Problem 23 - Precalculus with Limits

Question

Complete the table assuming continuously compounded interest. $$ \begin{array}{llll} \text { Isotope} & \text { Half-Life } & \text { Initial Quantity } & \text { Amount After } \\ \underline{\phantom{xxx}} & \ \text { Years } & \underline{\phantom{xxx}} & \text { 1000 Years } \\ \ ^{14}C&\quad5715 & \quad \space &\quad 2g\\\ \end{array} $$

Step-by-Step Solution

Answer

The initial quantity \(N_0\) of \(^{14}C\) isotope can be calculated using the formula \[N_0 = \frac{N}{e^{-λt}}\]. Upon substituting the given values and calculating, we get the initial quantity as \(\frac{2}{e^{-\frac{ln(2)}{5715} * 1000}}\) grams.

Step 1: Calculate decay constant

We need to calculate the decay constant (λ) first. This can be done using the formula \[λ = \frac{ln(2)}{T_{1/2}}\]. Here \(T_{1/2}\) is the half-life, which is given as 5715 years. Plugging in these values gives \(λ = \frac{ln(2)}{5715}\).

Step 2: Use Decay Formula to solve for the initial quantity

Once we have the decay constant, we can use the decay formula \[N = N_0 * e^{-λt}\] to solve for \(N_0\) (the initial quantity). We have \(N = 2g\) (the final amount after 1000 years), \(λ = \frac{ln(2)}{5715}\) (from Step 1), and \(t = 1000\) years. Plugging in these values and rearranging the equation for \(N_0\) gives \(N_0 = \frac{N}{e^{-λt}} = \frac{2}{e^{-\frac{ln(2)}{5715} * 1000}}\).

Step 3: Calculate the initial quantity

Calculate the fraction \(\frac{2}{e^{-\frac{ln(2)}{5715} * 1000}}\) to determine the initial quantity \(N_0\). This result will be the initial quantity of \(^{14}C\) isotope before 1000 years.

Key Concepts

Understanding Half-Life in Radioactive Decay

The term half-life is crucial when we’re dealing with the concept of radioactive decay or any process in which a quantity decreases by half over a consistent period of time. In simple terms, the half-life of a radioactive substance is the time it takes for half of the radioactive atoms to decay.

Imagine you have a piece of paper and you fold it in half. If the folding process represents a half-life, then each fold will reduce the paper’s size by half. Similarly, if you start with a 100-gram sample of a radioactive isotope, after one half-life has passed, you would only have 50 grams of it left; after two half-lives, just 25 grams would remain, and so on.

This concept is critical in various fields, from archaeology, using carbon dating, to medicine, in the form of radioactive tracers. It also applies to our exercise where the radioactive isotope Carbon-14 has a half-life of 5715 years. Knowing the half-life allows us to calculate other values, such as the decay constant, which is essential for determining the amount of substance remaining after a given period.

Decay Constant and Its Role in Exponential Decay

The decay constant, represented by the Greek letter λ (lambda), is an intrinsic property of a radioactive isotope and describes the rate at which it decays. It is tied directly to the half-life of the substance; while the half-life tells us how long it takes for half of the substance to decay, the decay constant gives us the rate of decay at any given moment.

The relationship between half-life and decay constant is given by the formula \[ \lambda = \frac{\ln(2)}{T_{1/2}} \], where \(\ln(2)\) is the natural logarithm of 2. The decay constant allows us to more accurately describe and calculate the expected amount of a radioactive substance over time, expressed through the exponential decay formula.

Using the example of Carbon-14, the decay constant was calculated in the exercise demonstrating that a smaller value of λ corresponds to a longer half-life, indicating that Carbon-14 decays slowly.

Exponential Decay and Its Mathematical Representation

When we talk about exponential decay, we’re dealing with a process where the quantity of a substance decreases at a rate proportional to its current value. This type of decay is represented mathematically by the exponential decay formula: \[ N = N_0 * e^{-\lambda t} \].

In this equation, \(N\) is the remaining quantity after time \(t\), \(N_0\) is the initial quantity, \(e\) is Euler's number (approximately 2.71828), \(\lambda\) is the decay constant, and \(t\) is time. In our exercise, we're given the final amount of the substance after 1000 years and the half-life of the substance. By using these parameters, we can work our way backward to find the initial amount \(N_0\) that would have led to the remaining 2 grams after a millennium.

This formula shows that as time increases, the value inside the exponent becomes more negative, leading to a smaller result for \(N\)—hence depicting the idea of decay. The understanding of exponential decay not only helps us in science but has applications in finance, population studies, and more, where growth or decay of any variable can be described exponentially.