The document explores the mathematical principles governing nuclear decay, focusing on deriving equations for chained decay. It emphasizes the differences between simple and multi-stage decay and provides detailed explanations suitable for young mathematicians. The exploration includes modeling techniques and practical applications, demonstrating how to predict the behavior of radioactive substances over time.

1. 1
Mathematical Exploration
NUCLEAR DECAY – A MATHEMATICAL PERSPECITVE
Exploring the Mathematical principles governing nuclear decay
Deriving the equation for chained decay
Erik Faust
Candidate Session Number: 003056 0008
British International School of Cracow, Poland
Word Count: 3208
Exam session: May 2015

2. 2
Table of contents
1. Introduction 2
2. Radioactive Decay as a Sequence 3
3. The Decay Equation 5
4. Products 10
5. Multiple Stage Decay 10
6. Modelling using Excel 15
7. Conclusion 20
8. Bibliography 21
I) Introduction
Radioactivity as a phenomenon is often misunderstood: if one says ‘Radioactive’, most people will
think about disastrous electrical plants, dangerous bombs and other forms of life-threatening details.
In my native Germany, members of the Green party have been campaigning for a decade to put an
end to nuclear energy. Only few think of the useful aspects of this unique actuality, although
radiotherapy is most promising of tools in the fight against cancer, and radioactive dating allows us
to identify the age of any historical item. But even fewer people see radioactivity as the natural
process that it actually is: A spontaneous mechanism, in which one nucleus decays into another. As
an aspiring Physicist and Engineer, Radioactivity is one my favourite topics in the realm of science. I
am fascinated at how we are able to predict exactly how many Nuclei will decay in a certain amount
of time, but not say for certain which Nuclei exactly will do so.
While I am familiar with the application and physical nature of radioactivity, my knowledge of the
mathematical principles governing it is limited. I know how to apply the equation of simple, two-
stage decay, and while the use of this formula is almost intuitive to me, in the course of this
exploration I will attempt to gain enough skill to be able to explain exactly how it came to be.
However, the decay of one ‘parent’ nucleus into another is not the whole story: often, radioactive
isotopes (This is the term used for unstable, decaying nuclei) form decay chains. In these chains, the
nucleus that was formed decays further to create another daughter nucleus. The IB Physics Higher
Level syllabus does not cover the mathematical principles that allow us to predict this type of decay,
but I am personally interested in how this second type of decay will vary from the first. Simple
radioactive decay can be described as follows: we start with a fixed number of parent nuclei, and
after a certain amount of time (the so-called half-life), half of the nuclei will have decayed. After
another half-life, again half of these nuclei will have decayed. Since in a decay chain, the number
intermediate nuclei is simultaneously affected by the decay from the parent and the decay to the
product, we can not apply this assumption any longer. This exploration will firstly derive a set of
equations to describe the number of parent, intermediate and daughter nuclei, and then investigate
the effect of changing the half-life of each of the nuclei.

3. 3
All existing papers on this topic explore and explain the derivation of these equations from a second
year university level or higher. Several Mathematical principles and certain simplifying steps are
being presumed at previous knowledge. This paper will approach this topic from a different angle: all
steps will be expressed in detail, in order to explain the simplest of chained decay equations for
young, aspiring Mathematicians to explore.
II) Radioactive Decay as a Sequence
Physically, radioactive decay is governed by the principle of a Half-Life. Half-Life, or 𝑇1/2 is defined as
the time at which 50% of the initial number of particles, 𝑁0, will have decayed. Let us assume that
𝑁0 = 100. Let us express the number of particles left in the sample, 𝑁 𝑇1/2
, as a sequence, where the
terms are determined by the number of Half-Life times that have already passed.
Let the first term, 𝑈1, be equal to 100. After one Half-Life,
𝑁 𝑇1/2
= 𝑈2=
𝑁0
2
=
100
2
= 50
After a next Half-Life, this number will be halved again. Therefore,
𝑈3=
𝑈2
2
=
50
2
= 25
We can therefore observe that the Term-To-Term rule for this Sequence is to divide by two. This
yields the Sequence
Figure 1
Number of
Half Lifes
NT1/2
Number of
Particles
1 100
2 50.0
3 25.0
4 12.5
5 6.25
6 3.13
7 1.56
8 0.781
9 0.391
10 0.195
The number of particles in the table is given to three significant figures
The nth Term, 𝑈 𝑛 is mathematically given by
𝑈 𝑛 = 𝑈1 × 𝑟 𝑛−1
Where 𝑈 𝑛 is the Nth Term, 𝑈1 is the First Term, r is the common ratio, and n is the term number. The
common ratio is 0.5, as each term is multiplied by 0.5 to yield the next

