Nucleur Physics Learning Material
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In this Physics tutorial, you will learn:

• What is natural radioactivity? How is it produced?
• How many types of radioactive decays are there?
• What particles are emitted during each type of radioactive decay? Why?
• What other particles are involved in both types of beta decay?
• What are the factors affecting radioactive decay?
• What is half-life? Why is it so important?
• What is decay rate? What does it indicate in nuclear radioactivity?
• How to calculate the number of undecayed particles and the rate of decay in a radioactive sample?

## Introduction

As explained in article 20.1, natural occurring radioactivity was discovered accidentally by Henry Becquerel in 1896, one year after Rontgen discovered X-rays as a form of EM radiation. As stated earlier, Becquerel knew that natural radioactivity (the ability of an atomic nucleus to emit radiation by its own that is without any external action) was not a form of radiations known up to that time, so it required further and deeper studies.

This article explains more in detail the phenomenon of radioactivity with all its forms. Likewise, the corresponding mathematical relations governing the radioactivity laws and quantities are provided here, so that readers will understand how to estimate the danger arisen from exposure to radioactive substances.

## More on Natural Radioactivity. Becquerel's Experiment

Becquerel's experiment on natural radioactivity consists on the following procedure:

The radioactive material emits radiation in the form of a narrow beam because of the collimator (a device for producing a parallel beam of rays or radiation). The traces resulted in the photograph plate were deflected from their linear path, as shown in the figure. The reason for this deflection of radioactive beam from their linear path is their interaction with the magnetic field after flowing out of collimator. These are not the standard X-rays, as X-rays rays cannot be deflected by magnetic field (as X-rays do not carry any electric charge to interact with magnetic field). Hence, obviously the radiation emitted by collimator is electrically charged. From this fact, we can obtain the definition of radioactivity as a phenomenon naturally occurring in materials:

Radioactivity is the phenomenon of emission of ionizing radiation or particles caused by the spontaneous disintegration of atomic nuclei.

When different radioactive materials were used in the experiment, different behavior of the corresponding beams was observed. Thus, in some cases there was a deflection due left, in other cases due right and in a few cases there was no deflection at all. This means that from the electric point of view, there are three types of radiations emitted by radioactive materials: positive, negative and neutral. Rutherford named them as alpha (α), beta (β) and gamma (γ) radiation respectively.

The term "radioactive" was first used by the Curie consorts (Pierre and Marie Curie), whose work on this field is invaluable. They managed to isolate some radioactive elements such as radium (Ra), uranium (U) and polonium (Po).

Now, let's explain more in detail each type of radiation mentioned above.

## Alpha Decay

In alpha (α) decay or disintegration, a heavy (massive) nucleus emits a helium (42He) nucleus and another daughter nucleus. For example, any of uranium isotopes such as (23892U) may emit an alpha particle and thus become a thorium isotope (23490Th). The mathematical relation in alpha decay is

AZX ⟶ A - 4Z - 2Y + 42He

Alpha particles were given this name prior to discovering what kind of particles they represent. However, now we know that alpha particles are nothing more but helium nuclei. An alpha particle is a very stable structure (we have explained that hydrogen and helium are very stable materials; indeed the Sun is mainly composed by hydrogen and helium elements).

Alpha particles detach from their parent nuclei because during the attempt to reduce the repelling electric forces, alpha particles, which are formed inside the nucleus, may find themselves in the periphery of nucleus and gain enough kinetic energy to leave it without any interference from an external source of energy that is to overcome the nuclear binding force. Alpha decay is schematically shown in the figure below. This process has a probabilistic nature; this means none of particles is favoured at start but everything depends on their actual arrangement inside the nucleus at a given instant. A probabilistic process always has a non-zero chance to occur, despite the conditions may be such that the event seems improbable. This occurs only in micro-world, not in real life. For example, the probability for an athlete to jump 10 m high without any aiding tool is zero as this exceeds the human physical capabilities but in micro-world nothing is improbable. A particle may overcome obstacles that may seem impossible - this is known as the "tunnel effect".

