Problem 34
Question
Use this scenario: A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day. The half-life of Radium-226 is 1590 years. What is the annual decay rate? Express the decimal result to four significant digits and the percentage to two significant digits.
Step-by-Step Solution
Verified Answer
The annual decay rate of Radium-226 is 0.04367 (decimal) or 0.044% (percentage).
1Step 1: Understand the Half-Life Concept
The half-life of a substance is the time it takes for half of the substance to decay. Here we are given the half-life of Radium-226 as 1590 years.
2Step 2: Use the Half-Life Formula
The formula for half-life in exponential decay is given by: \( N(t) = N_0 \times (\frac{1}{2})^{\frac{t}{T_{1/2}}} \), where \( N(t) \) is the remaining quantity of a substance, \( N_0 \) is the initial quantity, \( t \) is the time elapsed, and \( T_{1/2} \) is the half-life.
3Step 3: Find the Decay Constant
The decay constant \( \lambda \) can be found using the formula: \( \lambda = \frac{\ln(2)}{T_{1/2}} \), where \( T_{1/2} = 1590 \) years. Calculate \( \lambda \): \[ \lambda = \frac{\ln(2)}{1590} \approx 0.0004368 \].
4Step 4: Calculate the Annual Decay Rate
The annual decay rate can be calculated using the formula: \( r = 1 - e^{-\lambda} \). Since we want the annual rate, substitute \( t = 1 \): \[ r = 1 - e^{-0.0004368} \].
5Step 5: Compute the Annual Decay Rate
Calculate \( r \): \[ r = 1 - e^{-0.0004368} \approx 0.0004367 \]. This is the decimal result, representing the decay rate over one year.
6Step 6: Convert to Percent Form
To express the annual decay rate as a percentage: \( 0.0004367 \times 100 \approx 0.04367 \% \). Rounding to two significant figures, we get 0.044\%.
Key Concepts
Half-LifeDecay ConstantAnnual Decay RateRadium-226
Half-Life
The concept of half-life is essential in understanding exponential decay, such as in radioactive substances. It is defined as the time required for half of a given substance to decay. Imagine it as a gradual decline where every half-life period, your quantity of interest is reduced to 50%.
This measure is independent of the amount of material present; only the time to reach half of whatever is there matters. For Radium-226, we're dealing with a half-life of 1590 years, meaning that if you start with a certain amount today, in 1590 years, only half of that amount will remain.
This measure is independent of the amount of material present; only the time to reach half of whatever is there matters. For Radium-226, we're dealing with a half-life of 1590 years, meaning that if you start with a certain amount today, in 1590 years, only half of that amount will remain.
- Useful for Calculations: Helps predict how quickly a material decreases over time.
- Applications: Dating ancient objects, medical diagnoses, and treatment planning.
Decay Constant
The decay constant, often represented as \( \lambda \), is another crucial component when dealing with exponential decay. It provides a standardized measure of the rate at which the decay occurs. In essence, it's the probability per unit time for an atom to decay.
You can calculate the decay constant for Radium-226 using the formula:\[\lambda = \frac{\ln(2)}{T_{1/2}}\]where \( T_{1/2} = 1590 \) years for Radium-226. This results in a decay constant of approximately 0.0004368.
You can calculate the decay constant for Radium-226 using the formula:\[\lambda = \frac{\ln(2)}{T_{1/2}}\]where \( T_{1/2} = 1590 \) years for Radium-226. This results in a decay constant of approximately 0.0004368.
- Relationship with Half-life: Shows inverse relation; higher \( \lambda \) leads to shorter half-life.
- Significance: Describes how fast a substance undergoes decay, useful in diverse scientific scenarios.
Annual Decay Rate
The annual decay rate expresses how much of a substance decays over the course of one year. Unlike half-life, it gives insight into the percentage or fraction of the material that will have decayed after a full year.
To find the annual decay rate, use the formula:\[r = 1 - e^{-\lambda} \]where for Radium-226, \( \lambda = 0.0004368 \). Plug this into the formula to get:\[r = 1 - e^{-0.0004368} \approx 0.0004367\]This indicates an annual decay of around 0.044% when converted to a percentage.
To find the annual decay rate, use the formula:\[r = 1 - e^{-\lambda} \]where for Radium-226, \( \lambda = 0.0004368 \). Plug this into the formula to get:\[r = 1 - e^{-0.0004368} \approx 0.0004367\]This indicates an annual decay of around 0.044% when converted to a percentage.
- Purpose: Offers a complementary view to half-life by focusing on a fixed time frame.
- Usage: Often employed in environmental studies and health physics to assess potential exposure risks.
Radium-226
Radium-226 is a naturally occurring radioactive element found in uranium ores. It is a key player when discussing radioactive decay, known for its long half-life of about 1590 years.
This isotope undergoes decay to produce radon gas, which poses health risks. Due to its high energy radiation, Radium-226 was historically used in luminous paints and cancer treatments. However, its use is limited today due to safety concerns.
This isotope undergoes decay to produce radon gas, which poses health risks. Due to its high energy radiation, Radium-226 was historically used in luminous paints and cancer treatments. However, its use is limited today due to safety concerns.
- Importance in Science: Helps understand radioactive decay processes & harness energy.
- Safety Challenges: Requires careful handling and containment to prevent radon exposure.
Other exercises in this chapter
Problem 33
For the following exercises, refer to Table 8. $$\begin{array}{ccccccc}{x} & {1} & {2} & {3} & {4} & {5} & {6} \\ {f(x)} & {555} & {383} & {307} & {210} & {158}
View solution Problem 33
For the following exercises, solve for \(x\) by converting the logarithmic equation to exponential form. $$\log _{6}(x)=-3$$
View solution Problem 34
For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to f
View solution Problem 34
For the following exercises, use the definition of a logarithm to solve the equation. $$ 2 \log (8 n+4)+6=10 $$
View solution