4. 4
𝑟 =
𝑈 𝑛
𝑈 𝑛−1
⇔
𝑈2
𝑈1
=
50
100
= 0.5
Therefore,
𝑈 𝑛 = 𝑈1 × 𝑟 𝑛−1
= 100 × 0.5 𝑛−1
= 100 ×
0.5 𝑛
0.5
Where n is the time passed in units of Half Life.
Graphically represented, we can utilize Microsoft Excel to represent this Data in form of a dot
diagram
If we use a logarithmic scaling on the y-axis
0
20
40
60
80
100
120
0 2 4 6 8 10 12
Number
Time (Halflife)
Figure 2: Number of Particles vs time

5. 5
This may lead to the hypothesis that the number of particles, 𝑁(𝑡) can be expressed as an
exponential expression, as seemingly the relationship between 𝑁(𝑡) and 𝑡 is as follows
𝑙𝑜𝑔10 𝑁 = −𝑘𝑥 + 𝑙
Where k and l are positive constants.
But a sequence only yields values for 𝑈 𝑛, where the time is a multiple of the half-life. What if we
want to calculate the number of particles left after a time 𝑡, where 𝑡 is not a multiple of 𝑇1
2
?
III) The Decay Equation
In order to be able to calculate the number of particles left after any time 𝑡, we need to find an
expression for 𝑁(𝑡), where 𝑡 ∈ 𝑅+
, or 𝑡 = 0
From experimental data, it is known that the rate of decay is proportional to the number of particles
present in the sample. Mathematically this can be expressed as:
𝑑𝑁
𝑑𝑡
∝ 𝑁
Where
𝑑𝑁
𝑑𝑡
is the rate of change of the number of particles with respect to time and 𝑁 is the number
of particles present in the sample. If we define a real constant k, we can say that
𝑑𝑁
𝑑𝑡
= k × 𝑁
0.1
1
10
100
0 5 10 15 20 25
Number
Time (T(1/2))
Figure 3: Number of Particles with time

6. 6
In Physics however, constants are generally taken to have positive values. We know that the rate of
decay is proportional to the number of particles present in the sample. As this number 𝑁decreses
with time, which we can see from the Sequence we defined initially, the constant k would take a
negative value, or mathematically𝑘 < 0. For simplicity, we can define another constant 𝜆, which
takes values such that
−𝑘 = 𝜆
So that 𝜆 may take values greater than 0 (𝜆 > 0)
Therefore
𝑑𝑁
𝑑𝑡
= −𝜆 × 𝑁
Let us divide both sides by 𝑁. This is only possible, if 𝑁 ≠ 0. However, if 𝑁 were equal to 0, this
would mean there were no particles in the sample, which mean that no radioactive decay could
occur.
𝑑𝑁
𝑑𝑡
×
1
𝑁
= −𝜆
To separate the variables in this equation, let us multiply both sides by 𝑑𝑡, which yields
1
𝑁
× 𝑑𝑁 = −𝜆 × 𝑑𝑡
Now, we may take the indefinite integral of either side
∫
1
𝑁
𝑑𝑁 = ∫ −𝜆 𝑑𝑡
ln(𝑁) + 𝑐1= -𝜆 ×t +𝑐2
ln(𝑁) must be such that 𝑁 > 0, as the logarithm function is only defined for values of 𝑁 that are
greater than 0. As 𝑁 describes the number of particles, it is always larger than 0, and if 𝑁 = 0, no
decay would occur. Hence, antidifferentiaton is possible.
Where 𝑐1 and 𝑐2 are two arbitrary constants. Let us define 𝑐3which takes values such that
𝑐3 = 𝑐2 − 𝑐1
If we substitute 𝑐3
ln(𝑁) = 𝜆 ×t +𝑐3
We know that if two numbers are equal
𝑎 = 𝑏
Where a and b are any real numbers, laws of exponents state that the following is also true
𝑒 𝑎
= 𝑒 𝑏

