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Chapter 15: Problem 13

Two unstable isotopes \(A\) and \(B\) and a stable isotope \(C\) have the following decay rates per atom present: \(A \rightarrow B, 3 \mathrm{~s}^{-1} ; A \rightarrow C, 1 \mathrm{~s}^{-1} ; B \rightarrow C, 2 \mathrm{~s}^{-1}\). Initially a quantity \(x_{0}\) of \(A\) is present and none of the other two types. Using Laplace transforms, find the amount of \(C\) present at a later time \(t\).

Short Answer

Expert verified
\(C(t) = -x_0 e^{-4t} + 2x_0 e^{-2t}\)

Step by step solution

01

Define the Differential Equations

First, write down the differential equations for the quantities of isotopes A, B, and C. Let A(t), B(t), and C(t) be the amounts of isotopes A, B, and C at time t respectively:\[ \frac{dA(t)}{dt} = -4A(t) \]\[ \frac{dB(t)}{dt} = 3A(t) - 2B(t) \]\[ \frac{dC(t)}{dt} = A(t) + 2B(t) \]
02

Apply the Laplace Transform

Apply the Laplace Transform to both sides of the differential equations. Denote the Laplace transforms of A(t), B(t), and C(t) as \(\mathcal{A}(s)\), \(\mathcal{B}(s)\), and \(\mathcal{C}(s)\) respectively:\[ s\mathcal{A}(s) - x_0 = -4\mathcal{A}(s) \]\[ s\mathcal{B}(s) = 3\mathcal{A}(s) - 2\mathcal{B}(s) \]\[ s\mathcal{C}(s) = \mathcal{A}(s) + 2\mathcal{B}(s) \]
03

Solve for \(\mathcal{A}(s)\)

Rearrange the first equation to solve for \(\mathcal{A}(s)\):\[ s\mathcal{A}(s) + 4\mathcal{A}(s) = x_0 \]\[ (s+4)\mathcal{A}(s) = x_0 \]\[ \mathcal{A}(s) = \frac{x_0}{s+4} \]
04

Solve for \(\mathcal{B}(s)\)

Plug \(\mathcal{A}(s)\) into the second transformed equation to solve for \(\mathcal{B}(s)\):\[ s\mathcal{B}(s) + 2\mathcal{B}(s) = 3 \frac{x_0}{s+4} \]\[ (s + 2)\mathcal{B}(s) = 3 \frac{x_0}{s+4} \]\[ \mathcal{B}(s) = \frac{3x_0}{(s+4)(s+2)} \]
05

Solve for \(\mathcal{C}(s)\)

Substitute \(\mathcal{A}(s)\) and \(\mathcal{B}(s)\) into the third transformed equation to solve for \(\mathcal{C}(s)\):\[ s\mathcal{C}(s) = \frac{x_0}{s+4} + 2 \frac{3x_0}{(s+4)(s+2)} \]Simplify the right side:\[ s\mathcal{C}(s) = \frac{x_0(s+2) + 6x_0}{(s+4)(s+2)} \]\[ s\mathcal{C}(s) = \frac{x_0s + 2x_0 + 6x_0}{(s+4)(s+2)} \]\[ s\mathcal{C}(s) = \frac{x_0s + 8x_0}{(s+4)(s+2)} \]\[ \mathcal{C}(s) = x_0 \left( \frac{1}{s+4} + \frac{4}{(s+4)(s+2)} \right) \]
06

Inverse Laplace Transform

Find the inverse Laplace Transform to get \(C(t)\):Using partial fractions, the function can be split and then inverted:\[ \mathcal{C}(s) = x_0 \left( \frac{1}{s+4} + \frac{4}{(s+2)(s+4)} \right) \]\[ \frac{4}{(s+2)(s+4)} = \frac{2}{s+2} - \frac{2}{s+4} \]Therefore,\[ \mathcal{C}(s) = x_0 \left( \frac{1}{s+4} + 2\frac{1}{s+2} - 2\frac{1}{s+4} \right) \]\[ \mathcal{C}(s) = x_0 \left( - \frac{1}{s+4} + 2\frac{1}{s+2} \right) \]Taking inverse Laplace transform,\[ C(t) = -x_0 e^{-4t} + 2x_0 e^{-2t} \]

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Key Concepts

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

Differential Equations
Differential equations are mathematical expressions that relate a function to its derivatives. In the context of physics and radioactive decay, they describe the rate of change of quantities over time. For example, in the decay process of isotopes, the differential equations represent how each isotope changes as each decays into another over time.