When an alpha particle leaves the original nucleus, a more stable nucleus is formed. We have explained in the previous article that the proton-neutron ratio (or vice-versa) is an indicator on the nuclei stability. For example, in the alpha decay process shown below

21084Po ⟶ 20682Pb + 42He

the proton-neutron ratio of "parent" nucleus (Polonium, Po) is

ZPo/NPo = 84/210 - 84 = 84/126 = 0.667

and the proton-neutron ratio of "daughter" nucleus (Lead, Pb) is

ZPb/NPb = 82/206 - 82 = 82/124 = 0.661

Despite the change in ratio is small, it is sufficient to make the daughter nucleus shift from radioactive to stable region of the N vs Z graph given in the previous article. When an alpha decay takes place, the electric charge in the daughter nucleus bemomes smaller than in the parent nucleus. As a result, the binding energy in daughter nucleus is smaller too. This results in a more stable nucleus. Moreover, the nuclear mass also decreases, bringing a decrease in the stored energy in the daughter nucleus (recall the mass-energy equivalence).

From the law of energy conservation, it is obvious that this difference in energy between parent and daughter nuclei convers into kinetic energy of the daughter particle and helium nucleus (recall the law of conservation of momentum in explosions). Therefore, we may use the law of conservation of momentum to determine how fast the daughter nucleus and helium nucleus will move after an alpha-decay process does occur.

### Example 1

Calculate the energy released when a Seaborgium (263106Sg) nucleus experiences an alpha decay. Refer to the previous article for any useful information. Seaborgium nucleus is considered at rest and the two new particles move in opposite directions after the alpha decay takes place.

### Solution 1

First, it is useful to provide an overview of the situation. Since all particles possess some rest energy in the form of mass, which we can find through the mass-energy equivalence method, we can then find the change in energy by comparing them. This change in energy (which is the binding energy of daughter and helium nucleus when they were in the parent nucleus) represents the sum of kinetic energies of the new particles produced due to alpha decay, which corresponds to the energy released by the Seaborgium nucleus during this process.

Giving that the decay process that occurs in this reaction is

263106Sb → 259104Rf + 42He

and giving that atomic masses of these three materials are 266 u, 261 u and 4.003 u respectively, we obtain for the mass defect of this process:

∆m = m(263106Sb ) - [m(259104Rf) + m(42He )]
= 266 u - (261 u + 4 u)
= 1u

Since this value corresponds to 1.66054 × 10-27 kg, we obtain for the binding energy of parent nucleus:

Eb = ∆m ∙ c2
= (1.66054 × 10-27 kg) ∙ (3 × 108 m/s2 )2
= 1.49449×10-10 J

This energy corresponds to energy released during the alpha decay; it is in the form of kinetic energy.

## Beta Decay

In the experiment of Becquerel, there were some traces on the film plate in the opposite direction to those produced by alpha particles. This means such traces are produced by negatively charged particles, as alpha particles are positive (they are helium nuclei). The name beta (β) particles was given to those moving in opposite direction to alpha ones and the naturally occurring decay process of such particles was called beta decay. Later on, it was discovered that such particles are nothing more but electrons.

Many years later, a new elementary particle known as positron, which has the same mass as electron but is positively charge, it was observed that there are two types of beta decay: beta minus (β-) and beta plus (β + ).

Beta decay occurs when there is excess of any particle (neutron or proton) in the atomic nucleus a neutron or a proton. As a result, any of extra particles splits in two parts. Thus, in beta minus decay, a neutron splits into one proton and one electron according to the relation:

n = p + e-

while in a beta plus decay, a proton splits into one neutron and one positron according to the relation:

p = n + e+

where e+ stands for the positron.

Since all splitting processes involve release or absorption of energy (that is, a third element in the right side of decay equation), it was evident that in the above reactions something is missing. This was also proven from experiments (energy and momentum are not conserved if we don't consider a third particle in the above equations). Obviously, such particle must be also elementary, as it is not proton, neutron or electron but something else. From experiments, this new elementary particle involved in the beta decay process was identified. In fact, there are two new elementary particles involved in a beta decay (given that there are two types of beta decay). One is called antineutrino (ν ) and it is involved in the beta minus decay, i.e.

n = p + e- + ν

and the other one (involved in beta plus decay) is called neutrino (ν). Thus, the equation of beta plus decay becomes

p = n + e+ + ν

Antineutrino and neutrino do not contain any electric charge; they are involved in the process only for the energy conservation purpose. The discovery of these two elementary particles is attributed to Wolfgang Pauli - a German scientist who was the first to understand the need for a new particle to make the conservation laws (of energy and impulse) applicable in beta decay equations.