7. 7
Therefore, we can raise e to the power of both sides of the equation, which yields
𝑒ln(𝑁)
= 𝑒−𝜆×t +𝑐3
And as the natural logarithmic function is the inverse of the exponential function, this is equivalent to
𝑁 = 𝑒−𝜆×t +𝑐3
Using laws of exponents
𝑁 = 𝑒−𝜆×t
× 𝑒 𝑐3
As 𝑒 and 𝑐3 are both constants, let us define a fourth constant, 𝑐4, which is such that
𝑐4 = 𝑒 𝑐3
And substitute into the equation
𝑁 = 𝑒−𝜆×t
× 𝑐4
As 𝑁 is now expressed in terms of 𝜆, 𝑐4 and 𝑒, which are all constants, and t, which is a variable, we
can define a function 𝑁(𝑡) as follows:
𝑁(𝑡) = 𝑐4 × 𝑒−𝜆×t
Where 𝑁 > 0 and 𝑡 ≥ 0, and both take real values. The range of 𝑁(𝑡) depends on the number of
particles initially present, whereas the domain encompassed all real non-negative numbers.
To evaluate the constant 𝑐4, let us substitute 𝑡 = 0. We previously defined 𝑁0, the initial number of
particles present in the sample. Mathematically
𝑁(0) = 𝑁0
Let us substitute 𝑡 = 0
𝑁(0) = 𝑒−𝜆×0
× 𝑐4
𝑁(0) = 𝑒0
× 𝑐4
𝑁(0) = 1 × 𝑐4
𝑁(0) = 𝑐4
Substituting 𝑁0
𝑁0 = 𝑐4
Therefore, the function 𝑁(𝑡) can be defined as
𝑁(𝑡) = 𝑁0 × 𝑒−𝜆×t
Where 𝑁0 is the number of particles initially present in the sample, 𝜆 is an arbitrary constant taking a
positive value, t refers to the time elapsed since the beginning of decay and is such that 𝑡 ≥ 0 and 𝑁
is the number of particles left after time t.

8. 8
If we wanted to find 𝜆, we may substitute experimental values into the decay equation. Let us
assume that, as in the initial example, 𝑁0 = 100. The time taken for half the particles to decay was
found to be 2 years. Hence
𝑁0 = 𝑁(0) = 100
And
𝑁(2) = 𝑁 𝑇1/2
= 𝑈2=
𝑁0
2
=
100
2
= 50
Which yields the equation
𝑁(𝑡) = 𝑁0 × 𝑒−𝜆×t
50 = 100 × 𝑒−𝜆×2
Dividing both sides by 100
0.5 = 𝑒−𝜆×2
Taking the natural logarithm
ln(0.5) = ln(𝑒−𝜆×2
)
ln(0.5) = −2 × 𝜆
𝜆 = −
ln(0.5)
2
In this case, 𝜆 could therefore be estimated to three significant figures to be
𝜆 = −
ln(0.5)
2
≈ 0.347
So, for the worked example, the decay equation can be approximated as
𝑁(𝑡) ≈ 100 × 𝑒−0.347×t
Where 𝑡 is given in years. From this equation, we can calculate the number of particles left at any
point of time within the domain of the function.
Once more, we can plot this information on a graph

9. 9 And with logarithmic scaling
y = 100e-0.347x
0
20
40
60
80
100
120
0 2 4 6 8 10 12 14 16 18 20
Number
Time (years)
Figure 4: Number of Particles with respect to
time
y = 100e-0.347x
0.1
1
10
100
1000
0 2 4 6 8 10 12 14 16 18 20
Number
Time (years)
Figure 5: Number of Particles with respect to
time

