Nuclear Physics Tutorial 5 - Exponential Law of Decay


Radioactive Decay

Decay Constant

Exponential Decay
Nuclear Activity

Uses of Half-Life

Radioactive Dating
Rock Dating
Calculus Treatment


Radioactive Decay

It is important to understand that radioactive decay is entirely a random and unpredictable process.  If we look at any one nucleus, it might decay in ten seconds or ten million years.  There is no way of telling when the decay will happen, and there is certainly no way of speeding up the process.  Remember that chemical reactions involve the outer shell electrons; radioactive decay involves the nucleus.


However when we have many millions of millions of nuclei, we can apply statistical models of probability.  The radioactive decay law states that:


 The probability per unit time of a nucleus decaying is a constant, independent of the time.


The rate of disintegration of any nuclide at any time is directly proportional to the number of atoms left at the time:


This is calculus notation for rate of change in number.


The minus sign tells us that N decreases as time increases.



Decay Constant

There is a constant of proportionality called the radioactive decay constant, which is given the physics code l (Greek letter ‘l’, not to be confused with wavelength), and has the units s-1. 


So we can write:


The radioactive decay constant is defined as:


 the fraction of the total number of nuclei present that decays per unit time, provided that the time interval is small. 


For nuclides with relatively long half-lives, the decay constant will have units “per second”.  However some very unstable nuclides decay in microseconds, so the decay constant would have to be “per nanosecond”.


Worked Example

0.25 kg radium-226 emits alpha particles at a measured rate of 9.0 × 1012 s-1.  What is the decay constant of radium?  (No of atoms in a mole = 6.0 × 1023)


Work out the number of particles:

1 mole of radium atoms has a mass of 0.226 kg.


0.250 × 6.0 × 1023 = 6.64 × 1023 atoms



We know that the rate of decay is 9.0 × 1012 s-1. 


So we use DN/Dt = -lN


                        - 9 × 1012 s-1 = -l × 6.64 × 1023

l = 1.36 × 10-11 s-1

(The minus sign indicates a decay)


The unit Becquerel (Bq) is often used.  1 Bq = 1 count per second.


Question 1

A sample of living material contains carbon 14 with an activity of 260 Bq kg-1.  What is the decay constant? (The fraction that is made of carbon-14 is 1.4 × 10-12)




Exponential Decay

Over longer periods of time, the relationship above does not hold.  It can be shown by calculus methods (which we will look at later), that the decay follows the relationship:



[N – no of nuclei; N0 – original number of nuclei; e – exponential number, 2.718…; l - decay constant (s-1); t – time (s)]


This relationship is described as an exponential decay and the graph looks like this:



We should note the following:

  • The rate of decay represents the number of atoms remaining.  So we can use this graph for representation of the count rate, or ionisation current.

  • The rate of decay is the number of disintegrations per second, or the activity.  It is measured in Becquerels  (Bq).

  • 1 Bq = 1 disintegration per second.


Notice that the horizontal axis is calibrated in periods of time in which the decay goes from an initial value to half that value.  This period is called the half-life.  The formal definition of half-life is the time taken for the activity of a sample to decrease to half some initial value.  After 1 half life the activity is 50 %, between 1 and 2 half lives it falls from 50 % to 25 % and so on.  If we have a whole number of half-lives, we can do an iterative calculation of the fraction remaining, i.e. ½, ¼, 1/8, etc.


We can relate the half-life to the decay constant:


By definition


The formula is:

 N = N0e-lt. 


Rearranging gives us


T1/2 (pronounced "T-half") is the physics code for half-life.  So we can write:




Now take natural logarithms:





The minus signs cancel out:

So we can finally write:



This will work for any units of time, although we need to be consistent.


Maths Note

To work with these relationships, you must be familiar with the concept of logarithms (a way of expressing a number as a power of 10 or e).  You will see logarithms expressed as ‘lg’ or log10, which means log to the base 10, or ‘ln’ or loge, log to the base e.  The number e is the exponential number, 2.718...


The latter are known as natural logarithms, which are at the heart of exponential functions.




Worked Example

A radiographer has calculated that a patient is to be injected with 1 ´ 1018 atoms of iodine 131 to monitor thyroid activity.  The half-life is 8 days. 


(a)    the radioactive decay constant

(b)   the initial activity

(c)    the number of undecayed atoms of iodine 131 after 24 days.

(d)   The total activity after 3 days.


