands for the concentration
View solution Problem 96
Graph each function and its inverse on the same set of axes. $$ y=\left(\frac{1}{2}\right)^{x} ; y=\log _{1 / 2} x $$
View solution