10. 10
IV) Products
The paper so far has made it seem, as if the radioactive substance just ‘disappeared’ into nothing, as
radioactive products have not been mentioned. Mathematically speaking, this is a fine assumption,
as it allows us to predict the amount of so-called parent nuclei in our sample. However, in reality,
such a parent nucleus decays into a daughter nucleus.
We know now that
𝑁(𝑡) = 𝑁0 × 𝑒−𝜆×t
For the parent nucleus. We also know that one parent nucleus decay into one daughter nucleus. 𝑁2,
the number of daughter nuclei, assuming 𝑁2(0) = 0 , can therefore be expressed as
𝑁2(𝑡) = 𝑁0 − 𝑁0 × 𝑒−𝜆×t
𝑁2(𝑡) = 𝑁0 × (1 − 𝑒−𝜆×t
)
V) Two stage decay
In nature however, this product may also be a parent for another daughter nucleus. This makes the
whole problem more complicated. Therefore, let us first make sure all variables are well defined.
Let variables be as follows
Figure 6
Type of Nucleus First Parent Intermediate Daughter
Number of Particles 𝑁1 𝑁2 𝑁3
Decay constant 𝜆1 𝜆2 𝜆3
Reference Symbol A B C
And the decay occurs as follows
𝐴 → 𝐵 → 𝐶
The decay of 𝐴 will still be given by the equation
𝑁1(𝑡) = 𝑁1(0) × 𝑒−𝜆1×t
There are two factors affecting the rate of change of the intermediate 𝑁2, the growth due to 𝐴
decaying and the loss due to 𝐵 decaying to form 𝐶. Mathematically, we can express this as
𝑑𝑁2
𝑑𝑡
= 𝜆1 × 𝑁1 − 𝜆2 × 𝑁2(𝑡)
We can substitute for 𝑁1
𝑑𝑁2
𝑑𝑡
= 𝜆1 × 𝑁1(0) × 𝑒−𝜆1×t
− 𝜆2 × 𝑁2(𝑡)
Now lastly, we know that 𝐶 is growing with the same rate at which 𝐵 decays, which means that

11. 11
𝑑𝑁3
𝑑𝑡
= 𝜆2 × 𝑁2(𝑡)
Coming back to
𝑑𝑁2
𝑑𝑡
= 𝜆1 × 𝑁1(0) × 𝑒−𝜆1×t
− 𝜆2 × 𝑁2(𝑡)
We can add to both sides 𝜆2 × 𝑁2
𝑑𝑁2
𝑑𝑡
+ 𝜆2 × 𝑁2(𝑡) = 𝜆1 × 𝑁1(0) × 𝑒−𝜆1×t
And we multiply left and right side by 𝑒 𝜆2×t
𝑒 𝜆2×t
×
𝑑𝑁2
𝑑𝑡
+ 𝑒 𝜆2×t
× 𝜆2 × 𝑁2(𝑡) = 𝜆1 × 𝑁1(0) × 𝑒−𝜆1×t
× 𝑒 𝜆2×t
Now we may simplify
𝑒 𝜆2×t
×
𝑑𝑁2
𝑑𝑡
+ 𝑒 𝜆2×t
× 𝜆2 × 𝑁2(𝑡) = 𝜆1 × 𝑁1(0) × 𝑒(𝜆2−𝜆1)×t
This is the step in the derivation of Batemans equations, which challenged me personally most. It
requires quite an amount of being able to predict a simplification that will be helpful later.
We notice that the right hand side of the equation is given in terms of terms that we already know:
𝜆1, 𝜆2, and of course t which is the random variable as well as the constant 𝑁1(0). 𝑁2 is what we
eventually want to make the subject of the equation. This leaves the derivative expression
𝑑𝑁2
𝑑𝑡
as
what we want to eliminate.
Now, we need to make several conclusions about said expression, in order to simplify it
𝑒 𝜆2×t
×
𝑑𝑁2
𝑑𝑡
+ 𝑒 𝜆2×t
× 𝜆2 × 𝑁2(𝑡)
Firstly, it is a sum of two products,
𝑒 𝜆2×t
×
𝑑𝑁2
𝑑𝑡
As well as
𝜆2 × 𝑒 𝜆2×t
× 𝑁2(𝑡)
If we compare the two, we notice that
𝑑
𝑑𝑡
𝑁2(𝑡) =
𝑑𝑁2
𝑑𝑡
Furthermore

