Problem 22 - Precalculus with Limits

Question

Complete the table assuming continuously compounded interest. $$ \begin{array}{llll} \text { Isotope} & \text { Half-Life } & \text { Initial Quantity } & \text { Amount After } \\ \underline{\phantom{xxx}} & \ \text { Years } & \underline{\phantom{xxx}} & \text { 1000 Years } \\ \ ^{14}C&\quad5715 & \quad \space 6.5g&\quad \\ \end{array} $$

Step-by-Step Solution

Answer

The final remaining amount, \(N\), depends on the evaluation of the decay formula. Calculate this using a calculator or a computer program featuring scientific calculations to get a numerical answer.

Step 1: Understanding Exponential Decay

Exponential decay can be modeled by the formula \(N = N_0 * e^{(-λt)}\) where \(N\) is the remaining quantity after time \(t\), \(N_0\) is the initial quantity, \(e\) is the base of natural logarithm, and \(λ\) is the decay constant. To connect half-life to this equation, use the relationship \(λ = ln(2) / \text{half-life}\). First, calculate \(λ\) considering that the half-life of Carbon-14 is 5715 years: \(λ = ln(2) / 5715\)

Step 2: Plugging Values into the Decay Formula

Now that the value of \(λ\) is known, you can plug in the remaining values into the decay equation. Here, \(N_0 = 6.5g\), \(t = 1000 years\), and \(λ\) is the value calculated in the previous step. Substituting these values into the equation gives you the equation: \(N = 6.5 * e^{(- λ * 1000)}\)

Step 3: Calculate the Remaining Amount

Evaluate the right side of the equation determined in step 2 to find out the remaining amount \(N\) after 1000 years.

Key Concepts

Half-Life

The concept of half-life is quite fascinating and crucial in understanding radioactive decay. Half-life is the time required for half of a quantity of a substance to decay. In simpler terms, if you start with a certain amount of a radioactive isotope, half of it will remain after one half-life period. For Carbon-14, its half-life is 5715 years. This means that, after 5715 years, only half of the original amount of Carbon-14 remains.

• It is a constant value specific to each isotope.
• It helps in determining the decay rate of radioactive substances.
• It is essential in fields such as archaeology for carbon dating.

Knowing the half-life allows scientists to predict the amount of time it takes for a sample to decay to a certain amount by applying the formula for exponential decay.

Decay Constant

The decay constant, often represented by the Greek letter \(λ\), is a fundamental aspect of exponential decay. It relates the half-life of a substance to its decay rate. You can think of it as the probability of a single particle decaying per unit time.

• It is calculated by the formula: \(\lambda = \frac{\ln(2)}{\text{Half-Life}}\).
• This constant helps in expressing how fast a radioactive material will decay over some time.
• The larger the decay constant, the faster the rate of decay.

Once you have the decay constant, you can use it in the exponential decay formula \(N = N_0 \cdot e^{-\lambda t}\) to compute the remaining quantity of a substance after a certain time has passed.

Continuously Compounded Interest

Although it might seem unrelated, the concept of continuously compounded interest shares a mathematical bond with exponential decay. Imagine you have a bank account where the interest is added to your balance continuously rather than at fixed intervals. This is similar to how decay occurs in a constant flow.

• The formula used is \(A = P \cdot e^{rt}\) where \(A\) is the amount of money accumulated after time \(t\), \(P\) is the principal amount (initial quantity), \(r\) is the rate of interest, and \(e\) is the base of the natural logarithm.
• In the context of decay, \(r\) is replaced by \(-\lambda\).
• This principle helps in modeling the gradual and continuous decay of substances over time similarly to how money can grow with continuous interest.

By understanding this concept, you can see the parallels between financial growth and radioactive decay, both governed by the principle of continuous exponential change.