Introduction

Nuclear physics had its beginnings in 1896. In that year Henry Becquerel (1852-1908) made an important discovery in his studies of phosphorescence, he found that a certain mineral (which happened to contain uranium) would darken a photographic plate even when the plate was wrapped to exclude light. It was clear that the mineral emitted some new kind of radiation that, unlike X-rays, occurred without any external stimulus. This new phenomenon eventually came to be called radioactivity.

Henry Becquerel (1852-1908)

Soon after Becquerel’s discovery, Marie Curie (1867-1934) and her husband, Pierre Curie (1859-1906), isolated two previously unknown elements that were very highly radioactive. These were named polonium and radium. Other radioactive elements were soon discovered as well. The radioactivity was found in every case to be unaffected by the strongest physical and chemical treatments, including strong heating or cooling and the action of strong chemical reagents. It was clear that the source of radioactivity must be deep within the atom that it must emanate from the nucleus. It became apparent that radioactivity is the result of the disintegration or decay of an unstable nucleus.

Pierre and Marie Curie

Radioactive is one of area in Science that calculus could be applied. Radioactive decay is the process in which an unstable atomic nucleus loses energy by emitting ionizing particles and radiation. This decay, or loss of energy results in an atom of one type, called the nuclide transforming to an atom of a different type, called the daughter nuclide. For example: a carbon-14 atom (the ‘parent’) emits radiation and transforms to a nitrogen-14 atom (the ‘daughter’). This is random process on the atomic level, in that it is impossible to predict when a given atom will decay, but given a large number of similar atoms the decay rate, on average is predictable. The SI unit of radioactive decay is the Becquerel (Bq).

During radioactive decay, an unstable nucleus spontaneously decomposes to form a different nucleus, giving off radiation in the form of atomic particles or high energy rays. This decay occurs at a constant, predictable rate that is referred to as half-life. A stable nucleus will not undergo this kind of decay and is non-radioactive.

The trefoil symbol is used to indicate radioactive material

Application of Calculus in Radioactive Decay Law and Half-Life

How do we applied calculus in radioactive?

Let us say that in a sample of radioactive material there are N nuclei which have not decayed at a certain time, t. So what happens in the next brief period of time? How many nuclei will decay? Therefore, we can say that the number which will decay depends on total number of nuclei, N and also the length of the brief period of time. In other words the more nuclei there are the more will decay and the longer the time period the more nuclei will decay. Let us donate the number which will have decayed as dN and the small time interval as dt.

Generally, we could say that the number of radioactive nuclei which will decay during the time interval from t to (t + dt) must be proportional to N and dt. In symbol expression:

- dN α N.dt

The minus sign indicates that N is decreasing. Mathematically, it could be written as:

- dN = λN.dt (1)

In this equation, λ is a constant of proportionality called the decay constant, which is different for different isotopes. The greater λ is the greater the rate of decay and the more radioactive that isotope is said to be for a given number of nuclei. The number of decays that occur in the short time interval dt is designated dN because each decay that occurs corresponds to a decrease by one in the number N of nuclei present. That is, radioactive decay is “one-shot” process. This process means once a particular parent nucleus decays into its daughter, it cannot do it again.

If we take the limit ∆t → 0 in equation (1), dN will be small compared to N. We can determine N as a function of t by rearranging this equation to

N

dN= - λNdt (2)

So this equation describes the situation for any brief time interval, dt. In order to figure out what will happen for all periods of time, we simply add up what happens in each brief time interval. In other words, we integrate the equation. Expressing this formally, we could say that for the period of the time t = 0 to the time t = t, the number of radioactive nuclei will decrease from N0 to Nt, so that:

No

Ntln = - λt

No

Nt = e

tλ−

∴ Nt = N0 etλ− (3)

The final expression is known as the Radioactive Decay Law. It tells us that the number of radioactive nuclei will decrease in an exponential pattern with time with the rate of decreasing being controlled by the decay constant.

Before going further, lets we review again the steps of mathematics we used above:

1. First, we used integral calculus to figure out what was happening over a period of time by integrating what we knew would occur in a brief interval of time.

2. Secondly, we used a calculus relationship of

xlnx

dx =∫

where xln represents the natural logarithm of x.

3. Thirdly, we used the definition of logarithms which is

yxln =Then,

ey

x =

The Radioactive Decay Law tells us that the number of radioactive nuclei will decrease with time in an exponential pattern with the rate of decrease being controlled by the decay constant. The law could be shown in graphical figure as below:

The graph plots the number of radioactive nuclei at any time, Nt, against time, t. The number of radioactive nuclei decreases from N0 at time t = 0 in a rapid fashion initially and then became more slowly in the classic exponential manner. The influence of decay constant could be seen in following figure:

All the three curves are exponential in nature. The difference is only in decay constant. When the value of decay constant is small, the curve decreases relatively slowly. However, when the value of decay constant is large, the curve seems to be decreases rapidly.