12. 12
𝑑
𝑑𝑡
𝑒 𝜆2×t
= 𝜆2 × 𝑒 𝜆2×t
If we let
𝑁2(𝑡) = 𝑢
𝑒 𝜆2×t
= 𝑣
The expression can be written as
𝑒 𝜆2×t
×
𝑑𝑁2
𝑑𝑡
+ 𝑒 𝜆2×t
× 𝜆2 × 𝑁2(𝑡) = 𝑣 ×
𝑑
𝑑𝑡
(𝑢) + 𝑢 ×
𝑑
𝑑𝑡
(𝑣)
This is equivalent to the product rule, which states
𝑑
𝑑𝑡
(𝑢 × 𝑣) =
𝑑
𝑑𝑡
(𝑢) × 𝑣 +
𝑑
𝑑𝑡
(𝑣) × 𝑢
Our goal was to eliminate
𝑑𝑁2
𝑑𝑡
, which in terms of 𝑢 and 𝑣 is given by
𝑑
𝑑𝑡
(𝑢)
In the expression
𝑑
𝑑𝑡
(𝑢 × 𝑣) this differential term is not present anymore
Since our expression can be formulated in the form of the product rule, this implies that
𝑒 𝜆2×t
×
𝑑𝑁2
𝑑𝑡
+ 𝑒 𝜆2×t
× 𝜆2 × 𝑁2 =
𝑑
𝑑𝑡
(𝑒 𝜆2×t
× 𝑁2(𝑡))
Confirmation:
We can try to do this calculation in reverse to verify whether we these calculations were conducted
correctly
𝑑
𝑑𝑡
(𝑒 𝜆2×t
× 𝑁2)
Using the product rule
𝑑
𝑑𝑡
(𝑒 𝜆2×t
× 𝑁2(𝑡)) = 𝑁2 ×
𝑑
𝑑𝑡
(𝑒−𝜆2×t
) + 𝑒 𝜆2×t
×
𝑑
𝑑𝑡
(𝑁2)
= 𝜆2 × 𝑒−𝜆2×t
× 𝑁2 + 𝑒 𝜆2×t
×
𝑑𝑁2
𝑑𝑡
Does that look familiar? Maybe if we regroup a little
𝑑
𝑑𝑡
(𝑒 𝜆2×t
× 𝑁2) = 𝑒 𝜆2×t
×
𝑑𝑁2
𝑑𝑡
+ 𝑒 𝜆2×t
× 𝜆2 × 𝑁2
So we notice that the right side of this equation is indeed the left side of the one we derived
previously:
𝑒 𝜆2×t
×
𝑑𝑁2
𝑑𝑡
+ 𝑒 𝜆2×t
× 𝜆2 × 𝑁2(𝑡) = 𝜆1 × 𝑁1(0) × 𝑒(𝜆2−𝜆1)×t

13. 13
Confirmation End
Hence we can substitute
𝑑
𝑑𝑡
(𝑒 𝜆2×t
× 𝑁2(𝑡)) = 𝜆1 × 𝑁1(0) × 𝑒(𝜆2−𝜆1)×t
And now we can integrate both sides of the equation with respect to time to eliminate the
differential expression
∫ (
𝑑
𝑑𝑡
(𝑒−𝜆2×t
× 𝑁2(𝑡))) 𝑑𝑡 = ∫(𝜆1 × 𝑁1(0) × 𝑒(𝜆2−𝜆1)×t
) 𝑑𝑡
𝑒 𝜆2×t
× 𝑁2(𝑡) + 𝐾1 =
1
𝜆2 − 𝜆1
× 𝜆1 × 𝑁1(0) × 𝑒(𝜆2−𝜆1)×t
+ 𝐾2
Where𝐾1 and 𝐾2 are the constants of integration. As previously, we can define another constant 𝐾3
𝑒 𝜆2×t
× 𝑁2(𝑡) =
𝜆1
𝜆2 − 𝜆1
× 𝑁1(0) × 𝑒(𝜆2−𝜆1)×t
+ 𝐾3
If we assume that initially, the initial number of intermediate nuclei is 0 at time 𝑡 = 0
𝑁2(0) = 0
𝑒 𝜆2×t
× 𝑁2(0) =
𝜆1
𝜆2 − 𝜆1
× 𝑁1(0) × 𝑒(𝜆2−𝜆1)×0
+ 𝐾3
We can solve the equation for 𝐾3
𝑒 𝜆2×t
× 0 =
𝜆1
𝜆2 − 𝜆1
× 𝑁1(0) × 𝑒0
+ 𝐾3
0 =
𝜆1
𝜆2 − 𝜆1
× 𝑁1(0) × 1 + 𝐾3
−𝐾3 =
𝜆1
𝜆2 − 𝜆1
× 𝑁1(0)
𝐾3 = −
𝜆1
𝜆2 − 𝜆1
× 𝑁1(0)
And we can substitute this value back into the equation
𝑒 𝜆2×t
× 𝑁2(𝑡) =
𝜆1
𝜆2 − 𝜆1
× 𝑁1(0) × 𝑒(𝜆2−𝜆1)×t
+ 𝐾3
And obtain
𝑒 𝜆2×t
× 𝑁2(𝑡) =
𝜆1
𝜆2 − 𝜆1
× 𝑁1(0) × 𝑒(𝜆2−𝜆1)×t
−
𝜆1
𝜆2 − 𝜆1
× 𝑁1(0)
Factorization may make that look far less complicated