Both types of beta decay reactions occur for the purpose to make the daughter nucleus be closer to the stability zone, as shown in the graph below. Neutrino and antineutrino (referred to by the common name "leptons") are particles that do not naturally exist in the nucleus; they are created only in the moment at which the beta decay is taking place. Both reactions of beta decay occur in such a way that the daughter particle moves closer to the stability zone. Let's consider the hydrogen nucleus to illustrate this fact. This nucleus contains only one proton and nothing else. Applying the equation of beta plus decay, it results that a proton splits into one neutron and one positron (including one neutrino as well). Hence, based on this approach a hydrogen nucleus cannot exist. However, the existence of hydrogen nuclei is matter of fact. This is because an isolated proton is a stable structure and it does not split in two parts. Beta decay occurs only in heavy nuclei that have a large number of protons and this process occurs for the only purpose to produce more stable structures (i.e. to obtain new nuclei with less difference between the number of protons and neutrons).

It is obvious that since the number of protons in the nuclei during a beta decay changes (it increases by 1 in β- and decreases by 1 in β + decay), a new chemical element is obtained, as the atomic number Z (which indicates the number of protons in an element) changes. On the other hand, the atomic mass number A remains the same in both types of beta decay as the number of nucleons does not change. We simply have a proton-to-neutron or neutron-to-proton conversion. The general formula of elements involved in both types of beta decays is

AZX → AZ + 1Y + e- + ν

for beta minus decay and

AZX → AZ - 1Y + e+ + ν

for beta plus decay.

For example, a Carbon-14 nucleus (Carbon-14 has 6 protons and 8 neutrons in the nucleus, that is 6 + 8 = 14 nucleons in total) turns into a Nitrogen-14 nucleus after a beta minus decay (Nitrogen-14 contain 7 protons and 7 neutrons in the nucleus).

On the other hand, a Carbon-11 nucleus (Carbon-11 contains 6 protons and 5 neutrons in the nucleus) can turn into Boron-11 (Boron-11 contains 5 protons and 6 neutrons in the nucleus) by means of a beta plus decay, that is when a proton converts to neutron.

From the above examples, we reach the following conclusion:

Beta minus decay occurs when the parent nucleus has excess of neutrons, while beta plus decay occurs when the parent nucleus has excess of protons. In both cases, more stable nuclei are obtained after the beta decay process.

Tip to make it easy to remember the name of beta decay involved: Always look at whether an electron or positron is emitted during the decay. If there is an electron emitted, then we have a beta minus decay as electron is negative. On the other hand, since positron is positive, there is a beta plus decay when a positron is emitted.

The following figure shows how the two types of beta decay take place. ### Example 2

What type of radioactive decay takes place when Fluoroine-18 converts to Oxygen-18?

### Solution 2

Since the number of nucleons does not change, we have a beta decay involved (alpha decay makes the number of nucleons decrease by 4 as a helium nucleus is emitted during this process).

Now, we have to find what type of beta decay occurs in this case. From periodic table, we see that Fluorine (F) contains 9 protons (Z = 9) while Oxygen (O) contains 8 protons (Z = 8) in their respective nuclei. This means a proton is converted to neutron when Fluorine-18 turns to Oxygen-18. This is an example of beta plus decay, i.e. a positron is involved in the process (and a neutrino as well). The reaction that takes place in this process therefore is

189F → 188O + e+ + ν

We can write the symbol (01e ) instead of (e+ ) to indicate a positron. This is because positron is produced during a beta plus decay when the atomic number (number of protons therefore) decreases by 1 (one proton turns into one neutron and one positron), so, we want to show this fact on the right side of reaction. Likewise, we can use the symbol (0-1e ) to indicate and electron produced during a beta minus decay as this electron is obtained when the atomic number increases by 1, i.e. when one neutron turns into one proton and one electron. Thus, for example, instead of writing

189F → 188O + e+ + ν

we can write

189F → 188O + 01e

This notation is easier to understand as you can do the operation with subscripts and superscripts (indicating protons and nucleons respectively) in both sides to see whether the law of mass conservation is applied correctly (here we have 9 = 8 + 1 and 18 = 18 + 0).

