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Chapter 2: Problem 4

Suppose that the series is described schematically by \(W \Longrightarrow-\lambda_{1} X \Longrightarrow-\lambda_{2} Y \Longrightarrow-\lambda_{3} Z\) where \(-\lambda_{1},-\lambda_{2},\) and \(-\lambda_{3}\) are the decay constants for \(W, X\) and \(Y,\) respectively, and \(Z\) is a stable element. Let \(w(t), x(t), y(t),\) and \(z(t)\) denote the amounts of substances \(W, X, Y,\) and \(Z,\) respectively. A model for the radioactive series is $$\begin{aligned}&\frac{d w}{d t}=-\lambda_{1} w\\\&\begin{array}{l} \frac{d x}{d t}=\lambda_{1} w-\lambda_{2} x \\\\\frac{d y}{d t}=\lambda_{2} x-\lambda_{3} y \end{array}\\\&\frac{d z}{d t}=\lambda_{3} y\end{aligned}$$

Short Answer

Expert verified
Solve sequential differential equations using decay laws.

Step by step solution

01

Solve the Differential Equation for W

The first equation is \( \frac{d w}{d t} = -\lambda_{1} w \). This is a first-order linear differential equation with the solution \( w(t) = w_0 e^{-\lambda_{1} t} \), where \( w_0 \) is the initial amount of substance \( W \).
02

Solve the Differential Equation for X

The equation is \( \frac{d x}{d t} = \lambda_{1} w - \lambda_{2} x \). Substituting \( w(t) = w_0 e^{-\lambda_{1} t} \) gives \( \frac{d x}{d t} = \lambda_{1} w_0 e^{-\lambda_{1} t} - \lambda_{2} x \). This is a non-homogeneous linear differential equation. The solution can be obtained using the integrating factor method or Laplace transforms, resulting in \( x(t) = \frac{\lambda_{1} w_0}{\lambda_{2} - \lambda_{1}} \left(e^{-\lambda_{1} t} - e^{-\lambda_{2} t} \right) \), assuming \( \lambda_2 eq \lambda_1 \).
03

Solve the Differential Equation for Y

The differential equation for \( y(t) \) is \( \frac{d y}{d t} = \lambda_{2} x - \lambda_{3} y \). Substituting \( x(t) \) into the equation gives \( \frac{d y}{d t} = \lambda_{2} \frac{\lambda_{1} w_0}{\lambda_{2} - \lambda_{1}} \left(e^{-\lambda_{1} t} - e^{-\lambda_{2} t} \right) - \lambda_{3} y \). Solving this equation using the method of undetermined coefficients or variation of parameters provides the solution for \( y(t) \).
04

Solve the Differential Equation for Z

The equation for \( z(t) \) is \( \frac{d z}{d t} = \lambda_{3} y \). Assuming we have the expression for \( y(t) \), this equation can be integrated directly to find the expression for \( z(t) \). Since \( Z \) is a stable element, its amount accumulates as \( z(t) = \lambda_{3} \int y(t) \ dt \).
05

Conclusion: Formulate the Complete Solution

Having expressions for \( w(t) \), \( x(t) \), \( y(t) \), and \( z(t) \) in terms of time \( t \) and the initial amounts, the decay constants \( \lambda_1 \), \( \lambda_2 \), and \( \lambda_3 \), we can describe the dynamics of the radioactive decay series completely.

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Key Concepts

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

Differential Equations
In the realm of radioactive decay series, differential equations play a vital role. They are mathematical expressions used to describe the rate at which these decay processes occur.

The basic form of a differential equation in this context relates the rate of change of a substance to its current state. For example, the equation \( \frac{d w}{d t} = -\lambda_{1} w \) indicates that the rate at which substance \( W \) decays is directly proportional to its present amount. This type of equation is powerful because it captures the natural tendency of radioactive substances to decrease over time.

Solving a differential equation gives insights into how the quantities of substances change with time. We typically move from expressing how fast a substance decays to understanding its actual amount at any given moment. This transformation requires techniques such as separation of variables or integrating factors. These methods simplify the equation, making it possible to integrate and find a manageable solution.

Understanding these equations is crucial for accurately modeling real-world scenarios in which decay processes occur. They are foundational to predicting how substances transition and stabilize over time.
Decay Constants
Decay constants are a core component in modeling radioactive decay. Represented by the symbol \( \lambda \), a decay constant expresses the probability per unit time that an atom of a radioactive substance will decay.