14. 14
𝑒 𝜆2×t
× 𝑁2(𝑡) =
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
(𝑒(𝜆2−𝜆1)×t
− 1)
Dividing both sides by 𝑒 𝜆2×t
𝑁2(𝑡) =
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
(𝑒(𝜆2−𝜆1)×t
− 1) ×
1
𝑒 𝜆2×t
𝑁2(𝑡) =
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
(𝑒(𝜆2−𝜆1)×t
− 1) × (𝑒 𝜆2×t
)−1
𝑁2(𝑡) =
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
(𝑒(𝜆2−𝜆1)×t
− 1) × 𝑒−𝜆2×t
𝑁2(𝑡) =
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
(𝑒(𝜆2−𝜆1)×t
× 𝑒−𝜆2×t
− 𝑒−𝜆2×t
)
𝑁2(𝑡) =
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
(𝑒−𝜆1×t
− 𝑒−𝜆2×t
)
And this is the final equation for the number of nuclei in the intermediate sample, 𝑁2. 𝜆1 and 𝜆2 are
the respective integration constants, and 𝑡 is time, the random variable. 𝑁1(0) is the amount of
nuclei initially present in the parent sample. So the Two-step decay can e expressed by the system of
equations as follows.
𝑁1(𝑡) = 𝑁1(0) × 𝑒−𝜆1×t
𝑁2(𝑡) =
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
(𝑒−𝜆1×t
− 𝑒−𝜆2×t
)
𝑑𝑁3
𝑑𝑡
= 𝜆2 × 𝑁2(𝑡)
So we only need to solve for 𝑁3 to form the last of three equations describing the decay
𝑑𝑁3
𝑑𝑡
= 𝜆2 ×
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
(𝑒−𝜆1×t
− 𝑒−𝜆2×t
)
In order to integrate that last function, let us first expand
𝑑𝑁3
𝑑𝑡
= 𝜆2 ×
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
× 𝑒−𝜆1×t
− 𝑒−𝜆2×t
× 𝜆2 ×
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
And integrate both sides with respect to time
∫ (
𝑑𝑁3
𝑑𝑡
) 𝑑𝑡 = ∫ (𝜆2 ×
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
× 𝑒−𝜆1×t
− 𝑒−𝜆2×t
× 𝜆2 ×
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
) 𝑑𝑡
∫ (
𝑑𝑁3
𝑑𝑡
) 𝑑𝑡 = ∫ (𝜆2 ×
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
× 𝑒−𝜆1×t
) 𝑑𝑡 − ∫ (𝑒−𝜆2×t
× 𝜆2 ×
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
) 𝑑𝑡
Another case of an equation looking more complicated than it is