(a) We need to use:

t1/2 = 0.693



    We need to convert the 8 days into seconds.


l =        0.693            

         8 dy ´ 86400 s dy-1


= 1.00 ´ 10-6 s-1



(b) Use:


      DN/Dt = 1.00 ´ 10-6 s-1 ´ 1 ´ 1018 = 1 ´ 1012 Bq


(c) 24 days is 3 half-lives.  Therefore the number atoms remaining undecayed is 1/8 of the original.

N  =  1 ´ 1018 ¸ 8 = 1.25 ´ 1017

(d) 3 is not so easy.  We use A = A0e-lt


Þ A = 1 ´ 1012 Bq ´ e-(1.00 ´ 10-6 s-1 ´ 3 ´ 86400s)


Þ A = 1 ´ 1012 Bq ´ e-(0.2592)

        =  1 ´ 1012 Bq ´ 0.772 = 7.72 ´ 1011 Bq.

[Alternative method] If you are not so confident in the use of the natural logarithm, you can work out the number of half-lives a particular time gives.  In this case 3 days = 3/8 of a half life.


Activity = 1 ´ 1012 ´ 1/23/8 = 1 ´ 1012 ´ 1/1.30 = 7.7 ´ 1011 Bq


This is a perfectly valid alternative approach.


Question 2

What is the half life of radon-226?  l = 1.36 × 10-11 s-1




Nuclear Activity

Often we write in terms of the activity of a source.  This is the number of disintegrations per second, rather than the number of undecayed nuclei.  The activity can be written as:



If we know the activity of the sample, A, and the reference activity, A0, we can use the equation:




Uses of Half-Life

The half-life has important implications for the storage of radioactive waste.  Radioactive waste is some of the nastiest muck known to man, so it has to be stored carefully.  Isotopes with a short half-life have intense activities, whereas those with a long half-life have lower activities, but it takes much longer for the activity to decay to a reasonable level.  Either way, it’s not very nice.


Even when there have been several half-lives, there remains quite a considerable activity.  If in our example above the activity had decayed to 1/1000th of its original value, we would still have 109 disintegrations per second, which is quite a lot.


Another use of the decay equations is in radioactive dating.  Isotopes used are carbon-14, rubidium-87, and hydrogen 3.


Question 3

Strontium-90 is a beta emitter.  It is one of the radio-nuclides found in the fall out from an atomic bomb explosion.  It can be absorbed into the bone.  It emits beta particles and has a half life of 28 years.  What is the time needed for the activity to fall to 5 % of the original?


Question 4

A GM tube placed close to a radium source gives an initial average corrected count rate of 334 s-1

(a)     The GM tube detects 10 % of the radiation.  What is the initial activity?

(b)    Initially there were 1.5 ´ 109 nuclei in the sample.  What is the decay constant?

(c)     What is the half life of the radium in days?




Radioactive Dating

Carbon-14 has a half-life of 5700 years.  It decays by beta minus decay by this equation:



Carbon-14 is made by the collision of an energetic neutron with a nitrogen nucleus, which results in the ejection of a proton.  The energetic neutron is formed by cosmic ray activity.  Therefore carbon-14 is being made all the time. The rate of decay is balanced by the rate of formation, so the amount of carbon-14 remains almost constant.


The proportion of carbon-14 in living material is very low.  For each gram of carbon, the count-rate is about 140 min-1, 2.33 Bq.  The count rate is therefore 2300 Bq kg-1.  This is the reference point we use for 100 % activity.


If we know the activity of the sample, A, and the reference activity, A0, we can use the equation:


We can take natural logs to give us:


And this rearranges to:

-lt = ln A - ln A0


Since ln A0 is bigger than ln A, the right hand side of the equation is negative.  The left hand side is negative, so the two negative signs cancel.  This ensures that the time is positive.  We can only go forward in time.


To work out the decay constant, we need to use the equation:


In this case the half life is 5700 years, which we have to convert to seconds.  You do know how to do this, don't you?


Worked example

In an archaeological dig, remains are found of a warrior who was buried using a wooden boat for a coffin.  A 0.50 gram sample of the boat is found to have an activity of 3800 counts per hour.  What is the age of the boat? 


The count rate for living material is 2300 Bq kg-1.

The activity of carbon-14 in living material is 2300 Bq kg-1 and the half-life of carbon-14 is 5700 years.


Find out the activity of the sample in Becquerel.

A = 3800 h-1 ÷ 3600 s h-1 = 1.056 Bq


Convert this to activity per unit mass.


