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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

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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.

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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, 𝑁𝑇1/2, 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,

𝑁𝑇1/2= 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, 𝑈𝑛 is mathematically given by

𝑈𝑛 = 𝑈1 × 𝑟𝑛−1

Where 𝑈𝑛 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

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𝑟 =𝑈𝑛

𝑈𝑛−1

⇔𝑈2

𝑈1=

50

100= 0.5

Therefore,

𝑈𝑛 = 𝑈1 × 𝑟𝑛−1 = 100 × 0.5𝑛−1 = 100 ×0.5𝑛

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

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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

𝑙𝑜𝑔10𝑁 = −𝑘𝑥 + 𝑙

Where k and l are positive constants.

But a sequence only yields values for 𝑈𝑛, 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

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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

𝑁𝑑𝑁 = ∫ −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

𝑒𝑎 = 𝑒𝑏

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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 × 𝑒𝑐3

As 𝑒 and 𝑐3 are both constants, let us define a fourth constant, 𝑐4, which is such that

𝑐4 = 𝑒𝑐3

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.

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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) = 𝑁𝑇1/2= 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)

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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

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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

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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

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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 𝑒𝜆2×t

𝑒𝜆2×t ×𝑑𝑁2

𝑑𝑡+ 𝑒𝜆2×t × 𝜆2 × 𝑁2(𝑡) = 𝜆1 × 𝑁1(0) × 𝑒−𝜆1×t × 𝑒𝜆2×t

Now we may simplify

𝑒𝜆2×t ×𝑑𝑁2

𝑑𝑡+ 𝑒𝜆2×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

𝑒𝜆2×t ×𝑑𝑁2

𝑑𝑡+ 𝑒𝜆2×t × 𝜆2 × 𝑁2(𝑡)

Firstly, it is a sum of two products,

𝑒𝜆2×t ×𝑑𝑁2

𝑑𝑡

As well as

𝜆2 × 𝑒𝜆2×t × 𝑁2(𝑡)

If we compare the two, we notice that

𝑑

𝑑𝑡𝑁2(𝑡) =

𝑑𝑁2

𝑑𝑡

Furthermore

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𝑑

𝑑𝑡𝑒𝜆2×t = 𝜆2 × 𝑒𝜆2×t

If we let

𝑁2(𝑡) = 𝑢

𝑒𝜆2×t = 𝑣

The expression can be written as

𝑒𝜆2×t ×𝑑𝑁2

𝑑𝑡+ 𝑒𝜆2×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

𝑒𝜆2×t ×𝑑𝑁2

𝑑𝑡+ 𝑒𝜆2×t × 𝜆2 × 𝑁2 =

𝑑

𝑑𝑡(𝑒𝜆2×t × 𝑁2(𝑡))

Confirmation:

We can try to do this calculation in reverse to verify whether we these calculations were conducted

correctly

𝑑

𝑑𝑡(𝑒𝜆2×t × 𝑁2)

Using the product rule

𝑑

𝑑𝑡(𝑒𝜆2×t × 𝑁2(𝑡)) = 𝑁2 ×

𝑑

𝑑𝑡(𝑒−𝜆2×t ) + 𝑒𝜆2×t ×

𝑑

𝑑𝑡(𝑁2)

= 𝜆2 × 𝑒−𝜆2×t × 𝑁2 + 𝑒𝜆2×t ×𝑑𝑁2

𝑑𝑡

Does that look familiar? Maybe if we regroup a little

𝑑

𝑑𝑡(𝑒𝜆2×t × 𝑁2) = 𝑒𝜆2×t ×

𝑑𝑁2

𝑑𝑡+ 𝑒𝜆2×t × 𝜆2 × 𝑁2

So we notice that the right side of this equation is indeed the left side of the one we derived

previously:

𝑒𝜆2×t ×𝑑𝑁2

𝑑𝑡+ 𝑒𝜆2×t × 𝜆2 × 𝑁2(𝑡) = 𝜆1 × 𝑁1(0) × 𝑒(𝜆2−𝜆1)×t

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Confirmation End

Hence we can substitute

𝑑

𝑑𝑡(𝑒𝜆2×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

𝑒𝜆2×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

𝑒𝜆2×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

𝑒𝜆2×t × 𝑁2(0) =𝜆1

𝜆2 − 𝜆1× 𝑁1(0) × 𝑒(𝜆2−𝜆1)×0 + 𝐾3

We can solve the equation for 𝐾3

𝑒𝜆2×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

𝑒𝜆2×t × 𝑁2(𝑡) =𝜆1

𝜆2 − 𝜆1× 𝑁1(0) × 𝑒(𝜆2−𝜆1)×t + 𝐾3

And obtain

𝑒𝜆2×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

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𝑒𝜆2×t × 𝑁2(𝑡) =𝜆1 × 𝑁1(0)

𝜆2 − 𝜆1(𝑒(𝜆2−𝜆1)×t − 1)

Dividing both sides by 𝑒𝜆2×t

𝑁2(𝑡) =𝜆1 × 𝑁1(0)

𝜆2 − 𝜆1(𝑒(𝜆2−𝜆1)×t − 1) ×

1

𝑒𝜆2×t

𝑁2(𝑡) =𝜆1 × 𝑁1(0)

𝜆2 − 𝜆1(𝑒(𝜆2−𝜆1)×t − 1) × (𝑒𝜆2×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

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

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

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

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

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

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

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.