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23.2Radioactivity

At the end of the nineteenth century and in the early part of the twentieth century, it was discovered that most nuclides are unstable. This means that the nuclei emit particles and energy. This discovery was mainly due to the work of Henri Becquerel (1852–1908), Marie Sklodowska-Curie (1867–1934) and Pierre Curie (1859–1906).

Radioactive decay

An unstable nucleus is one that randomly it cannot be predicted when or which nucleus will decay and spontaneously the decay cannot be prevented from happening or forced to happen emits particles that carry energy away from the nucleus.

The emission of particles and energy from a nucleus is called radioactivity the phenomenon where particles and energy are emitted by an unstable nucleus. It was soon realised that three distinct emissions take place. The emissions are called alpha particles, beta particles and gamma rays. They have different ionising power (i.e. ability to knock electrons off atoms) and penetrating power (distance travelled through matter before they are stopped).

Alpha particles and alpha decay

An example of alpha decay a decay of a nucleus producing an alpha particle is uranium decaying into thorium:

U 92 238→Th 90 234+α 2 4

An alpha particle is emitted from the parent nucleus the decaying nucleus (uranium) creating the daughter nucleus the nucleus produced by radioactive decay (thorium).

The alpha particles were shown to be identical to nuclei of helium in an experiment by E. Rutherford and T. Royds in 1909. They collected the gas that the alpha particles produced when they came in contact with electrons and then investigated its spectrum. The spectrum was found to be identical to that of helium gas. Alpha particles have a mass that is about four times the mass of the hydrogen atom and an electric charge equal to +2e.

Notice the balancing of the total proton and nucleon numbers on both sides of the reaction equation as we discussed in Section 23.1.

The energy released in this decay can be calculated with what we learned in Section 23.1. The atomic masses are U = 238.0508 u, Th = 234.0436 u, He = 4.0026 u. The mass difference is then

238.0508−234.0436−4.0026=0.0046 u

The energy released is 0.0046 × 931.5 = 4.3 MeV.

Two other examples of alpha decay are:

Ra 88 224→Rn 86 220+α 2 4

Po 84 212→Pb 82 208+α 2 4

Why does a nucleus emit alpha particles (helium nuclei) rather than other nuclei (or even single protons or neutrons)? A partial reason is the fact that helium-4 has a very large binding energy and so the decay is energetically possible. For most nuclei, emitting a proton would require energy rather than releasing it, and so it does not happen (but there are exceptions to this statement).

Beta particles and beta decay

In beta minus decay a decay of a nucleus producing an electron and an anti-neutrino, a neutron in the decaying nucleus turns into a proton, emitting an electron and an anti-neutrino. An example is the nucleus of thorium decaying into a nucleus of protactinium:

Th 90 234→Pa 91 234+e−1 0+ν¯0 0

The ‘beta minus particle’ is just the electron. It was called beta minus before experiments showed it was identical to the electron. Note again how the proton and nucleon numbers balance in the reaction equation.

Also note that unlike alpha decay, where two particles are produced, here we have three. The third is called the ‘anti-neutrino’, ν¯. The bar over the symbol indicates that this is an anti-particle. (It was predicted theoretically and subsequently verified in experiments that all particles in nature have a corresponding anti-particle of the same mass but opposite electric charge.)

Notice that in beta minus decay, one neutron in the parent nucleus gets converted into a proton. This explains why the proton number of the daughter nucleus increases by one, whereas the nucleon number stays the same.

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Two other examples of beta minus decay are:

Pb 82 214→Bi 83 214+e−1 0+ν¯0 0

K 19 40→Ca 20 40+e−1 0+ν¯0 0

Another type of beta decay is beta plus decay a decay of a nucleus producing a positron and a neutrino. Instead of emitting an electron the nucleus emits its anti-particle, the positron, which is positively charged. The third particle is the neutrino. Two examples of beta plus decay are:

Na 11 22→Ne 10 22+e+1 0+ν 0 0

N 7 13→C 6 13+e+1 0+ν 0 0

It is tempting to think that the electron in beta minus decay (and the positron in beta plus decay) was emitted from within the nucleus, where the electron found itself. It can be shown that an electron cannot exist within a nucleus. The electron and the antineutrino were created through the action of the weak nuclear force in a process that is beyond the level of this book.

Gamma rays and gamma decay

In gamma decay a decay of a nucleus producing a short wavelength photon a nucleus emits a gamma ray, in other words a photon of high-frequency electromagnetic radiation:

U 92 238→U 92 238+γ 0 0

Unlike alpha and beta decay, in gamma decay the nucleus does not change identity. It just moves from a higher to a lower nuclear energy level. The wavelength of the photon emitted is given by:

λ=h c E

just as with atomic transitions. Here E is the energy of the emitted photon. In contrast to the photons in atomic transitions, which can correspond to visible light, these photons have very small wavelength (smaller than 10−12 m), and they are called gamma rays.