15. 15
𝑁3(𝑡) = −
1
𝜆1
× 𝜆2 ×
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
× 𝑒−𝜆1×t
+
1
𝜆2
× 𝜆2 ×
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
× 𝑒−𝜆2×t
Group like terms and simplify
𝑁3(𝑡) =
𝜆2 × 𝑁1(0)
𝜆1 × (𝜆2 − 𝜆1)
× 𝑒−𝜆2×t
−
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
× 𝑒−𝜆1×t
𝑁3(𝑡) =
𝑁1(0)
𝜆2 − 𝜆1
× (
𝜆2 × 𝑒−𝜆2×t
𝜆1
− 𝜆1 × 𝑒−𝜆1×t
)
There is actually a simpler method to find and expression for 𝑁3(𝑡). Since we know that a Nucleus
decays into one other nucleus, and 𝑁1(0) Nuclei were initially present, there must still be 𝑁1(0)
Nuclei after a certain amount of time t. This means the sum of 𝑁1(𝑡), 𝑁2(𝑡) and 𝑁3(𝑡) equals 𝑁1(0).
We could therefore also express 𝑁3(𝑡) as follows and simplify.
𝑁3(𝑡) = 𝑁1(0) − 𝑁1(𝑡) − 𝑁2(𝑡)
But this should yield the same result as before. We have therefore completed our set of three
equations
𝑁1(𝑡) = 𝑁1(0) × 𝑒−𝜆1×t
𝑁2(𝑡) =
𝜆1 × 𝑁1(0)
𝜆2 − 𝜆1
(𝑒−𝜆1×t
− 𝑒−𝜆2×t
)
𝑁3(𝑡) =
𝑁1(0)
𝜆2 − 𝜆1
× (
𝜆2 × 𝑒−𝜆2×t
𝜆1
− 𝜆1 × 𝑒−𝜆1×t
)
VI) Modelling using MS Excel
At this point I made an excel spreadsheet to be able to model a few conditions. I took the example
used previously
𝑁1(0) = 100
And I took the half-life again to be two years
𝑁(𝑡) = 𝑁0 × 𝑒−𝜆×t
50 = 100 × 𝑒−𝜆×2
0.5 = 𝑒−𝜆×2
ln(0.5) = ln(𝑒−𝜆×2
)
ln(0.5) = −2 × 𝜆
𝜆 = −
ln(0.5)
2

16. 16
Just as a first trial, I took
𝜆2 = − ln(0.5)
With the spreadsheet as follows
The formula for the first column, as written in Excel, is
=$C$2*$F$2^-($A$2*C7)
Second column
=(($A$2*$C$2)/($B$2-$A$2))*($F$2^(-$A$2*C7)-$F$2^(-$B$2*C7))
And third column
=(($B$2*$C$2)/($A$2($B$2-$A$2)))*$F$2^(-$B$2*C7)-(($A$2*$C$2)/($B$2-$A$2))*$F$2^(-$A$2*C7)
This yields the table
time(years) Parent Intermediate Daughter
0 100.000 0.000 0.000
1 70.711 20.711 8.579
2 50.000 25.000 25.000
3 35.355 22.855 41.789
4 25.000 18.750 56.250
5 17.678 14.553 67.770
6 12.500 10.938 76.562
7 8.839 8.058 83.104
8 6.250 5.859 87.891
9 4.419 4.224 91.356
10 3.125 3.027 93.848
11 2.210 2.161 95.629
12 1.563 1.538 96.899
13 1.105 1.093 97.802
14 0.781 0.775 98.444
15 0.552 0.549 98.898
16 0.391 0.389 99.220
17 0.276 0.275 99.448
18 0.195 0.195 99.610
19 0.138 0.138 99.724

17. 17
20 0.098 0.098 99.805
21 0.069 0.069 99.862
22 0.049 0.049 99.902
23 0.035 0.035 99.931
24 0.024 0.024 99.951
25 0.017 0.017 99.965
26 0.012 0.012 99.976
27 0.009 0.009 99.983
28 0.006 0.006 99.988
29 0.004 0.004 99.991
30 0.003 0.003 99.994
And as a Graph
If we increase 𝜆2 to 10 times that value, what will happen? We are decreasing the Half-life of this
Nuclide
0.000
20.000
40.000
60.000
80.000
100.000
120.000
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
Nuclei(number)
Time (years)
Figure 8: Chained Decay, Example 1
Parent
Intermediate
Daughter