In such a notation, we don't write the neutrino or antineutrino as they do not contain any charge or particle, but their presence is implied.

During beta decay processes, energy and momentum are both conserved. Since in energy sharing process after the decay three particles are involved (daughter nucleus, beta particle and neutrino/antineutrino), but we have only two laws of conservation (of energy and momentum), then the energy sharing is not well-determined. Likewise, the binding energy of daughter nucleus varies based on the specific situation. This energy is the sum of energies of beta particle, daughter nucleus and neutrino/antineutrino. Given that a number of factors are involved in this energy sharing, the amount of energy each particle shares is not determined, despite the laws of conservation are applied in each case. As a result, the beta particle produced has not a determined energy after the decay.

### Example 3

Calculate the energy obtained during the β decay in which Carbon-14 turns into Nitrogen-14. Take the mass of Carbon-14 as 14.003242 u and that of Nitrogen-14 as 14.003074 u.

### Solution 3

This radioactive decay is written as

146C → 147N + 0-1e

The C-14 atom is neutral; this means 6 electrons revolve around its nucleus. When it turn into N-14, the new atom is not neutral anymore, it is positively charged as it still contains 6 electrons but now it has 7 protons. Actually, the mass of daughter nucleus (N-14) plus 6 electrons revolving around it plus the mass of electron radiated gives the mass of neutral N-14 atom. Giving that the mass of neutral N-14 atom is 14.003074 u, the change in mass (mass defect) in this case is

∆m = m(C - 14) - m(N - 14)
= 14.003242 u - 14.003074
= 0.000168 u
= 1.68 × 10-4 u

When converted to kg, this value becomes

∆m = (1.68 × 10-4 u) ∙ (1.66054 × 10-27 kg/u)
= 2.7897 × 10-31 kg

Hence, the energy produced during this beta (minus) decay process is

E = ∆m ∙ c2
= (2.7897 × 10-31 kg) ∙ (3 × 108 )2
= 2.51 × 10-14 J

This energy is carried by beta particles in the form of kinetic energy, as the daughter particle is much heavier that the beta particle produced (which is an electron). As a result, the daughter particle (N-14) is assumed as non-moveable and therefore, all the energy obtained during the decay goes to the electron in the form of kinetic energy.

## Gamma Decay

Gamma (γ) decay occurs when radiation emitted by the collimator in Becquerel experiment does not deviate but goes straight. This means the particles obtained do not carry any electric charge, i.e. they are neither protons, nor electrons. Such a decay takes place when the parent nucleus has a very high energy (we say it is excited). Hence, to become more stable, it emits a high-energy photon. Such a photon is known as "gamma particle". Since gamma particles are electrically neutral, they do not interact with electric or magnetic field and therefore, they go straight when flowing out of collimator. The figure below shows an example of gamma decay occurring in a Helium-3 nucleus. When gamma decay takes place, nucleons go closer to each other, so they have less energy than prior to decay. This energy is released in the form of light energy of the photon emitted during this process. Therefore, nothing changes in the number and type of nucleons; the only change occurs in their closeness, which results only in change in energy, not in structure of matter. This means we have neutral He-3 atoms both before and after gamma decay; the only difference is that the atom after the decay is less energetic, i.e. it is less excited.

## Half-Life

So far, we have explained various radioactive decays but without giving any information about the amount and rate of their occurrence. For different reasons, at a certain moment a nucleus may be in a state where a decay is possible. The conditions in which a decay occurs are not determined, so the radioactive decay is a pure causal event. Therefore, we rely on theory of probability to explain the rate of radioactive decay processes.