Other examples of gamma decay are:

Ni 28 60→Ni 28 60+γ 0 0

Mg 12 24→Mg 12 24+γ 0 0

The identification of gamma rays with photons was made possible through diffraction experiments in which gamma rays from decaying nuclei were directed at crystals. The wavelengths were measured from the resulting diffraction patterns.

Check yourself 4

A nucleus of Cs 55 137 decays into a nucleus of Ba 56 137. What kind of decay is this?

Properties of alpha, beta and gamma radiations

Table 23.2 summarises the properties of alpha, beta and gamma radiations. Notice that alpha particles are the most ionising and gamma rays the most penetrating.

Characteristic | Alpha particle | Beta particle | Gamma ray
Nature | Helium nucleus | (Fast) electron | Photon
Charge | +2e | −e | 0
Mass | 6.64 × 10−27 kg | 9.1 × 10−31 kg | 0
Penetrating power | A few cm of air | A few mm of metal | Many cm of lead
Ions per mm of air for 2 MeV particles | 10 000 | 100 | 1
Detection | Affects photographic film | Affects photographic film | Affects photographic film
Is affected by electric and magnetic fields | Is affected by electric and magnetic fields | Is not affected by electric and magnetic fields

Table 23.2: Properties of alpha, beta and gamma radiations.

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Decay series

The changes in the proton and nucleon numbers of a nucleus when it undergoes radioactive decay can be represented in a diagram of nucleon number against proton number. A radioactive nucleus such as thorium (Z = 90) decays first by alpha decay into the nucleus of radium (Z = 88). Radium, which is also radioactive, decays into actinium (Z = 89) by beta decay. Further decays take place until the resulting nucleus is stable. The set of decays that takes place until a given nucleus ends up as a stable nucleus is called the decay series the series of decays until a nucleus reaches a stable nuclide of the nucleus. Figure 23.2 shows the decay series for thorium. Successive decays starting with thorium end with the stable nucleus of lead (Z = 82, A = 208).

Figure 23.2: The decay series of thorium (Z = 90, A = 232). One alpha decay reduces the nucleon number by 4 and the proton number by 2. One beta minus decay increases the proton number by one and leaves the nucleon number unchanged. The end result is the nucleus of lead (Z = 82, A = 208).

Check yourself 5

Part of the decay series of polonium-215 includes polonium-211. By what series of decays does this happen?

Worked example 23.5

A nucleus Z A X decays by alpha decay followed by two successive beta minus decays.

Find the proton and nucleon numbers of the resulting nucleus.

Answer

The decay equation for alpha decay is:

Z A X→Z−2 A−4 Y+2 4 α

Then we have two beta decays:

Z−2 A−4 Y→Z−1 A−4 Y′+−1 0 e+ν¯ and Z−1 A−4 Y′→Z A−4 X+−1 0 e+ν¯

The nucleon number doesn’t change in beta minus decay, but the proton number increases by one for each decay (since in beta minus decay a neutron turns into a proton).

So the proton number of the resulting nucleus is Z and the nucleon number is A − 4.

The law of radioactive decay

Radioactive decay is random and spontaneous. By random we mean that we cannot predict which unstable nucleus in a sample will decay or when there will be a decay. It is spontaneous because we cannot affect the rate of decay of a given sample in any way. Although we cannot predict or influence when a particular nucleus will decay, we know that the number of nuclei that will decay per second is proportional to the number of nuclei in the sample that have not yet decayed.

The law of radioactive decay states that the rate of decay is proportional to the number of nuclei that have not yet decayed:

Δ N Δ t∝N

A consequence of this law is that the number of radioactive nuclei decreases exponentially.

Consider the beta minus decay of thallium (the parent nucleus) into lead (the daughter nucleus):

Tl 81 208→Pb 82 208+e−1 0+ν¯0 0

The isotope of lead is stable and does not decay. Figure 23.3a shows how the number of thallium nuclei decreases with time. Initially (at t = 0) there are 1.6 × 1022 nuclei of thallium. This corresponds to a mass of about 6 g of thallium. After 3 min the number of thallium nuclei left is half of the initial number (0.8 × 1022). After another 3 min, the number is one-quarter of the initial number (0.4 × 1022). After yet another 3 min the number of thallium nuclei is one-eighth of the initial number (0.2 × 1022).

