Ch 2.2 Exponential Decay and Radioactivity

Radioactive Decay

  • Historians, geologists, archaeologists, palaeontologists and many others use dating procedures to establish theories within their disciplines.
  • Unstable elements emit \( \alpha \)-particles, \( \beta \)-particles, or photons while decaying into isotopes of other elements.

Compartment Diagram

  • We consider the process in terms of a compartment model.
  • There is an output term over time, but no input term.

Background

  • The disintegration of one nucleus is a random event, so for small numbers of nuclei we could use probability functions to model the amount of material as a function of time.
  • For large numbers of nuclei we can assume that a proportion of nuclei will decay in any time interval.
  • Thus we model process as continuous with fixed decay rate.

Assumptions

  • Amount of element present is large enough so that we are justified in ignoring random fluctuations.
  • Process is continuous in time.
  • Rate of decay for an element is fixed.
  • No increase in mass of the body of material.
  • Rate of change of mass of decayed material proportional to mass of nuclei.

Word Equation

  • Start with word equation:

\[ \small{ \begin{Bmatrix} \mathrm{amount \, of } \\ \mathrm{radioactive \, material } \\ \mathrm{at \, time \,} t + \Delta t \end{Bmatrix} = \begin{Bmatrix} \mathrm{amount \, of } \\ \mathrm{radioactive \, material } \\ \mathrm{at \,time \,} t \end{Bmatrix} - \begin{Bmatrix} \mathrm{amount \, of } \\ \mathrm{radioactive \, material } \\ \mathrm{ decayed \, in \, time \,} \Delta t \end{Bmatrix} } \]

  • Subract middle term to left side, and for small \( \Delta t \), we obtain

\[ \small{ \begin{Bmatrix} \mathrm{rate \, of \, change \, of} \\ \mathrm{mass \, of \,radioactive } \\ \mathrm{material \, at \, time \,} t \end{Bmatrix} = - \begin{Bmatrix} \mathrm{rate \, of \, change \, of} \\ \mathrm{mass \, of \,decayed } \\ \mathrm{material \, at \, time \,} t \end{Bmatrix}} \]

Formulation of Differential Equation

  • Let \( N(t) \) = mass (grams) of radioactive nuclei at time \( t \).
  • Rate of change of mass of decayed material is proportional to mass of nuclei.
  • Proportionality constant \( k \) determined by experiment.

\[ \small{ \begin{Bmatrix} \mathrm{rate \, of \, change \, of} \\ \mathrm{mass \, of \,radioactive } \\ \mathrm{material \, at \, time \,} t \end{Bmatrix} = - \begin{Bmatrix} \mathrm{rate \, of \, change \, of} \\ \mathrm{mass \, of \,decayed } \\ \mathrm{material \, at \, time \,} t \end{Bmatrix} = - \{kN\} } \]

  • Corresponding initial value problem (IVP):

\[ \small{\frac{dN}{dt} = - kN, \,\,\, N(0) = n_0 } \]

Analytical Solution for IVP

  • General IVP has the form:

\[ \small{\frac{dN}{dt} = - kN, \,\,\, N(t_0) = n_0 } \]

  • Solve IVP using separation of variables:

\[ \small{ \begin{align*} \int \frac{dN}{N} &= - k \int dt \\ \ln N & = -kt + C \\ N(t) &= e^{-kt+C} = e^{-kt}e^C = A e^{-kt}\\ N(t) & = \left(n_0 e^{kt_0}\right) e^{-kt} = n_0 e^{-k(t-t_0)} \end{align*}} \]

  • Note that when \( \small{ t_0 = 0, \, N(t) = n_0 e^{-kt} } \).

Example: Analytical Solution

From Listing 2.2 in book:

  • \( k = 2.0 \)
  • \( N(0) = 10^5 \)

Our IVP is then

\[ \small{\frac{dN}{dt} = - 2N, \,\,\, N(0) = 10^5 } \]

The solution is given by

\[ \small{ N(t) = 10^5 e^{-2t} } \]

  • See Fig 2.3 in book.

Numerical Solution: Runge-Kutta Method

title

\[ \begin{align} \frac{dy}{dt} & = f(t,y), \,\, y(t_0) = y_0 \\ k_1 &= hf(t_i,y_i) \\ k_2 &= hf\left(t_i + 0.5h, y_i + 0.5k_1 \right) \\ k_3 &= hf\left(t_i + 0.5h, y_i + 0.5k_2 \right) \\ k_4 &= hf\left(t_i + h, y_i + k_3 \right) \\ y_{i+1} & = y_i + \frac{1}{6} (k_1 + 2k_2 + 2k_3 + k_4) \\ t_{i+1} &= t_i + h \end{align} \]

Numerical Solution: RK4

title

\[ \begin{align} \frac{dy}{dt} & = f(t,y), \,\, y(t_0) = y_0 \\ a &= hf(t_i,y_i) \\ b &= hf\left(t_i + 0.5h, y_i + 0.5a \right) \\ c &= hf\left(t_i + 0.5h, y_i + 0.5b \right) \\ d &= hf\left(t_i + h, y_i + c \right) \\ y_{i+1} & = y_i + \frac{1}{6} (a + 2b + 2c + d) \\ t_{i+1} &= t_i + h \end{align} \]

Numerical Solution: RK4

rk4plot <- function(f,x0,y0,h,n){
  x <- rep(0,n)
  y <- rep(0,n)
  x[1] <- x0
  y[1] <- y0
  for(i in 1:n) {
    a <- h*f(x[i], y[i])
    b <- h*f(x[i] + h/2, y[i] + a/2)
    c <- h*f(x[i] + h/2, y[i] + b/2)
    d <- h*f(x[i] + h, y[i] + c)
    y[i+1] <- y[i] + 1/6*(a + 2*b + 2*c + d)
    x[i+1] <- x[i] + h
  }
  return(plot(x,y,type="l", col="blue"))
}

RK4 Solution Plot

f <- function(x,y){-2*y}
rk4plot(f,0,10^5,0.1,40)

plot of chunk unnamed-chunk-2

title

Plot matches Fig 2.3.

