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Chapter 3: Problem 6

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by \(3 \%\). If the rate of decay is proportional to the amount of the substance present at time \(t,\) find the amount remaining after 24 hours.

Short Answer

Expert verified
Approximately 88.9 mg of the substance remains after 24 hours.

Step by step solution

01

Understand the Problem

We are given an initial mass of 100 mg of a radioactive substance that decreases by 3% after 6 hours. We need to find the amount remaining after 24 hours, assuming the decay rate is proportional to the current amount.
02

Write the Exponential Decay Model

The decay process follows the exponential decay formula: \(A(t) = A_0 e^{-kt}\), where \(A(t)\) is the amount at time \(t\), \(A_0\) is the initial amount, and \(k\) is the decay constant.
03

Set Up Initial Conditions

At \(t = 0\), \(A_0 = 100\) mg. After 6 hours, the amount decreases by 3%, so \(A(6) = 97\) mg. Substitute these into the decay equation to find \(k\).
04

Solve for Decay Constant \(k\)

Substitute \(A(6) = 97\) into the formula: \[ 97 = 100 e^{-6k} \] Divide both sides by 100:\[ 0.97 = e^{-6k} \] Taking the natural log of both sides:\[ \ln(0.97) = -6k \]Solve for \(k\):\[ k = -\frac{\ln(0.97)}{6} \].
05

Calculate Amount After 24 Hours

Now that we have the decay constant \(k\), we find \(A(24)\): \[ A(24) = 100 e^{-24k} \] Substitute \(k\) from previous step and compute \(A(24)\).
06

Compute the Final Amount

Using the known value of \(k\) in the formula, calculate: \[ A(24) = 100 e^{-24 \times \frac{-\ln(0.97)}{6}} \]This gives approximately 88.9 mg after performing the calculations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactive Decay
Radioactive decay is a fascinating process and serves as a perfect example to illustrate exponential decay. It happens when unstable atomic nuclei lose energy by emitting radiation. As a result, the substance decreases in amount over time. This is like a ticking clock, with each tick representing a gradual loss.

Key features of radioactive decay include:
  • Randomness: It occurs spontaneously and unpredictably in time.
  • Stable End: Eventually, the substance becomes stable after emitting enough radiation.
  • Half-Life: This term refers to the time it takes for half of the substance to decay. It is a constant and helps in predicting how much remains after a set period.
In our exercise, 3% reduction in 6 hours demonstrates how decay negatively impacts the quantity of the substance. Understanding this concept helps predict future amounts accurately by using models like exponential functions.
Exponential Functions
Exponential functions are widely used to model situations involving exponential growth and decay. They have a standard form: \[ A(t) = A_0 e^{-kt} \]Here, \(A_0\) is the initial amount, \(A(t)\) is the amount at time \(t\), and \(k\) is the decay constant. This constant determines the rate at which the substance decays.

When dealing with radioactive decay, the key aspect of an exponential function is its continuous and smooth curve. This accurately reflects the gradual loss of the substance over time. The formula helps:
  • Relate current and future amounts.
  • Calculate the decay constant when the initial and recent amounts are known.
  • Predict changes accurately in practical and natural situations.
In the given problem, modeling the decay with exponential functions allows for precise predictions of how much of the substance will remain after 24 hours.
Differential Equations
Differential equations are powerful tools used to model how quantities change over time. They are integral in representing physical phenomena such as exponential decay. For radioactive decay, the rate of change is proportional to the current amount of the substance. This idea is captured in the differential equation:\[ \frac{dA}{dt} = -kA \]where \(A\) is the amount at time \(t\) and \(k\) is the decay constant. This equation suggests the rate of decay is directly linked to how much of the substance is present at that time.

Key points about this equation include:
  • Influence: As \(A\) decreases, the rate of decay gets slower since less substance is left.
  • Solution: The general solution is the exponential decay function we use: \(A(t) = A_0 e^{-kt}\).
  • Adaptation: It can be adapted to different scenarios beyond radioactive decay, such as population dynamics, cooling processes, and more.
Solving such differential equations provides deeper insight and allows for precise calculations, like determining the remaining amount of a radioactive substance after 24 hours as shown in our problem.

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Most popular questions from this chapter

A 200 -volt electromotive force is applied to an \(R C\) -series circuit in which the resistance is 1000 ohms and the capacitance is \(5 \times 10^{-6}\) farad. Find the charge \(q(t)\) on the capacitor if \(i(0)=0.4 .\) Determine the charge and current at \(t=0.005 \mathrm{s}\) Determine the charge as \(t \rightarrow \infty\)

Suppose a small single-stage rocket of total mass \(m(t)\) is launched vertically, the positive direction is upward, the air resistance is linear, and the rocket consumes its fuel at a constant rate. In Problem 22 of Exercises 1.3 you were asked to use Newton's second law of motion in the form given in (17) of that exercise set to show that a mathematical model for the velocity \(v(t)\) of the rocket is given by $$\frac{d v}{d t}+\frac{k-\lambda}{m_{0}-\lambda t} v=-g+\frac{R}{m_{0}-\lambda t}$$ where \(k\) is the air resistance constant of proportionality, \(\lambda\) is the constant rate at which fuel is consumed, \(R\) is the thrust of the rocket, \(m(t)=m_{0}-\lambda t, m_{0}\) is the total mass of the rocket at \(t=0,\) and \(g\) is the acceleration due to gravity. (a) Find the velocity \(v(t)\) of the rocket if \(m_{0}=200 \mathrm{kg}, R=2000 \mathrm{N}\) \(\lambda=1 \mathrm{kg} / \mathrm{s}, g=9.8 \mathrm{m} / \mathrm{s}^{2}, k=3 \mathrm{kg} / \mathrm{s},\) and \(v(0)=0\) (b) Use \(d s / d t=v\) and the result in part (a) to find the height \(s(t)\) of the rocket at time \(t\)

A dead body was found within a closed room of a house where the temperature was a constant \(70^{\circ} \mathrm{F}\). At the time of discovery the core temperature of the body was determined to be \(85^{\circ}\) F. One hour later a second measurement showed that the core temperature of the body was \(80^{\circ} \mathrm{F}\). Assume that the time of death corresponds to \(t=0\) and that the core temperature at that time was \(98.6^{\circ} \mathrm{F}\) Determine how many hours elapsed before the body was found. [Hint: Let \(t_{1}>0\) denote the time that the body was discovered.

Consider the competition model defined by $$\begin{aligned}&\frac{d x}{d t}=x(2-0.4 x-0.3 y)\\\&\frac{d y}{d t}=y(1-0.1 y-0.3 x),\end{aligned}$$ where the populations \(x(t)\) and \(y(t)\) are measured in thousands and \(t\) in years. Use a numerical solver to analyze the populations over a long period of time for each of the following cases: (a) \(x(0)=1.5, \quad y(0)=3.5\) (b) \(x(0)=1, \quad y(0)=1\) (c) \(x(0)=2, \quad y(0)=7\) (d) \(x(0)=4.5, \quad y(0)=0.5\)

(a) Consider the initial-value problem \(d A / d t=k A, A(0)=A_{0}\) as the model for the decay of a radioactive substance. Show that, in general, the half-life \(T\) of the substance is \(T=-(\ln 2) / k\) (b) Show that the solution of the initial-value problem in part (a) can be written \(A(t)=A_{0} 2^{-t / T}\) (c) If a radioactive substance has the half-life \(T\) given in part (a), how long will it take an initial amount \(A_{0}\) of the substance to decay to \(\frac{1}{8} A_{0} ?\)
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