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The time of 3.0 min is called the half-life time after which the activity is reduced by a factor of 2 of thallium. It is the time after which the number of the radioactive nuclei is reduced by a factor of 2. The blue curve in Figure 23.3b shows how the number of lead nuclei increases with time.

Figure 23.3: a The number of thallium nuclei decreases exponentially. b As thallium decays (red curve) lead is produced and so the number of lead nuclei increases (blue curve).

Check yourself 6

The half-life of an element is one month. Does this mean that, after two months, all the radioactive material will have decayed away?

A concept that is useful in experimental work is that of decay rate or activity the rate of decay (number of decays per second) A: this is the number of decays per second. We cannot easily measure how many unstable nuclei are present in a sample, but we can detect the decays. The unit of activity is the becquerel (Bq): 1 Bq is equal to one decay per second.

Activity obeys the same exponential decay law as the number of nuclei. In a time equal to the half-life the activity is reduced by a factor of 2. This is shown in Figure 23.4.

Figure 23.4: The activity of thallium decreases exponentially with time. You can also use this graph to determine half-life.

It is best to define the half-life in terms of activity:

KEY POINT

Half-life is the interval of time after which the activity of a radioactive sample is reduced by a factor of 2.

Check yourself 7

There are 100 × 1025 nuclei of a radioactive isotope. After how many half-lives will 75 × 1025 nuclei decay?

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Provided the half-life is not too long, a graph of activity against time can be used to determine the half-life. In Figure 23.4 the activity approaches zero as the time increases. In practice, however, this is not the case. The detector that measures activity from the radioactive sample under study also measures the activity from natural sources. As a result, the activity does not approach zero; it approaches the activity due to all other sources of background radiation radiation from natural sources such as cosmic rays and radioactive material in rocks. These background sources include cosmic rays from the sun, radioactive material in rocks and the ground, radiation from nuclear weapons testing grounds, and so on.

The effect of background radiation can be seen in the activity curve of Figure 23.5a. This shows a background rate of 40 Bq. By subtracting this value from all data points we get the graph in Figure 23.5b. Using the corrected graph we get a half-life of 6.0 min. Using Figure 23.5a without correcting for the background gives a half-life of 6.9 min, which is inaccurate by 15%.

Figure 23.5: a An activity curve that includes a background rate of 40 Bq. This curve cannot be used to measure half-life as it is. The background needs to be subtracted as shown in b.

Activity measures the total number of decays per second of a radioactive sample. In practice, an instrument, such as a Geiger–Muller tube, will be used to measure this activity. Obviously, the instrument will only record those α, β or γ particles that enter the tube and not all the particles emitted by the sample. The reading of the instrument is called count rate and, like activity, is measured in Bq or in counts per minute. Count rate is also exponentially decreasing with time and may also be used to measure half-lives.

Worked example 23.6

An isotope has a half-life of 20 min. Initially there are 1024 g of this isotope. Determine the time after which 128 g are left.

Answer

Find the fraction remaining: 128 1024=1 8.

This corresponds to three half-lives, as 1 2×1 2×1 2=1 8.

Since the half-life is 20 min, three half-lives is:

3×20=60 min

After 60 min, 128 g are left.

Worked example 23.7

The count rate for a sample is initially 80 counts per minute. After one half-life the count rate becomes 48 counts per minute.

aWhat is the background count rate?

bWhat is the count rate after an additional two half-lives go by?

Answer

aLet the initial count rate from the sample alone be C 0 and the background rate be b. Then 80 = C 0 + b. After one half-life, the count rate from the sample alone becomes C 0 2 and so 48=C 0 2+b. Solving, we find b = 16 counts per minute (and C 0 = 64 counts per minute).

bAfter three half-lives from the initial time the count rate from the sample alone will be 64 → 32 → 16 → 8 counts per minute. With the background, the count rate will then be 8 + 16 = 24 counts per minute.

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Worked example 23.8

The activity of a sample is 15 decays per minute. The half-life is 30 min. Predict the time when the activity was 60 decays per minute.

Answer

One half-life before the sample was given to us the activity was 30 decays per minute, and one half-life before that it was 60 decays per minute.

So the activity was 60 decays per minute two half-lives earlier, which is 60 minutes earlier.

Half-life and probability

The meaning of a half-life can also be understood in terms of probability. Any given nucleus has a 50% chance of decaying within a time interval equal to the half-life. If a half-life goes by and the nucleus has not decayed, the chance that it will decay in the next half-life is still 50%. This is shown as a tree diagram in Figure 23.6.

Figure 23.6: Tree diagram for nuclear decay.