18. 18
Not much actually changes, the local maximum of the intermediate function decreases and it occurs
at an earlier value of time
And if we make the half-life very short, increase 𝜆2 to 100
The intermediate functions’ local maximum is very small. This makes sense, as it almost instantly
decays to form the daughter.
0.000
20.000
40.000
60.000
80.000
100.000
120.000
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
Nuclei(number)
Time (years)
Figure 9: Chained Decay, Example 2
Parent
Intermediate
Daughter
0.000
20.000
40.000
60.000
80.000
100.000
120.000
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
Nuclei(number)
Time (years)
Figure 10: Chained Decay, Example 3
Parent
Intermediate
Daughter

19. 19
And decreasing the half-life of the first Parent?
This peak can not be seen quite well on this graph, so we can change the scaling of the x axis
0.000
20.000
40.000
60.000
80.000
100.000
120.000
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
Nuclei(number)
Time (years)
Figure 11: Chained Decay, Example 4
Parent
Intermediate
Daughter
0.000
20.000
40.000
60.000
80.000
100.000
120.000
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
Nuclei(number)
Time (years/10)
Figure 12: Chained Decay, Example 5
Parent
Intermediate
Daughter

20. 20
Overall we seen that the shape of the functions stays the same approximately, but the local maxima
as well as rates of change are alternated.
VII) Conclusion & Evaluation
From the very start of working on this exploration, I chose to work quite independently and wanted
to, as much as possible, make this mathematical paper my own. As such, I began to first explain the
more simple Mathematical principles governing radioactive decay. During my working progress, I
became more familiar with how to actually convey these principles. On challenge for me personally
was to make this paper accessible for someone without a large previous knowledge on the topic of
Physics. Many of the sources that I consulted had the limitation of taking for granted certain physical
knowledge. In my exploration, I aimed to explain these presumptions.
After having explored and explained my existing knowledge, I consulted several sources to pursue
the mathematical expression of three-stage decay. While many of these papers varied in their
approach, they had one thing in common: neither detailed the mathematical progress which
eventually resulted in the set of three equations, only the key principles were outlined. All existing
papers on this topic approach the issue from a second-year university level or higher. Collecting,
collating and analyzing these papers to serve as a base of this exploration gave me some experience
in reading and understanding Mathematical papers, written at quite a high level.
Individual steps in this derivation allowed me to apply several of my personally favourite
Mathematical topics: a thorough knowledge of Calculus is required. Another key ingredient to
deriving these equations is a thorough amount of attention paid to every single step. With long
working such as shown in this exploration, as simple error will cause a large divergence from the
desired result. Further, this attention is required to notice certain Mathematical simplifications and
tricks, as for example on page 11. Putting this exploration to paper has been a challenge for me, but I
am looking forward to studying more advanced Mathematics and Physics at University as much as
never before.
I was convinced to conduct this entire exploration very much independently, and make it very much
my own. As such, I decided to explore a real-world phenomenon from a Mathematical perspective
rather than research and explain a Mathematical phenomenon. This created more of a challenge, as I
had to research mathematical theories independently from my chosen topic, and apply them with a
view to radioactive decay. At the same time, I am very glad I went out of my way, as I have become
more familiar with more advanced Mathematics, had to figure out key working on my own, and now
know a great deal more about radioactivity.

21. 21
VIII) Bibliography
"Radioactive decay and the Bateman equation." Lecture, Introduction to Nuclear Science from Simon
Fraser University , January 19, 2011.
"Sequential Decay." CUA Faculty. January 1, 2007. Accessed March 3, 2015.
http://faculty.cua.edu/sober/635/Sequential.pdf.
"The Energy From Thorium Foundation." The Energy From Thorium Foundation. Accessed March 3,
2015. http://energyfromthorium.com/tech/physics/decay2/.
Dobson, Katy, and Alan Slomson. "Solving Differential Equations by Separating Variables."Mathcentre
Community Project, January 1, 2010.
Pressyanov, Dobromir S. "Short Solution of the Radioactive Decay Chain Equations." American
Journal of Physics, 2001, 444-45. Accessed March 3, 2015.
Reid, Alastair. "Simulating Decay Chains." Wordpress, January 1, 2012, 18-20.
Watterson, J.. "Radioactivity - the radioactive decay chain." Lecture, Lecture 22 from CERN, , October
1, 2007.