Problem 77 - Algebra 2 and Trigonometry

Question

The decay constant of radium is \(-0.0004\) per year. The amount of radium, \(A,\) present after \(t\) years in a sample that originally contained 1 gram of radium is \(A=e^{-0.0004 t} .\) a. Express \(-0.0004 t\) in terms of \(A\) and \(e .\) b. Solve for \(t\) in terms of \(A\) .

Step-by-Step Solution

Answer

a. \(-0.0004t = \ln(A)\); b. \(t = \frac{\ln(A)}{-0.0004}\).

Step 1: Analyze the Given Equation

The given equation describes the amount of radium present after time \( t \):\[ A = e^{-0.0004t} \]Here, the decay constant is given as \(-0.0004\). Our task is to express this exponent, \(-0.0004t\), in terms of \(A\).

Step 2: Express the Exponent in Terms of A

We need to rewrite \(-0.0004t\) in terms of \(A\) and \(e\). From the equation \( A = e^{-0.0004t} \), take the natural logarithm of both sides:\[ \ln(A) = \ln(e^{-0.0004t}) \]Using the property that \( \ln(e^x) = x \), we obtain:\[ \ln(A) = -0.0004t \]Therefore, \(-0.0004t = \ln(A)\).

Step 3: Solve for t in Terms of A

Now that we have \(-0.0004t = \ln(A)\), we can solve for \(t\) by isolating it on one side:\[ -0.0004t = \ln(A) \]Divide both sides by \(-0.0004\) to get:\[ t = \frac{\ln(A)}{-0.0004} \]

Step 4: Conclusion

We have derived the expression for \(-0.0004t\) in terms of \(A\):\[-0.0004t = \ln(A)\]And, solved for \(t\) in terms of \(A\):\[ t = \frac{\ln(A)}{-0.0004} \].

Key Concepts

Decay Constant

The decay constant is a fundamental parameter in the study of exponential decay, especially in the context of radioactive substances like radium. It is denoted by the Greek letter lambda (\( \lambda \)), and in our example, the value is \(-0.0004\) per year. The decay constant represents the probability per unit time that a given atom will decay.

When we say a decay constant is \(-0.0004\), it means that each atom in our sample has a 0.04% chance of decaying in any given year. This might seem small, but over a long time period, it significantly impacts how much of the material remains.

The decay constant is negative in our equation because we are considering the decay of a substance; thus, it represents a decrease over time. In practical terms, understanding the decay constant allows scientists and engineers to predict how long a radioactive sample will last or the rate at which it will decay.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a mathematical function that is the inverse of the exponential function with base \( e \) (approximately 2.71828). In the context of the decay equation \( A = e^{-0.0004t} \), applying the natural logarithm to both sides allows us to "undo" the exponential and simplify the expression to solve for \( t \).

When we have an equation of the form \( e^x = y \), taking the natural logarithm of both sides gives us \( x = \ln(y) \).

In our exercise, using \( \ln \) is a crucial step. When we apply it to \( A = e^{-0.0004t} \), we get \( \ln(A) = \ln(e^{-0.0004t}) \). Simplifying further, since \( \ln(e^x) = x \), this gives us \( \ln(A) = -0.0004t \). This transformation is essential for solving for \( t \), which is a common method in handling exponential decay problems.

Radioactive Decay Equation

The radioactive decay equation, often written as \( A = A_0 e^{-\lambda t} \), models how the amount of a radioactive substance decreases over time. In our exercise, the equation is presented in a slightly simplified form where the original amount \( A_0 \) is 1 gram, which means the equation is \( A = e^{-0.0004t} \).

This equation tells us how many grams of radium remain after a certain amount of time, \( t \). The exponent \( -0.0004t \) encapsulates both the effect of time and the decay constant in determining how fast the decay happens.

In general, the decay equation is a powerful tool because it allows scientists to predict the future state of a radioactive material. By altering or solving for different variables, such as \( A \) or \( t \), one can glean insights into how the decay process unfolds over time. Understanding and manipulating this equation is key for areas like nuclear physics, archaeology (carbon dating), and nuclear energy management.