The decay constant is characteristic of individual radionuclide. Some like Uranium-238 have a small value and the material therefore decays quite slowly over a long period of time. Other nuclei such as Technetium-99m have a relatively large decay constant and they decay far more quickly. It is possible to consider the Radioactive Decay Law from another perspective by plotting the logarithm of Nt against time. In other words, we can plot from the expression

No

Ntln = - λt

In the form of,

No lnt-Nt ln += λ

From this expression, we could notice that this is simply an equation of the form y = mx + c where m = -1 and c = ln No. As a result of this equation, it show that it is a straight line with slope of -1 as shown in following figure,

The rate of decay of any isotope is often specified by giving its half-life rather than the decay constant λ. The half life of an isotope is defined as the time it takes for half the original amount of isotope in a given sample to decay. It expresses the length of time it takes for the radioactivity of a radioisotope to decrease by a factor of two. From a graphical point of view, it could be say that when:

Nt 2

No=

The time take is the half-life:

Half-life does not express how long a material will remain radioactive but simply the length of time for its radioactivity to be halved. Examples of the half-life of some radioisotopes are given in the following table. Notice that some of these have a relatively short half-life. These tend to be the ones used for medical diagnostic purposes because they do not remain radioactive for very long following administration to a patient and hence result in a relatively low radiation dose.

Radioisotope Half Life (approx.)

81mKr 13 seconds

99mTc 6 hours

131I 8 days

51Cr 1 month

137Cs 30 years

241Am 462 years

226Ra 1620 years

238U 4.51 x 109 years

Relationship between the Decay Constant and Half-Life

From the statement above, we could conclude that there is a relation between the decay constant and half-life. As for example, when the decay constant is small, the half-life should be long and same goes when the decay constant is large, the half-life is short. Why this happened?

This could be explained by using the definition of Half-life and applying it to the Radioactive Decay Law.

The law tells us that any time, t:

Nt = N0 etλ−

Meanwhile the definition of half-life tells us that:

Nt 2

0N=

When,

tt2

1=

Therefore the equation of Radioactive Decay Law could be rewrite by substituting Nt and t as follows:

20N

= N0et

21λ−

Thus,

et

21

2

1 λ−=

et

212

1 λ−− =

21t- ln λ=−

21

21t ln λ=2

21t . λ=6930

λ6930.

t2

1 =

And

21

6930

t

.=λ

These last two equations express the relationship between the decay constant and the half-life. They are very useful when solving numerical questions relating to radioactivity and usually form the first step in solving a numerical problem.

Let’s Do It

1. Find the radioactivity of 1 g sample of 226Ra given that t1/2: 1620 years and Avogadro's number: 6.023 x 1023.

Answer:

We can start answer by calculating the Decay Constant from the Half-Life using the following equation:

14

21

1028.41620

693.0693.0 −−×=== yeart

λ

1111036.1 −−×=∴ sλ

Note that the length of a year used in converting from 'per year' to 'per second' above is 365.25 days to account for leap years. In addition the reason for converting to units of 'per second' is because the unit of radioactivity is expressed as the number of nuclei decaying per second.

Secondly we can calculate that 1 g of 226Ra contains:

Nuclei.)g)(.(

MassNumber

)mass)(sNumber'Avogadro(N 21

23

1072226

1100236 ×=×==

Thirdly we need to express the Radioactive Decay Law in terms of the number of nuclei decaying per unit time. We can do this by differentiating the equation as follows:

)t(eNN λ−=0

teNte.Ndt

dN λλλλ −−=−−=∴00

Ndt

dN λ−=∴

Ndt

dN λ=∴

The reason for expressing the result above in absolute terms is to remove the minus sign in that we already know that the number is decreasing.

We can now enter the data we derived above for λ and N:

)107.2)(1036.1(dt

dN 2111 ××=∴ −

10106.3dt

dN ×=∴ decays per second

So the radioactivity of our 1 g sample of radium-226 is approximately 1 Ci.

This is not a surprising answer since the definition of the curie was originally conceived as the radioactivity of 1 g of radium-226!

2. The half-life of 99mTc is 6 hours. After how much time will 1/16th of the radioisotope remain?

Answer:

Starting with the relationship we established earlier between the Decay Constant and the Half Life we can calculate the Decay Constant as follows:

1

21

hr1155.06

693.0

t

693.0 −===λ

Now applying the Radioactive Decay Law,

t0t eNN λ−=

It also can be write it in the form:

t

0

t eN

N λ−=

The question tells us that N0 has reduced to 1/16th of its value, that is:

16

1

N

N

0

t =

Therefore

)t1155.0(e16

1 −=

which we need to solve for t. One way of doing this is as follows:

)t1155.0(e116 −=−

t1155.016ln −=−∴

hours241155.0

16lnt ==

So it will take 24 hours until 1/16th of the radioactivity remains.