Problem 6 - Active Calculus

Question

The mass of a radioactive sample decays at a rate that is proportional to its mass. a. Express this fact as a differential equation for the mass \(M(t)\) using \(k\) for the constant of proportionality. b. If the initial mass is \(M_{0}\), find an expression for the mass \(M(t)\). c. The half-life of the sample is the amount of time required for half of the mass to decay. Knowing that the half-life of Carbon-14 is 5730 years, find the value of \(k\) for a sample of Carbon-14. d. How long does it take for a sample of Carbon-14 to be reduced to one- quarter its original mass? e. Carbon-14 naturally occurs in our environment; any living organism takes in Carbon14 when it eats and breathes. Upon dying, however, the organism no longer takes in Carbon-14. Suppose that you find remnants of a pre-historic firepit. By analyzing the charred wood in the pit, you determine that the amount of Carbon-14 is only \(30 \%\) of the amount in living trees. Estimate the age of the firepit.

Step-by-Step Solution

Answer

The differential equation is \( \frac{dM}{dt} = -kM \). The expression for mass is \( M(t) = M_{0}e^{-kt} \). The value of \( k \) is approximately \( 1.2104 \times 10^{-4} \text{years}^{-1} \). It takes about 11460 years to decay to one-quarter. The firepit is approximately 10000 years old.

Step 1 - Set up the differential equation

The mass of the radioactive sample decays at a rate proportional to its mass. This can be mathematically expressed using the differential equation \[ \frac{dM}{dt} = -kM \] where \( k \) is the constant of proportionality.

Step 2 - Solve the differential equation

To solve the differential equation \( \frac{dM}{dt} = -kM \), separate the variables and integrate:\[ \int \frac{1}{M} dM = -k \int dt \]This gives \[ \ln M = -kt + C \]Exponentiating both sides, we find \[ M(t) = M_{0}e^{-kt} \]where \( M_{0} \) is the initial mass.

Step 3 - Find the decay constant \( k \)

The half-life \( T_{1/2} \) of Carbon-14 is 5730 years. Using the relation \[ M(5730) = \frac{1}{2}M_{0} \], we can write \[ \frac{M_{0}}{2} = M_{0}e^{-k \cdot 5730} \].Dividing both sides by \( M_{0} \) and simplifying, we get \[ \frac{1}{2} = e^{-k \cdot 5730} \], Taking the natural logarithm of both sides, we find \[ -5730k = \text{ln}\frac{1}{2} \] or \[ k = \frac{-\text{ln}(1/2)}{5730} \] \[ k = \frac{\text{ln} 2}{5730} \] \[ k \approx 1.2104 \times 10^{-4} \text{years}^{-1} \].

Step 4 - Time to decay to one-quarter

To find the time \( t \) when the sample is reduced to one-quarter of its original mass, we use \[ M(t) = \frac{M_{0}}{4} = M_{0}e^{-kt}. \] Dividing both sides by \( M_{0} \) and simplifying, we get \[ \frac{1}{4} = e^{-kt} \], Taking the natural logarithm of both sides, we find \[ -kt = \text{ln}\frac{1}{4} \], \[ t = \frac{-\text{ln}(1/4)}{k} \], \[ t = \frac{\text{ln} 4}{1.2104 \times 10^{-4}} \], \[ t \approx 11460 \text{ years} \].

Step 5 - Age of the firepit

If the amount of Carbon-14 is only 30% of the original amount, we use \[ M(t) = 0.3M_{0} = M_{0}e^{-kt}. \] Dividing both sides by \( M_{0} \) and simplifying, we get \[ 0.3 = e^{-kt} \], Taking the natural logarithm of both sides, we find \[ -kt = \text{ln}0.3 \], \[ t = \frac{\text{ln}(1/0.3)}{k} \], \[ t \approx \frac{1.204}{1.2104 \times 10^{-4}} \approx 10000 \text{ years} \].

Key Concepts

differential equations

Differential equations are mathematical equations that relate a function with its derivatives. They are used to model various real-world phenomena by describing how a system changes over time.
In the context of radioactive decay, we use a differential equation to represent how the mass of a radioactive sample decreases over time. The key idea is that the rate of decay is proportional to the current mass.
For a mass \(M(t)\), we can express this with the differential equation: \ \frac{dM}{dt} = -kM \ Here, \ \frac{dM}{dt} \ represents the rate of change or decay of the mass \(M(t)\),
and \(k\) is the constant of proportionality that defines the decay rate.

radioactive decay

Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. This process transforms the original nucleus into a different state, often into a different element.
The decay process follows an exponential pattern, depicted by the equation: \ M(t) = M_{0}e^{-kt} \ Here, \(M(t)\) is the mass of the radioactive sample at time \(t\), and \(M_{0}\) is the initial mass of the sample.
The term \(e^{-kt}\) indicates that the mass decreases exponentially over time, with \(k\) being the decay constant. This formula is derived by solving the differential equation \ \frac{dM}{dt} = -kM \ which expresses that the rate of decay is proportional to the current mass.

half-life calculations

The half-life of a radioactive substance is the time required for half of the sample to decay. It is a unique characteristic of each radioactive isotope.
We can use the half-life to find the decay constant \(k\). For example, the half-life of Carbon-14 is 5730 years. The formula to calculate \(k\) is: \ k = \frac{\text{ln}(2)}{T_{1/2}} \ Here, \(T_{1/2}\) is the half-life. Plugging in the half-life of Carbon-14, we have: \ k = \frac{\text{ln}(2)}{5730} \ This value of \(k\) helps in determining how the mass decays over time.

Carbon-14 dating

Carbon-14 dating, also known as radiocarbon dating, is a method for determining the age of an object containing organic material. This technique utilizes the known half-life of Carbon-14, which is 5730 years.
When an organism dies, it no longer absorbs Carbon-14, and the Carbon-14 it contains starts to decay at a predictable rate. By measuring the remaining amount of Carbon-14, we can estimate the age of the sample.
For instance, if a sample contains only 30% of its original Carbon-14, we can set up the equation: \ 0.3M_{0} = M_{0}e^{-kt} \ Solving for \(t\), we determine the sample's approximate age, in this case, nearly 10,000 years. This technique is invaluable in archaeology and other fields studying ancient organic materials.