2. Radioactive Decay and Half-Life

Background Theory

A sample of radioactive atoms often decay according to  equation (1) below

((d )/dt) 
        N[t]  =  - λ N[t]

where N[t] is the radioactive number of atoms in the sample at time t.  Radioactive atoms are atoms that disintegrate (decay) into other atoms and various forms of radiation (alpha, beta, and gamma).  λ is the "decay constant" and this constant has units of 1/time since the units on the RHS of equations (1) must be the same as the LHS.(See an appendix for a derivation of equation (1).)

	The rate of decay R[t] is the number of atoms decaying per unit time (with units for example,  "counts/min") so R[t] is defined as

R[t]  =  ((d )/dt) N[t]

which is same as the LHS of equation (1).  R[t] is what is usually directly observed in experiments instead of N[t] since it is difficult to count the number of atoms directly N[t].  R[t] is what is measured with the usual Geiger counter.

Combining equation (1) and (2) we easily obtain

R[t]  = - λ N[t]

Equation (3) or at least its consequences are what we use to interpret the experimental data.  Notice the minus sign on the RHS of equation (3) which means R[t] is negative.  While maybe surprising at first this is to be expected.  Note that in the limit

R[t] = ((N[t + Δt] - N[t])/Δ)t

Even though Δt is positive, N[t+Δt] is smaller than N[t] as atoms are decaying so R[t] should be negative.

Solution to Equation (1):

Possibly you have seen the solution to equation (1) in your introductory physics class or in Atomic and Nuclear Physics.  In any case the solution  is (see the appendix for a derivation).

N[t]  =  
      (Subscript[N, 0]) (e^(- λ t))

where (Subscript[N, 0])is a constant.  The "decay time" τ is a constant inversely related to λ

τ  =  (1/λ)

so the solution equation (4) could also be written

N[t]  =  
      (Subscript[N, 0]) (e^(-  ((t/τ))))

The units of τ are time (for example, "minutes").  In addition to λ and τ there is a third (more commonly used) parameter called the "half-life" (Subscript[t,  (1/2)]).  The "half-life" (Subscript[t,  (1/2)]) is defined as the TIME t  that half the original number atoms have NOT decayed (or half the original number atoms have decayed which is the same thing).  Specifically, the definition of (Subscript[t,  (1/2)]) means that

N[t=(Subscript[t,  (1/2)])] = ((Subscript[N, 0])/2).

and with equation (4)

N[t=(Subscript[t,  (1/2)])] = \(N\_0\ e\^\(\(-\ λ\)\ t\_\(1\/2\)\)\)

Combining equations (7) and (8) yields

((Subscript[N, 0])/2) =  
      (Subscript[N, 0]) (e^(- λ (Subscript[t, (1/2)])))

which might seem a little complicated at first but equation (9) allows us to obtain a simple relationship between λ and the half-life (Subscript[t, (1/2)])as we now see.  
	After cancelling the (Subscript[N, 0]) on both sides of eqt. (8) we obtain

\(\(\(\ \)\(1\/2 = \ \ \(e\^\(\(-\ λ\)\ t\_\(1\/2\)\) = \ \ 1\/e\^\(\(-\ λ\)\ t\_\(1\/2\)\)\)\)\)\)

we also used a property of exponential functions.  Finally inverting we obtain

2 =   (e^( λ (Subscript[t, (1/2)])))

Taking the NATURAL logarithm (Log base e) of both sides of equation. (11) yields

\(\(\(\ \)\(Log[
        2] = \ \(Log[e\^\(\(\ \)\(λ\ t\_\(1\/2\)\)\)]\  = \ λ\ t\_\(1\/2\)\)\)\)\)

where we used that the idea that the Exponential function is the inverse of the natural Log function.  Solving for the half-life using equation (12) yields

(Subscript[t, (1/2)])  =  
      Log [2]/λ

Now you might know

Log[2. ]

or you can have Mathematica do the calculation.  Equation (13) thus appears

\(\(\(\ \)\(t\_\(1\/2\)\  = \ \(0.6931471805599453`\/λ = \ 
        0.6931471805599453`\ τ\)\)\)\)

where equation (5) was used for the last equality.  Thus there is a simple relationship between the half-life (Subscript[t, (1/2)]), the decay time τ, and the decay constant λ.