In this exercise, we defined differential equations for the amounts of isotopes A, B, and C at any time t. They were:
  • \(\frac{dA(t)}{dt} = -4A(t)\)
  • \(\frac{dB(t)}{dt} = 3A(t) - 2B(t)\)
  • \(\frac{dC(t)}{dt} = A(t) + 2B(t)\)
These equations show how single isotopes 'A' decay into 'B' and 'C', and how 'B' additionally decays into 'C'.
Radioactive Decay
Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation. Each type of radiation decay is characterized by a fixed decay constant, which determines the rate of decay.

In our exercise, isotope A decays into B and C with certain rates. Isotope B is also unstable and decays further into C. We described this mathematically using decay rates:
  • A \(\rightarrow\) B: 3 s-1
  • A \(\rightarrow\) C: 1 s-1
  • B \(\rightarrow\) C: 2 s-1
Rates per second (s-1) means for each atom of A or B, it has respectively 3, 1, and 2 chances per second to decay into B or C. When we solve these decay processes mathematically, we can see how over time these quantities change, leading to more stable isotopes.
Partial Fractions
Partial fractions are a method used in calculus to break down complex rational expressions into simpler parts. This technique is particularly useful in finding the inverse Laplace transforms.

For example, in step 6 of our solution, we used partial fractions to decompose the equation:
\[ \frac{4}{(s+2)(s+4)} = \frac{2}{s+2} - \frac{2}{s+4} \]

By simplifying the right-hand side, we make it easier to transform back from the Laplace domain to the time domain. This ultimately makes it more straightforward to understand how the amounts of isotopes evolve over time.

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Most popular questions from this chapter

Find the form of the solutions of the equation $$ \frac{d y}{d x} \frac{d^{3} y}{d x^{3}}-2\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\left(\frac{d y}{d x}\right)^{2}=0 $$ which have \(y(0)=\infty\) (You will need the result \(\int^{z} \operatorname{cosech} u d u=-\ln (\operatorname{cosech} z+\operatorname{coth} z)\).)

The two functions \(x(t)\) and \(y(t)\) satisfy the simultaneous equations $$ \begin{aligned} &\frac{d x}{d t}-2 y=-\sin t \\ &\frac{d y}{d t}+2 x=5 \cos t \end{aligned} $$ Find explicit expressions for \(x(t)\) and \(y(t)\), given that \(x(0)=3\) and \(y(0)=2\). Sketch the solution trajectory in the \(x y\)-plane for \(0 \leq t<2 \pi\), showing that the trajectory crosses itself at \((0,1 / 2)\) and passes through the points \((0,-3)\) and \((0,-1)\) in the negative \(x\)-direction.

The first few terms of a series \(u_{n}\), starting with \(u_{0}\), are \(1,2,2,1,6,-3\). The series is generated by a recurrence relation of the form $$ u_{n}=P u_{n-2}+Q u_{n-4} $$ where \(P\) and \(Q\) are constants. Find an expression for the general term of the series and show that the series in fact consists of two other interleaved series given by $$ \begin{array}{r} u_{2 m}=\frac{2}{3}+\frac{1}{3} 4^{m} \\ u_{2 m+1}=\frac{7}{3}-\frac{1}{3} 4^{m} \end{array} $$ for \(m=0,1,2, \ldots\)

Find the roots of the auxiliary equation for the following. Hence solve them for the boundary conditions stated. (a) \(\frac{d^{2} f}{d t^{2}}+2 \frac{d f}{d t}+5 f=0 \quad\) with \(f(0)=1, f^{\prime}(0)=0\). (b) \(\frac{d^{2} f}{d t^{2}}+2 \frac{d f}{d t}+5 f=e^{-t} \cos 3 t \quad\) with \(f(0)=0, f^{\prime}(0)=0\).

A solution of the differential equation $$ \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+y=4 e^{-x} $$ takes the value 1 when \(x=0\) and the value \(e^{-1}\) when \(x=1\). What is its value when \(x=2 ?\)
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