Problem 31 - Calculus Early Transcendentals

Question

Radon-222 is a radioactive gas with a half-life of 3.83 days. This gas is a health hazard because it tends to get trapped in the basements of houses, and many health officials suggest that homeowners seal their basements to prevent entry of the gas. Assume that \(5.0 \times 10^{7}\) radon atoms are trapped in a basement at the time it is sealed and that \(y(t)\) is the number of atoms present \(t\) days later. (a) Find an initial-value problem whose solution is \(y(t) .\) (b) Find a formula for \(y(t) .\) (c) How many atoms will be present after 30 days? (d) How long will it take for \(90 \%\) of the original quantity of gas to decay?

Step-by-Step Solution

Answer

(a) Solve \( \frac{dy}{dt} = -ky \) with \( y(0) = 5.0 \times 10^7 \). (b) \( y(t) = 5.0 \times 10^7 \cdot e^{-0.1808t} \). (c) ~\(1.3 \times 10^6\) atoms after 30 days. (d) ~12.66 days for 90% decay.

Step 1: Understand the half-life formula

The decay of radioactive substances is modeled by the equation \(y(t) = y_0 \cdot e^{-kt}\), where \(y_0\) is the initial quantity of the substance, \(t\) is time, and \(k\) is the decay constant. The half-life (\(t_{1/2}\)) is given by the formula \(t_{1/2} = \frac{\ln 2}{k}\). For Radon-222, \(t_{1/2} = 3.83\) days.

Step 2: Find the decay constant

We rearrange the half-life formula to find \(k\): \(k = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{3.83}\). Calculating this gives \(k \approx 0.1808\).

Step 3: Set up the initial-value problem

To set up the initial-value problem, use \(y(0) = 5.0 \times 10^7\). The differential equation is \(\frac{dy}{dt} = -ky\) with the initial condition \(y(0) = 5.0 \times 10^7\).

Step 4: Solve the differential equation

The form of the differential equation is \(\frac{dy}{dt} = -0.1808y\). Solving this by integrating gives \(y(t) = 5.0 \times 10^7 \cdot e^{-0.1808t}\).

Step 5: Calculate the number of atoms after 30 days

Substitute \(t = 30\) into \(y(t) = 5.0 \times 10^7 \cdot e^{-0.1808 \cdot 30}\). Solving this gives \(y(30) \approx 1.3 \times 10^6\) atoms remaining after 30 days.

Step 6: Determine time for 90% decay

If 90% of the atoms have decayed, 10% remain. Therefore, set \(y(t) = 0.1 \cdot 5.0 \times 10^7\) and solve \(5.0 \times 10^7 \cdot e^{-0.1808t} = 0.5 \times 10^6\). Taking the natural logarithm of both sides results in \(-0.1808t = \ln(0.1)\), giving \(t \approx \frac{\ln(0.1)}{-0.1808}\). Calculating this gives \(t \approx 12.66\) days.

Key Concepts

Half-life

The concept of half-life is crucial in understanding radioactive decay. It is the time required for half of the radioactive atoms in a sample to decay. When dealing with radioactive substances like Radon-222, the half-life is a constant that helps predict how quickly the substance will break down over time.

In mathematical terms, the half-life (\( t_{1/2} \)) is related to the decay constant (\( k \)) by the equation:

• \( t_{1/2} = \frac{\ln 2}{k} \)

Radon-222 has a half-life of 3.83 days, meaning every 3.83 days, half of the original Radon-222 atoms will have decayed. This predictable decay makes it easier to determine the remaining quantity of a radioactive substance over time.

Decay Constant

The decay constant (\( k \)) is a fundamental component in the equation of radioactive decay. It represents the probability per unit time that a given atom will decay. The decay constant allows us to express the rate of decay in a differential equation.

To find the decay constant for Radon-222, use the half-life formula:

• \( k = \frac{\ln 2}{t_{1/2}} \)

Given that the half-life of Radon-222 is 3.83 days, the decay constant becomes:\( k \approx 0.1808 \).

This constant is then used in the exponential decay equation \( y(t) = y_0 \cdot e^{-kt} \) to model the number of atoms present at time \( t \).

Differential Equation

In this problem, we use a differential equation to model the process of radioactive decay. A differential equation describes how a quantity changes over time and can be solved to predict future behavior of a system, such as the remaining atoms of a radioactive substance.

For Radon-222, the differential equation is given by:

• \( \frac{dy}{dt} = -ky \)

This equation indicates that the rate of decay (\( \frac{dy}{dt} \)) is proportional to the number of atoms present at time \( t \) and the decay constant \( k \). By solving this differential equation, we derive an expression for \( y(t) \), the number of atoms remaining after time \( t \). The solution takes the form \( y(t) = y_0 \cdot e^{-kt} \), showing exponential decay over time.

Initial-Value Problem

An initial-value problem involves solving a differential equation with a specific initial condition. It ensures that the solution accurately fits the scenario being modeled by specifying the starting point of the system.

For Radon-222 trapped in a basement, the initial-value problem uses:

• Initial condition: \( y(0) = 5.0 \times 10^7 \)
• Differential equation: \( \frac{dy}{dt} = -0.1808y \)

This tells us that at time \( t = 0 \), the number of Radon-222 atoms is initially \( 5.0 \times 10^7 \). Solving this initial-value problem leads to an explicit expression \( y(t) = 5.0 \times 10^7 \cdot e^{-0.1808t} \), which can be used to predict the number of atoms at any future time \( t \). This approach is instrumental in calculating the atom count after 30 days or determining how long it will take for a certain percentage of the substance to decay.