Graphical Results:

It might help to look at a specific numerical example to see the expected results.  Suppose for example, the half-life of a particular isotope is 5 days.  Then using equation (14) we compute

λ  =  
      (0.693147/( 5 ))

and

τ  =  ((5 )/0.693147)

Using this numerical value for λ=0.138629 in equation (4) and graphing the result
for 100 radioactive atoms initially (an unrealistic example)

(Subscript[N, 0]) = 100;

Natoms[t_]  :=  
      (Subscript[N, 0]) (E^(- λ t))

Note we use the function Natoms[t] instead of N[t] because the N[t] function is reserved or builtin to Mathematica as the "Numerical Approximation" function.

Plot[Natoms[t], {t, 0, 20},  
      AxesLabel -> {"time t", "Natoms[t]"}]

One way to interpret this graph is to graph the Log[Natoms[t]] versus time since from equation (4) we expect

Log[Natoms[t]] = Log[(Subscript[N, 0])] - λ t

so the slope of the graph should be  - λ.=0.138629

<<  Graphics`Graphics`

ExpGraph = 
        Plot[Log[Natoms[t]], {t, 0, 20}, GridLines -> Automatic, 
          AxesLabel -> {"t", "Log[Natoms]"}];

Forming the slope numerically

(((2.5 - 3.9))/((15 - 5. )))

which is close to λ=0.138   A more accurate calculation yields

(Log[Natoms[15. ]] - Log[Natoms[5. ]])/(15.  - 5. )

The actual experiment involves the measurement of the radioactivity R[t] which from equation (3) appears

R[t_]  := - λ Natoms[t]

Plot[R[t], {t, 0,  20}, 
      AxesLabel -> {"time t", "R[t]"}]

Notice that R[t] is negative as expected.  Usually the absolute value | R[t]}| is recorded in measurements (the rate counts/sec is usually data collected as positive) and this graph would appear positive

Plot[Abs[R[t]], {t, 0,  20}, 
      AxesLabel -> {"time t", "|R[t]|"}]

Interpretation of the  Actual Decay Experiment:

You might expect (from the title) that we are going to measure the half-life (Subscript[t, (1/2)]) of a radioactive isotope in this laboratory.  We will proceed combining equations (3) and (4)

\(\(\(\ \ \)\(R[
        t]\  = \(-\ λ\)\ \ N\_0\ e\^\(\(-\ λ\)\ t\)\)\)\)

and using this at two different times (Subscript[t, 1]) and (Subscript[t, 2]).  We thus write

\(\(\(\ \ \ \)\(R[
        t\_1]\  = \(-\ λ\)\ \ N\_0\ e\^\(\(-\ λ\)\ t\_1\)\)\)\)

and

\(\(\(\ \ \ \)\(R[
        t\_2]\  = \(-\ λ\)\ \ N\_0\ e\^\(\(-\ λ\)\ t\_2\)\)\)\)

Divide equation (16) by equation (17) and obtain

\(\(\(\ \)\(R[t\_1]\/R[t\_2]\  = \ \(e\^\(\(-\ λ\)\ t\_1\)\/\(\(\ \)\(e\^\(\(-\ λ\)\ t\_2\)\)\) = \ e\^\(\(-\ λ\)\ \((\ t\_1 - \ t\_2\ )\)\)\)\)\)\)

or inverting

\(\(\(\ \)\(R[t\_2]\/R[t\_1]\  = \ e\^\(\(\ \)\(λ\ \((\ t\_1 - \ t\_2\ )\)\)\)\)\)\)

Take the natural logarithm of both sides

\(\(\(\ \)\(Log[
        R[t\_2]\/R[t\_1]]\  = \(λ \((\ 
            t\_1 - \ t\_2)\) = \(\(0.6931471805599453`\/t\_\(1\/2\)\) \((\ 
            t\_1 - \ t\_2\ )\)\(\ \)\)\)\)\)\)

where equation (14) was used.  Finally solving for the half-life

\(\(\(\ \)\(t\_\(1\/2\) = \ \(0.6931471805599453`\ \((\ t\_1 - \ t\_2\ )\)\)\/\(\(\ \)\(Log[R[t\_2]\/R[t\_1]]\)\)\)\)\)

The half-life  (Subscript[t, (1/2)]) can be obtained from two (or more) measurements of the radioactivity R[t] at different times (Subscript[t, 1]) and (Subscript[t, 2]).