In a radioactive decay series, decay constants are essential because they differ for each substance involved in the decay chain. For instance, \( \lambda_{1} \) characterizes how rapidly substance \( W \) decays, while \( \lambda_{2} \) and \( \lambda_{3} \) describe the decay rates of \( X \) and \( Y \), respectively.

The unique decay constant of each substance means the transition between elements occurs at different rates. In practice, this allows scientists and engineers to predict the quantities of different substances present at any time. It helps determine the half-life of a substance, which is the time required for half of the substance to decay.

Understanding decay constants enables accurate modeling and planning in various fields, such as nuclear medicine and radiocarbon dating. These constants provide the necessary precision to forecast the behavior of radioactive elements over time.
Exponential Functions
Exponential functions are intimately related to the process of radioactive decay. They describe mathematical functions in which the rate of change of a quantity is proportional to the current value of the quantity itself. This relationship is key in solving differential equations arising from decay processes.

An exponential function typically takes the form \( w(t) = w_0 e^{-\lambda_{1} t} \), representing how the amount of substance \( W \) decreases over time. The base \( e \) is a mathematical constant approximately equal to 2.71828, providing a natural foundation for these expressions.

Exponential decay explains why radioactive substances diminish rapidly at first and then more slowly over time. Each passing moment, there's a consistent percentage decrease rather than a fixed amount.

The versatility of exponential functions makes them valuable in various decay-related applications. They adeptly capture the essence of decay processes because they can model continuous and uninterrupted transitions from one substance to another. This quality is why exponential functions are a mathematical favorite in understanding and predicting the behavior of decaying systems.

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Most popular questions from this chapter

(a) \((i)\) If \(s(t)\) is distance measured down the plane from the highest point, then \(d s / d t=v .\) Integrating \(d s / d t=16 t\) gives \(s(t)=8 t^{2}+c_{2} .\) Using \(s(0)=0\) then gives \(c_{2}=0 .\) Now the length \(L\) of the plane is \(L=50 / \sin 30^{\circ}=100 \mathrm{ft} .\) The time it takes the box to slide completely down the plane is the solution of \(s(t)=100\) or \(t^{2}=25 / 2,\) so \(t \approx 3.54 \mathrm{s}\). \((i i)\) Integrating \(d s / d t=4 t\) gives \(s(t)=2 t^{2}+c_{2} .\) Using \(s(0)=0\) gives \(c_{2}=0,\) so \(s(t)=2 t^{2}\) and the solution of \(s(t)=100\) is now \(t \approx 7.07 \mathrm{s}\). \((\text {iii})\) Integrating \(d s / d t=48-48 e^{-t / 12}\) and using \(s(0)=0\) to determine the constant of integration, we \(\operatorname{obtain} s(t)=48 t+576 e^{-t / 12}-576 .\) With the aid of a CAS we find that the solution of \(s(t)=100,\) or $$\begin{aligned} &100=48 t+576 e^{-t / 12}-576 &\text { or } \quad 0=48 t+576 e^{-t / 12}-676 \end{aligned}$$ is now \(t \approx 7.84 \mathrm{s}\). (b) The differential equation \(m d v / d t=m g \sin \theta-\mu m g \cos \theta\) can be written $$m \frac{d v}{d t}=m g \cos \theta(\tan \theta-\mu)$$. If \(\tan \theta=\mu, d v / d t=0\) and \(v(0)=0\) implies that \(v(t)=0 .\) If \(\tan \theta< \mu\) and \(v(0)=0,\) then integration implies \(v(t)=g \cos \theta(\tan \theta-\mu) t < 0\) for all time \(t\). (c) since \(\tan 23^{\circ}=0.4245\) and \(\mu=\sqrt{3} / 4=0.4330,\) we see that \(\tan 23^{\circ}<0.4330 .\) The differential equation is \(d v / d t=32 \cos 23^{\circ}\left(\tan 23^{\circ}-\sqrt{3} / 4\right)=-0.251493 .\) Integration and the use of the initial condition gives \(v(t)=-0.251493 t+1 .\) When the box stops, \(v(t)=0\) or \(0=-0.251493 t+1\) or \(t=3.976254 \mathrm{s}\). From \(s(t)=-0.125747 t^{2}+t\) we find \(s(3.976254)=1.988119 \mathrm{ft}\). (d) With \(v_{0} >0, v(t)=-0.251493 t+v_{0}\) and \(s(t)=-0.125747 t^{2}+v_{0} t .\) Because two real positive solutions of the equation \(s(t)=100,\) or \(0=-0.125747 t^{2}+v_{0} t-100,\) would be physically meaningless, we use the quadratic formula and require that \(b^{2}-4 a c=0\) or \(v_{0}^{2}-50.2987=0 .\) From this last equality we find \(v_{0} \approx 7.092164 \mathrm{ft} / \mathrm{s} .\) For the time it takes the box to traverse the entire inclined plane, we must have \(0=-0.125747 t^{2}+7.092164 t-100 .\) Mathematica gives complex roots for the last equation: \(t=\) \(28.2001 \pm 0.0124458 i .\) But, for $$0=-0.125747 t^{2}+7.092164691 t-100$$ the roots are \(t=28.1999 \mathrm{s}\) and \(t=28.2004 \mathrm{s} .\) So if \(v_{0} >7.092164,\) we are guaranteed that the box will slide completely down the plane.

