Problem 438 - Calculus Volume 1

Question

True or False? Justify your answer with a proof or a counterexample. If the half-life of seaborgium-266 is 360 \(\mathrm{ms}\) , then \(k=(\ln (2) / 360\)

Step-by-Step Solution

Answer

True; the decay constant \( k \) is indeed \( \frac{\ln(2)}{360} \).

Step 1: Understand Half-life

The half-life of a radioactive element is the time it takes for half of a given amount of the substance to decay. We use the decay formula which involves an exponential decay constant to describe this process.

Step 2: Exponential Decay Formula

The formula for radioactive decay is given as \[ N(t) = N_0 e^{-kt} \]where \( N(t) \) is the amount of substance at time \( t \), \( N_0 \) is the initial amount, and \( k \) is the decay constant.

Step 3: Derive the Half-life Equation

At half-life, \( N(t) = N_0 / 2 \). Plugging this into the decay formula gives\[ \frac{N_0}{2} = N_0 e^{-kt_{1/2}} \]which simplifies to \[ e^{-kt_{1/2}} = \frac{1}{2} \].

Step 4: Solve for Decay Constant \(k\)

We take the natural logarithm of both sides: \[ -kt_{1/2} = \ln \left( \frac{1}{2} \right) \]and solve for \( k \):\[ k = -\frac{\ln \left( \frac{1}{2} \right)}{t_{1/2}} = \frac{\ln(2)}{t_{1/2}} \].

Step 5: Substitute Given Half-Life

Given that \( t_{1/2} = 360 \text{ ms} \) for seaborgium-266, substitute into the formula:\[ k = \frac{\ln(2)}{360} \].

Step 6: Conclusion

Therefore, the statement is true: \( k = \frac{\ln(2)}{360} \) is correct given the definition of half-life and decay constant.

Key Concepts

Radioactive Decay

Radioactive decay refers to the process by which an unstable atomic nucleus loses energy by emitting radiation, such as alpha particles, beta particles, or gamma rays. This transformation results in the conversion of an element into another substance over time. The key aspect to remember here is that radioactive decay is a random process, but it has a predictable average rate when you look at a large number of atoms.
One important measure of this decay process is the half-life, which is the time it takes for half of the radioactive substance to decay. This concept is universal to all radioactive substances, meaning the half-life is consistent for any given element or isotope, regardless of the amount you start with.
Some handy points about radioactive decay:

• The rate of decay is exponential, meaning it decreases over time in a consistent proportional fashion.
• With each half-life that passes, the quantity of the substance reduces by half again.
• Each radioactive isotope has a distinct half-life.

The decay process continues until the material becomes a stable element, which no longer undergoes radioactive decay. Understanding this will provide insight into the concept of how we use it in calculations, such as determining the decay constant.

Exponential Decay Constant

The exponential decay constant, often represented as "k," plays a pivotal role in the description of radioactive decay. It serves as a measure of how quickly a reactive substance undergoes decay. The decay constant provides a link between the half-life of a substance and the rate equation for its decay.
To see it in action, we use the formula: \[N(t) = N_0 e^{-kt}\] where:

• \(N(t)\) is the remaining quantity of substance at time \(t\).
• \(N_0\) is the initial quantity of the substance.
• \(k\) is the exponential decay constant.

During half-life calculations, we set the remaining quantity \(N(t)\) to half the initial amount \(N_0/2\) and derive the relationship: \[e^{-kt_{1/2}} = \frac{1}{2}\] Solving for \(k\), we can express it in terms of the half-life \(t_{1/2}\):\[k = \frac{\ln(2)}{t_{1/2}}\] This allows us to calculate \(k\) knowing the half-life, enabling prediction of the rate at which any given radioactive substance decays. By understanding \(k\), scientists and researchers can determine the time scales involved in radioactive decay processes.

Natural Logarithm

The natural logarithm, represented by \( \ln\), is a fundamental concept in mathematics, often used in the analysis of exponential growth and decay processes. Natural logarithms have a base of \( e \), which is an irrational constant approximately equal to 2.71828. You may encounter this concept frequently as you work with equations related to exponential growth or decay.
Why is the natural logarithm important in radioactive decay? When examining exponential decay, particularly in solving for values like the decay constant \(k\), natural logarithms allow us to handle exponents and logarithmic conversions simplistically:

• In the half-life calculation, by rewriting \[e^{-kt_{1/2}} = \frac{1}{2}\] as\(-kt_{1/2} = \ln \left( \frac{1}{2} \right)\),we simplify the equation using logarithms.
• This manipulation makes it easier to solve for \(k\) and understand the relationship between dynamic quantities such as the half-life \(t_{1/2}\) and the natural logarithm \( \ln(2) \).

Natural logarithms make complex mathematical relationships easier to solve and understand, providing a much-needed bridge between straightforward arithmetic and the complex exponential behaviors found in nature, like radioactive decay.