A = 1.056 s-1 ÷ 0.5 × 10-3 kg = 2111 Bq kg-1


Now we need to work out the decay constant.

l = 0.693 ÷ (5700 y × 365 dy × 24 h × 3600 s) = 3.86 × 10-12 s-1


Now use:

-lt = ln A - ln A0


-3.86 × 10-12  s-1 × t = ln (2111 Bq kg-1) - ln (2300 Bq kg-1)


-3.86 × 10-12 s-1  × t =7.655 - 7.741 = -0.0857

Notice that the minus signs cancel:

t = 0.0857 ÷ 3.86 × 10-12 = 2.2 × 1010 s (700 years)


Normally you would leave your answer in seconds.  The answer is given to 2 significant figures as the "activity in living material" data item is to 2 significant figures.


The calculation above assumes that the level of carbon-14 is the same in the year 1300 as it is now.  If the cosmic ray activity were less then, the amount of carbon 14 would be reduced.  The warrior's remains would be more recent.



Rock Dating

Carbon-14 dating is only useful for objects under about 50000 years old.  It's no good for dating rocks.  The isotope potassium-40 is useful.  There are two modes of decay for potassium-40:

  1. Normal beta minus decay;

  2. Electron Capture.


For beta minus, the decay is:


The half life for this is about 1250 million years.



For electron capture, the electron is captured from the inner shell:



About 1 in 8 decays is of the latter type.  So out of 9 potassium-40 atoms, 8 will decay to calcium, while 1 decays to argon.  The problem with the decay to calcium is that calcium is as common as muck, so we wouldn't be able to tell which calcium atom came from the potassium decay, and which came from other sources.


However, Argon is an inert gas, and does not react with any crystals.  Argon atoms remain trapped in the rock, so if we knew how many argon atoms there were, we could work out how many potassium atoms had decayed.  So if we counted 12 × 106 argon atoms, we could say that these resulted from 108 × 106 potassium 40 atoms.  So if we have 500 × 106 potassium 40 atoms left, we can say that we would have started off with 108 × 106 + 500 × 106 = 608 × 106 atoms.


Question 5

If potassium-40 has a half-life of 1250 million years, what is the age of the sample of rock discussed in the paragraph above?



Calculus Treatment of Radioactive Decay

You may have seen the calculus derivation of the exponential equation for discharge of a capacitor.  Exponential decay of unstable nuclei is similar.


Maths note

A differential equation is where there is a function combined with a derivative, for example:


Differential equations describe the way that values change when a system changes by a constant fraction.  The solution is not always easy, but a mathematical technique called the separation of variables is useful.  To do this we:

1. Multiply both sides by dt;

2. Divide both sides by x.


This gives us:

Then we need to integrate both sides to get rid of the derivative. 


The problem here is that the power rule doesn't work:


which is impossible.  Instead there is another rule:

So we do the first integration:

And the second (using the power rule since k = kx0):


C and D are constants of two different values.  We can, for the sake of this argument, roll the two constants into one by writing:


a = D - C


So we rewrite the equation as:

ln (x)  = kt + a


Now we bring in the exponential number e (= 2.718...) on both sides to get rid of the natural logarithm (ln is a logarithm to the base e).    The terms become exponents.  (An exponent is a power to which a base number is raised, i.e. eexponent.)  We rewrite the equation as:


e(ln(x)) = e(kt + a) = ekt × ea


The term ea is simply a constant.  Since e(ln(x)) = x, so we now write the equation as:



Note that when two numbers add up, they multiply when they become exponents.  This is the converse of when logarithms of two multiplied numbers add up.


If the constants C and D are zero, ea = e0 = 1.  Therefore the expression becomes:



We know that:


We can rewrite this in calculus notation:


We can apply the method of separation of variables to the differential equation above.  In this case the term l is the fractional amount by which the charge decays.  So to solve this differential equation, we multiply by dt and divide by N.  So our equation becomes:

Now we integrate both sides:


On the right hand side, there doesn't appear to be anything to integrate with reference to t.  The term l is a constant.  However we can rewrite the equation as:



The trick here is that 1 = t0 so we can rewrite the equation further:


So we can carry out the integration, applying the rules in the Maths Note above:



We can combine the constants by saying:

a = D - C


So we now write:


To get rid of the natural logarithm, we make both sides an exponent of the exponential number e (=2.718...).  Now eln(N) = N, so our equation now becomes:


Now when t = 0, we can see that:


So far we have haven't attempted to define what ea is.  But we can easily see that it is number at time 0.  This has the physics code N0, so we can rewrite the equation as:


This is our final result.