Data Collection-Lab Notebook:

Below is a list of some items you should include in your lab notebook and your lab report.  You might think of other additional items as well.  

The count rate is the

Count Rate= ((number of counts)/(observation time interval T))

The half-life of Barium-137 isotope in this experiment is the order of a few minutes so you should NOT count for very long times.  If your observation time interval is too long how would this effect your experiment?  Do some trial runs (using pairs of data points) to get a rough idea of the half-life in this experiment.

There is always some background radiation due to Cosmic rays, radioactive particles in building materials like granite, etc. and unless you do the experiment in a lead shielded room the effect of the background radiation must be included.  This background radiation obviously does not come from the radioactive material Barium-137 used in this lab and measurements of the background radiation should be done before you get your sample.

It is plausible that the background radiation should be subtracted from the measured sample count rate to get an  effective count rate.  Why?

Experiment Date [mm/dd/yr]:                                     

Time interval of observations [sec]:  T =                              

Background Radiation [Counts/sec]:  B =                                  

The "elapsed time" in the table below is the time at the beginning of a particular  measurement  of the count rate from the start of the experiment.  The elapsed time should be taken with a stop watch or your writst watch.

RAW DATA TABLE

({{Elapsed Time

          [sec], # Counts,  Count rate   A

          [counts/sec] }, {, , }, {, , }, {, , }, {, , }, {, , }, {, , }})

You might want to collect more data points than indicated in the table above.

The data can be transformed using the following data table.   The "effective count rate" 
R[t] = A-B is the same as the "count rate corrected for the background radiation".

				 DATA ANALYSIS

({{Elapsed Time

          [sec], Effective

          Rate R[t], Variance

          σ  =  (Sqrt[R])[t],     Log[
            R]    ,    Log[R + σ]  }, {, , , , }, {, , , , }, { , , , , }, {, , , , }, {, , , , }, {, , , , }})

Laboratory Report

Some things you should include in your lab report; you might think of others.

1. Use Mathematica to graph the effective count rate R[t] versus time t.  On a separate plot, graph the Log R versus the time.  Use Mathematica (with the PlotStyle option Axes-Label) to label the axes of the graph.   Also include the option for GridLines.

2. Use pairs of data points to calculate the half-life for this experiment.  Calculate the average of the half-lives for the pairs of data points and the standard deviation.

3.Find the best fit straight line through the data using Mathematica (using the Fit function).  Determined the half-life, decay rate, and decay constant for this experiment.

4. Use Mathematica to graph the data together with error bars indicating the experimental uncertainty.  Do this for both R versus t and LogR versus t.  Compare the size of the experimental uncertainty of the data in both graphs.

5. Determine the experimental uncertainty in the half-life you reported in step #3 above.
6. Recall you did an experiment on the statistics of counting where you calculated the average and standard deviation after N (20) measurements.  Can you do this sort of experiment with the Ba-137m sample?  Explain your answer in detail (why or why not).

7. Discuss the length of the Observation Time Interval T  in equation (22).  Could you take T as 5 minutes long?  Could you take T as 3 minutes long?  Could you take T as 1 minute long?  Could you take T as 30 seconds long? Could you take T as 10 seconds long?  Could you take T as 1 second long?  What is the smallest value T could be in this experiment?  What is the largest value T could be in this experiment? Explain your reasoning. HINT:  Recall the discussion of "resolving time" and also the discussion of "half-life".