Half-Life

  • Half-life \( \tau \) is the time required for half of the radioactive nuclei to decay.
  • The half-life \( \tau \) is more commonly known than the value of the rate constant \( k \) for radioactive elements.
  • The half-life \( \tau \) can be used to determine \( k \).

Finding Rate Constant k using Half-Life

  • Recall general solution \( \small{N(t) = n_0 e^{-k(t-t_0)} } \)

  • Let \( t = \tau \) denote the half-life. Then

\[ \small{ \begin{align*} N(t+\tau) & = 0.5N(t) \\ n_0 e^{-k(t+\tau-t_0)} &= 0.5 n_0 e^{-k(t-t_0)} \\ n_0e^{-kt}e^{-k\tau}e^{kt_0} &= 0.5 n_0 e^{-kt}e^{kt_0 } \\ e^{-k\tau} &= 0.5 \\ -k\tau & = \ln\left(2^{-1}\right) = -\ln(2) \\ k & = \frac{\ln(2)}{\tau} \end{align*}} \]

  • Note that \( k \) and \( \tau \) are independent of \( n_0 \) and \( t_0 \).

Example: Carbon 14

  • Half-life of \( ~^{14} \) C is

\[ \tau \cong 5568 \, \mathrm{years} \]

  • The decay rate is then

\[ \small{ \begin{align*} k & = \frac{\ln(2)}{\tau} \\ & = \frac{\ln(2)}{5568} \\ & \cong 0.0001245 \end{align*}} \]

  • Can use R to compute \( k \):
log(2)/5568
[1] 0.0001244876
  • Use log(x) for \( \ln(x) \).
  • Use log10(x) for \( \log(x) \).
  • Use log(x,b) for \( \log_b(x) \).

Example 2.4: Lascaux Cave Paintings

  • The Lascaux Cave paintings are believed to be prehistoric.
  • Using Geiger counter, a decay rate of \( ~^{14} \) C in charcoal fragments from cave measured 1.69 disintegrations per minute per gram of carbon.
  • In comparison, for living tissue in 1950, the measurement was 13.5 disintegrations per minute per gram of carbon.
  • How long ago was the radioactive carbon formed?

Example 2.4: Use IVP Model

  • Let \( \small{N(t)} \) = amount of \( \small{~^{14}} \) C per gram in the charcoal at time \( t \).
  • Let \( t_0 = 0 \) denote the current time.
  • Let \( \small{T} < 0 \) denote the time at which carbon was formed.
  • Assume our previous IVP model applies here:

\[ \small{ \frac{dN}{dt} = - kN, \,\,\, N(0) = n_0, \,\,\, t > T } \]

  • The solution to the IVP for \( \small{~^{14}} \) C is

\[ \small{ N(t) = n_0 e^{-kt}, \,\,\, k \cong 0.0001245 } \]

Example 2.4: Solve for T

  • From \( \small{N(t) = n_0 e^{-kt}} \), we substitute \( \small{ t } \) = \( \small{T} \) and use algebra to solve for \( \small{T} \) as follows:

\[ \small{ \begin{aligned} N(T) & = n_0 e^{-kT} \\ e^{-kT} &= \frac{N(T)}{n_0} \\ -kT &= \ln\left(\frac{N(T)}{n_0} \right) \\ T &= -\frac{1}{k} \ln\left(\frac{N(T)}{n_0} \right) \end{aligned} } \]

Example 2.4: Determine T

  • From \( \small{N(t) = n_0 e^{-kt}} \), we have

\[ \small{N'(t) = -k n_0 e^{-kt} = -k N(t)} \]

  • From previous slide,

\[ \small{ T = -\frac{1}{k} \ln\left(\frac{N(T)}{n_0} \right) } \]

  • The values of \( \small{N(T)} \) and \( n_0 \) are not known. However,

\[ \small{ \frac{N'(T)}{N'(0)} = \frac{- k N(T)}{-k N(0)} = \frac{N(T)}{n_0} } \]

Example 2.4: Determine T

  • Recall that a decay rate of \( ~^{14} \) C in charcoal fragments from cave measured 1.69, and for living tissue it was 13.5. Then

\[ \small{ \frac{1.69}{13.5} = \frac{N'(T)}{N'(0)} = \frac{N(T)}{n_0} } \]

  • Thus

\[ \small{ T = -\frac{1}{k} \ln\left(\frac{N(T)}{n_0} \right) = -\frac{1}{k} \ln\left(\frac{1.69}{13.5}\right) \cong 16,692 } \]

-1/0.0001245*log(1.69/13.5)
[1] 16690.45

Discussion

  • The accuracy of the process depends on having prior knowledge of the exact ratio of \( ~^{14} \) C to the total C in the atmosphere, but this has changed over the years.
  • There is a basic sinusoidal variation with an 8,000 year period.
  • Volcanic eruptions and industrial smoke emit only \( ~^{12} \) C, from materials that are older than 100,000 years.
  • Nuclear testing has resulted in a 100% increase in ratio in certain places, and are now factored into dating process.
  • For recent past, a lead isotope with half-life of 22 years is used for dating, while Uranium-238, with half-life of billions of years, is used to date the Earth.