Problem 10 - Calculus and its applications

Question

Carbon- 14 has a decay rate of \(0.012097 \%\) per year. The rate of change of an amount \(N\) of carbon- 14 is given by $$\frac{d N}{d t}=-0.00012097 N$$ where \(t\) is the number of years since decay began. a) Let \(N_{0}\) represent the amount of carbon-l4 present at \(t=0\). Find the exponential function that models the situation. b) Suppose \(200 \mathrm{~g}\) of carbon- 14 is present at \(t=0 .\) How much will remain after 800 yr? c) After how many years will half of the \(200 \mathrm{~g}\) of carbon-l4 remain?

Step-by-Step Solution

Answer

a) \(N(t) = N_0 e^{-0.00012097t}\); b) \(177.83\) g remains after 800 years; c) Half-life is approximately 5730 years.

Step 1: Understanding the Differential Equation

The given differential equation is \( \frac{dN}{dt} = -0.00012097 N \). It describes the rate of change of the carbon-14 amount. This forms a standard exponential decay model.

Step 2: Solving the Differential Equation

To solve \( \frac{dN}{dt} = -0.00012097 N \), we separate variables and integrate: \( \int \frac{1}{N} dN = \int -0.00012097 \, dt \). This gives \( \ln|N| = -0.00012097t + C \). Exponentiating both sides gives \( N(t) = N_0 e^{-0.00012097t} \), where \( N_0 \) is the initial amount at \( t = 0 \).

Step 3: Defining the Exponential Function

From the solution, the exponential function becomes \( N(t) = N_0 e^{-0.00012097t} \). This function models the decay of carbon-14 over time.

Step 4: Initial Conditions and Substituting Known Values

Given \( N_0 = 200 \) grams at \( t = 0 \), the function becomes \( N(t) = 200 e^{-0.00012097t} \). This is our specific model for the amount of carbon-14.

Step 5: Calculating Amount after 800 Years

To find the amount remaining after 800 years, substitute \( t = 800 \) into the model: \( N(800) = 200 e^{-0.00012097 \times 800} \). Calculating this gives \( N(800) \approx 177.83 \) grams.

Step 6: Finding Half-Life (Half of 200g Remaining)

The half-life condition is \( \frac{N_0}{2} = N_0 e^{-0.00012097t} \). Simplifying, this gives \( \frac{1}{2} = e^{-0.00012097t} \). Taking the natural logarithm of both sides, \( \ln(0.5) = -0.00012097t \). Solving for \( t \), we find \( t \approx \frac{\ln(0.5)}{-0.00012097} \approx 5730 \) years.

Key Concepts

Carbon-14 Dating

Carbon-14 dating is a fascinating method used by scientists to determine the age of an artifact or object that contains carbon material. This process is important in archaeology, geology, and other sciences that require dating materials up to several tens of thousands of years old.
Carbon-14, a radioactive isotope of carbon, is naturally found in the atmosphere and absorbed by living organisms. Upon death, the organism stops absorbing carbon-14, and the existing amount starts to decay at a known rate. By measuring the remaining amount of carbon-14 in a sample, scientists can calculate the time that has elapsed since the organism's death.

• The principle behind carbon-14 dating relies on radioactive decay, which transforms carbon-14 into nitrogen-14 over time.
• This decay process follows a first-order kinetics, modeled by a differential equation that describes the decay rate.
• The key challenge is measuring how much carbon-14 remains and then applying the decay rate to find the elapsed time, known as radiocarbon dating.

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. These equations are incredibly useful in modeling a variety of physical processes, such as population growth, heat transfer, and radioactivity.
In the context of carbon-14 dating, the differential equation given is \( \frac{dN}{dt} = -0.00012097 N \). This illustrates an exponential function where the rate of decay is proportional to the remaining quantity of carbon-14.
To solve such an equation, the method of separation of variables is often applied:

• First, separate the variables: \( \frac{1}{N} dN = -0.00012097 \, dt \).
• Next, integrate both sides: \( \int \frac{1}{N} dN = \int -0.00012097 \, dt \).
• This results in the solution: \( \ln|N| = -0.00012097t + C \).
• Exponentiate to solve for \( N(t) \), resulting in: \( N(t) = N_0 e^{-0.00012097t} \).

Ultimately, this describes how the quantity of carbon-14, \( N(t) \), decreases exponentially over time.

Half-Life Calculation

The concept of half-life is crucial in understanding exponential decay processes, particularly in radioactive decay like that of carbon-14. A half-life is the amount of time it takes for half of a radioactive sample to decay.
For carbon-14, the half-life is a key factor. It allows researchers to date artifacts based on how much of the radioactive isotope remains in a sample.
To calculate the half-life, the initial amount \( N_0 \) and the decay constant \( r \) (in this case, \( 0.00012097 \)) are used. The relationship is based on the formula:
\[ \frac{N_0}{2} = N_0 e^{-rt} \]
Simplifying gives:
\[ \frac{1}{2} = e^{-rt} \]
Taking the natural logarithm of both sides:
\[ \ln(0.5) = -rt \]
Finally, solving for \( t \), we find:
\[ t = \frac{\ln(0.5)}{-r} \approx 5730 \text{ years} \]
This calculation helps identify the time it takes for half the amount of carbon-14 to decay, which is approximately 5730 years.