The Preparation of the Ba 137m Sample

The Barium-137 isotope has (as said before) a half-life of the order of minutes and since the half-life is so short most of a sample disappears after about 30 minutes.  For this reason, the Ba-137 sample must be prepared in the laboratory. 
	The Ba-137 sample is prepared from an isogenerator containing Cs-137 (Cesium) bound on an ion exhange medium.  Roughly 95% of the Cs-137 atoms decay into Ba-137m emitting a β particle (beta).  The Cs-137 has a half-life of roughly 30 years. The other 5% of the Cs-137 decay directly into the stable isotope of Ba-137.  
	The isotope Ba-137m is the important isotope created in the isogenerator and interests us here. The isotope Ba-137m decays via a γ ray (gamma) into the isotope Ba-137 in a few minutes (we do not want to say how long as this is what is measured in this experiment).
	The isotope Ba-137m is eluted (washed) out of the isogenerator with EDTA liquid with drops appearing in the metal plachett you have as part of your radioactive sample.  Rubber gloves should be used when operating the isogenerator to avoid contamination.  Also, care should be taken to avoid spills of the liquid.  Again, after 30 minutes most of the Ba-137 has decayed into Ba-137 a stable isotope and after this time little above background radiation is emitted.

An Example of Graphing Data

Suppose you have collected the following data where t is time in minutes and R[t] is counts/min.:

({{t, |R[t]|}, {0.5, 3.9}, {1.0, 3.0}, {1.5, 2.4}, {2.0, 1.8}, {2.5, 1.4}, {3.0, 1.1}, {3.5, 0.87}, {4.0, 0.67}, {4.5, 0.53}, {5.0, 0.41}, {5.5, 0.32}, {6.0, 0.25}, {6.5, 0.19}, {7.0, 0.15}})

You can give the data above a name (for example, "DecayData") using

DecayData=({{0.5, 3.9}, {1.0, 3.0}, {1.5, 2.4}, {2.0, 1.8}, {2.5, 1.4}, {3.0, 1.1}, {3.5, 0.87}, {4.0, 0.67}, {4.5, 0.53}, {5.0, 0.41}, {5.5, 0.32}, {6.0, 0.25}, {6.5, 0.19}, {7.0, 0.15}})

The output from Mathematica is a list of data which may be graphed using the ListPlot function.  The simplest graph appears

ListPlot[DecayData]

If you want to include axis labels, then

\(ListPlot[DecayData, 
      AxesLabel -> {"time t", "|R[t]|"}, 
PlotStyle -> PointSize[0.02]]\)

We also used a PlotStyle option called PointSize to increase the size of the data points to 2% of the width of the graph.

The Log-linear graphing function is not usually automatically loaded by Mathematica so to include this package we use

<< Graphics`Graphics`

\(LogLinearGraph = 
      LogListPlot[DecayData, 
        AxesLabel -> {"time t", "Log|R[t]|"}, 
PlotStyle -> PointSize[0.02], 
        GridLines -> Automatic]\)

The option GridLines->Automatic draws the lines of the graph.

The half-life can easily be computed using equation (29)

(Subscript[t, (1/2)]) =  ((0.693 ( (Subscript[t, 1]) -  (Subscript[t, 2]) ))/( Log[R[(Subscript[t, 2])]/R[(Subscript[t, 1])]]))

For example, using a pair of data points

(Subscript[t, (1/2)]) =  ((0.693 ( 2.0 - 2.5 ))/( Log[(1.4/1.8)]))

Note that the natural Log of the R[ (Subscript[t, 1])] at (Subscript[t, 1]) = 2.0  is

Log[1.4]

If you look at the Log-linear graph above what you see above  (Subscript[t, 1]) = 2.0  is the data point label 1.4 so Log-linear graph paper is a bit peculiar.  The actual numerical data is graphed on paper that has a Log scale.

An alternative to the above proceedure might make things clearer.

LogDecayData=({{0.5, Log[3.9]}, {1.0, Log[3.0]}, {1.5, Log[2.4]}, {2.0, Log[1.8]}, {2.5, Log[1.4]}, {3.0, Log[1.1]}, {3.5, Log[0.87]}, {4.0, Log[0.67]}, {4.5, Log[0.53]}, {5.0, Log[0.41]}, {5.5, Log[0.32]}, {6.0, Log[0.25]}, {6.5, Log[0.19]}, {7.0, Log[0.15]}})

\(ListPlot[LogDecayData, 
      AxesLabel -> {"time t", "LogR[t]"}, 
PlotStyle -> 
        PointSize[0.02], 
\ \ \ GridLines -> 
        Automatic]\)

Now the numerical value above  (Subscript[t, 1]) = 2.0 has the correct numerical value

Log[2.0]