Let's suppose we have a number N unstable atomic nuclei of type A, which can transform into other nuclei according the chain

A → B + C

This does not mean that all N nuclei of the type A immediately transform into products B and C, rather, here we are discussing only about the probability (possibility) for this process to occur. This probability obviously depends on the time interval we are interested to and a number of other internal factors, which cannot be controlled and manipulated. Since every event takes place inside the nuclei A, it is obvious that the decay process does not depend on external factors such as state of matter, temperature, pressure, chemical structure, etc. The only factor that may affect this process is any possible interaction of nucleons with particles coming from outside the nucleus. In this way, entire process will depend only on the nuclei properties and the instant in which the decay takes place is purely casual.

Let's denote by N0 the number of nuclei of A-type present in a radioactive sample at a given instant, which we consider as the initial stage of process (t = 0). Obviously, after some time t, a number of radioactive decays have taken place, so the number of nuclei N remained in the original sample is N < N0. The question that arises here is: what is the relationship between the number of decayed nuclei and the time interval involved? In other words, our duty is to find the form of the function

N(t) = f(N0,t)

that gives the relationship between the number of nuclei still not decayed and the time elapsed given the original number of particles present in the sample.

This relationship relies on the supposition that every individual decay process is purely casual and independent from the other decays occurring in the rest of radioactive nuclei. Obviously, there is a linear relationship between the number ΔN of decayed particles and the interval Δt of this event's occurrence, which represents the time elapsed since the beginning of process. It is obvious that the number of decayed nuclei is ΔN = N(t) - N0 where N(t) is the number of undecayed nuclei at the time instant t. Hence, we can write

∆N ∝ N0 ∙ ∆t

From mathematics, it is known that we must multiply the right side of a proportion by a constant in order to turn it into an equation, Here, this constant if given by (). Thus, we obtain

∆N = -λ ∙ N0 ∙ ∆t

The negative sign before λ is used to make the right side of the above equation negative since the left side is negative as well (ΔN = N(t) - N0 and N(t) < N0).

The constant λ known as the constant of radioactive decay is an intrinsic property of material itself (unit: s-1) and it includes all factors affecting the decay of a given radioactive nucleus.

The equation for the number N(t) of undecayed nuclei at a given instant t is obtained by rearranging the above equation for N. Thus, we have

∆N = -λ ∙ N0 ∙ ∆t
∆N/N0 = -λ ∙ ∆t

Integrating both sides of the above equation in order to include all nuclei involved, we obtain after reducing the intervals to infinitely small ones to increase the accuracy of result (the symbol "Δ" is replaced by "d" in such cases, as explained in Electromagnetism in Section 16):

∆N/N0 = -λ ∙ ∆t
dN/N0 = -λ ∙ dt
dN/N0 = -λ ∙ dt
dN/N0 = -λ ∙ dt
ln⁡N/N0 = -λ ∙ t

Hence, we obtain for the number of undecayed nuclei in a radioactive sample as a function of time:

N(t) = N0 e-λ ∙ t

### Example 4

The radioactive decay constant for Bismuth-214 is 4330 s-1. How many Bismuth-214 nuclei are still present in a sample of Bismuth-214 after 500 microseconds if initially there were 100,000 nuclei in the sample?

### Solution 4

Clues:

t = 500 μs = 5 × 10-4 s
λ = 4330 s-1 = 4.33 × 103 s-1
N0 = 100 000 = 105
N(t) = ?

Using the equation

N(t) = N0 e-λ ∙ t

we obtain for the number of undecayed nuclei in the sample after substituting the known values:

N(500 μs) = 105 ∙ e(-4.33 × 103 s-1 ) ∙ (5 × 10-4 s)
= 10000 ∙ e-2.165
= 105 ∙ 0.11475
= 11475 nuclei

An important quantity used in equations of radioactive decay is half-life period, T1/2 (often referred to as simply "half-life"). It represents the time needed for the decay of half of the original radioactive nuclei in a sample. Like radioactive decay constant λ, half-life period T1/2 is a property that depends on the type of radioactive material too. Half-life is very useful as it makes the calculations much easier. The following table gives the value of half-life period for some radioactive isotopes.