From \(d A / d t=10-A / 100\) we obtain \(A=1000+c e^{-t / 100} .\) If \(A(0)=0\) then \(c=-1000\) and \(A(t)=\) \(1000-1000 e^{-t / 100}\).

We are looking for a function \(y(x)\) such that \\[y^{2}+\left(\frac{d y}{d x}\right)^{2}=1\\]. Using the positive square root gives \\[\frac{d y}{d x}=\sqrt{1-y^{2}} \Longrightarrow \frac{d y}{\sqrt{1-y^{2}}}=d x \Longrightarrow \sin ^{-1} y=x+c\\]. Note that when \(c=c_{1}=0\) and when \(c=c_{1}=\pi / 2\) we obtain the well known particular solutions \(y=\sin x\) \(y=-\sin x, y=\cos x,\) and \(y=-\cos x .\) Note also that \(y=1\) and \(y=-1\) are singular solutions.

A model is $$\begin{aligned}&\frac{d x_{1}}{d t}=(4 \mathrm{gal} / \mathrm{min})(0 \mathrm{lb} / \mathrm{gal})-(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{200} x_{1} \mathrm{lb} / \mathrm{gal}\right)\\\&\begin{array}{l}\frac{d x_{2}}{d t}=(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{200} x_{1} \mathrm{lb} / \mathrm{gal}\right)-(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{150} x_{2} \mathrm{lb} / \mathrm{gal}\right) \\\\\frac{d x_{3}}{d t}=(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{150} x_{2} \mathrm{lb} / \mathrm{gal}\right)-(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{100} x_{3} \mathrm{lb} / \mathrm{gal}\right)\end{array}\end{aligned}$$ or $$\begin{aligned}\frac{d x_{1}}{d t} &=-\frac{1}{50} x_{1} \\\\\frac{d x_{2}}{d t} &=\frac{1}{50} x_{1}-\frac{2}{75} x_{2} \\\\\frac{d x_{3}}{d t} &=\frac{2}{75} x_{2}-\frac{1}{25} x_{3}\end{aligned}$$ Over a long period of time we would expect \(x_{1}, x_{2},\) and \(x_{3}\) to approach 0 because the entering pure water should flush the salt out of all three tanks.

(a) \(\mathrm{By}\) inspection \(y=x\) and \(y=-x\) are solutions of the differential equation and not members of the family \(y=x \sin \left(\ln x+c_{2}\right).\) (b) Letting \(x=5\) and \(y=0\) in \(\sin ^{-1}(y / x)=\ln x+c_{2}\) we get \(\sin ^{-1} 0=\ln 5+c\) or \(c=-\ln 5 .\) Then \(\sin ^{-1}(y / x)=\ln x-\ln 5=\ln (x / 5) .\) Because the range of the arcsine function is \([-\pi / 2, \pi / 2]\) we must have $$\begin{array}{c} -\frac{\pi}{2} \leq \ln \frac{x}{5} \leq \frac{\pi}{2} \\ e^{-\pi / 2} \leq \frac{x}{5} \leq e^{\pi / 2} \\ 5 e^{-\pi / 2} \leq x \leq 5 e^{\pi / 2}. \end{array}$$ The interval of definition of the solution is approximately [1.04, 24.05].
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