IsotopeHalf-life period (T1/2)
Uranium-2384.5 billion years
Plutonium-23912.110 years
Carbon-145730 years
Litium-80.838 seconds
Bismuth-21420 minutes
Hydrogen-3 (tritium)12.32 years

We can find the relationship between λ and T1/2 by substituting t = T1/2 and N = N0/2 in the equation of radioactive decay. In this way, we obtain

N(t) = N0 e-λ ∙ t
N0/2 = N0 e-λ ∙ T1/2

Simplifying N0 from both sides, we obtain

e-λ ∙ T1/2 = 1/2
e-λ ∙ T1/2 = 2-1
eλ ∙ T1/2 = 2
ln⁡(eλ ∙ T1/2 ) = ln⁡2
λ ∙ T1/2 = ln⁡2

Since ln 2 ≈ 0.693, we obtain

λ = 0.693/T1/2

Another important quantity commonly used in quantitative approach of situations involving radioactive decay is the decay rate R(t) = ΔN/Δt. It shows how fast a radioactive decay occurs and varies by time as the speed of radioactive decay decreases by time. Thus, since

∆N = -λ ∙ N0 ∙ ∆t

we obtain for the initial rate of radioactive decay:

R0 (t) = ∆N/∆t = -λ ∙ N0

and for the rate R(t) of radioactive decay at a given time t:

R(t) = R0 e-λ ∙ t

The unit of radioactive decay rate (if measured in number of decays/second) is known as Becquerel, Bq.

The relationship between undecayed nuclei and the time elapsed is given by an inverse function (an inverse function has the form y(t) = C/x where C is a constant). The graph of such functions is a hyperbola, as shown in the figure below, where the data regarding radioactive decay of Magnesium-28 are shown. ### Example 5

Based on the graph shown above for radioactive action of Magnesium-28, calculate:

1. Half-life period of Mg-28
2. The number of undecayed nuclei after four half-life cycles if initially there were 1200 nuclei in the sample
3. How long does it take to be still 10 nuclei undecayed in the sample?
4. What is the rate of radioactive decay for Mg-28 after 60 days from the beginning of process?

### Solution 5

Clues:

N0 = 1200
T1/2 = ?
N(t = 4·T1/2) = ?
t(N = 10) = ?
R(t) = ?

1. From the graph, we see that the radioactive nuclei in the vertical axis are not given as a number but as a part of whole. We see that half of the initial sample (which correspond to the value 0.5 in the vertical axis) correspond to 20 days in the (horizontal) time axis. Thus, T1/2 = 20 days.
2. Four half-life cycles correspond to t = 4 · T1/2 = 4 · 20 days = 80 days. Since we cannot find the number of undecayed nuclei at this instant as the graph units are not divided so minutely, we find the number of undecayed nuclei at t = 80 days through calculations. Thus, since the number of undecayed nuclei halves by every half-life cycle, we have for the remaining nuclei after 4 half-cycles:
N(80 days) = N0/24
= 1200/16
= 75 nuclei
3. We must find the constant of radioactive decay λ first, and then calculate the number of undecayed nuclei at the given time. We have
λ = ln2/T1/2
= 0.693/20 days
= 0.03465 day-1
Hence, the time in which N(t) = 10, is calculated by
∆N = -λ ∙ N0 ∙ ∆t
∆t = -∆N/λ ∙ N0
= -N(t) - N0/λ ∙ N0
= -10 - 1200/(0.03465 day-1 ) ∙ 1200
= --1190/41.58 days
= 28.62 days
4. The initial rate R(t) of radioactive decay for Mg-28 (we can skip the negative sign, as it only indicates that the number of nucleons dercreases), is
R0 (t) = λ ∙ N0
= (0.03465 day-1 ) ∙ 1200
= 41.58 nuclei/day
Hence, the rate of radioactive decay at 60 days after the beginning of process is
R(t) = R0 e-λ ∙ t
R(60) = 41.58 ∙ e(-0.03465) ∙ (60)
= 41.58 ∙ e-2.079
= 41.58 ∙ 0.125
≈ 5.2 nuclei/day

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