The probability that a nucleus will have decayed by the second half-life is the sum of the probability that it decays in the first half-life and the probability that it decays during the second half-life:

1 2+(1 2×1 2)=3 4=0.75 or 75%

Check yourself 8

The half-life of a radioactive element is 5 minutes. A nucleus of this element has not decayed after the passage of 5 minutes. What is the probability that this nucleus will decay in the next 5 minutes?

nature of science

Accidental discovery

The discovery of radioactivity is an example of an accidental discovery. Henri Becquerel, working in Paris in 1896, believed that minerals made phosphorescent (emitting light) by visible light might give off X-rays. His idea was to wrap a photographic plate in black paper and place on it a phosphorescent uranium mineral that had been exposed to bright sunlight. But the sun did not shine, and he stopped the experiment, placing the wrapped plate and the mineral in a drawer. A few days later he developed the photographic plate, expecting to see only a very weak image. To his surprise, there was a very strong image. Becquerel concluded that this image was formed by a new kind of radiation that had nothing to do with light. The radiation came from the uranium mineral.

Science in context

Radioactivity has many practical applications. A common household device is a smoke detector. The device contains a very small quantity of a radioactive element. The radiation produced when this element decays ionises the surrounding air molecules, and a small current is established in a circuit. When smoke enters the device, the radiation is absorbed by the larger smoke particles and the current is reduced, triggering an alarm.

Radioactive dating is another important application. Using isotopes with very long half-lives, it has been established that earth and moon rocks are equally old, about 4.5 billion years, providing stringent tests of theories about the formation of the solar system. Use of carbon-14, with a half-life of 5730 years, has made possible the determination of the age of ancient objects. Living things (trees or animals) contain a fixed amount of carbon-14 as long as they are alive.

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After death, the amount of carbon-14 is reduced according to the exponential decay law. Measuring the content of carbon-14 in the object therefore allows a determination of its age. So if a bone, say, contains half the amount of carbon-14 it would have in a living bone, its age must be 5730 years.

Radioactivity can also be used to find a leak in a water pipe buried into the ground: a small amount of a radioactive isotope that emits gamma rays is inserted in the pipe. Water collects at the point of the leak, and therefore that area becomes a strong gamma ray emitter. The gamma rays are penetrating enough to go through the soil, so they can be detected at the surface, thus locating the leakage point. An alpha particle emitter isotope would not be appropriate since the alpha particles would be absorbed by the soil. (Some beta emitting isotopes can also be used.) The half-life of the isotope used must be a few days: this is long enough for detectable emissions to reach the surface but short enough to limit exposure and contamination to people and animals that drink the water.

Nuclear physics has found significant applications in medicine for example in the use of radioisotopes for the treatment of cancer as well as in imaging using radioisotope tracing. The injection of radioactive materials in the human body carries risks, and so the choice of isotope and the dosage must be carefully calculated to maximize the intended effect and minimize damage to the rest of the body. Other imaging techniques involving important nuclear physics phenomena include positron emission tomography (PET scans), magnetic resonance imaging (MRI) and many other applications.

TEST YOUR UNDERSTANDING

(You can look at the periodic table for names of elements.)

23Describe what is meant by radioactivity.

24Radioactive decay is described as random and spontaneous. Explain these terms in the context of radioactivity.

25Write down the equation for the beta minus decay of H 1 3.

26Write down the equation for the beta plus decay of Mg 12 23.

27Bismuth (Bi 83 210) decays by beta minus decay, followed by gamma emission. State the equation for the reaction and the proton and nucleon numbers of the nucleus produced.

28Plutonium (Pu 94 239) decays by alpha decay. State the equation for this reaction and name the nucleus plutonium decays into.

29By what sequence of decays would a nucleus X Z A end with an isotope of X?

30Small, stable nuclei tend to have equal numbers of protons and neutrons. A small (Z < 20) unstable nucleus has fewer neutrons than protons. What is the likely decay of this nucleus?

31Small, stable nuclei tend to have equal numbers of protons and neutrons. A small (Z < 20) unstable nucleus has more neutrons than protons. What is the likely decay of this nucleus?

32Alpha decay occurs mainly for large, heavy nuclei, but one exception to this is the decay of beryllium (Be 4 8). Write down the equation for the decay and calculate the energy released. The atomic mass of beryllium is 8.0053 u.

33What sequence of decays of a nucleus X Z A will end with a nucleus with proton number Z – 3 and nucleon number A – 4?

34What sequence of decays of a nucleus X Z A will end with a nucleus with proton number Z – 3 and nucleon number A – 8?

35What sequence of decays of a nucleus X Z A will end with a nucleus with proton number Z – 4 and nucleon number A – 4?

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36Outline what is meant by the activity of a radioactive sample.

37What is the activity in Bq of a sample that has 3600 decays per minute?

38A radioactive source has a half-life of 3.0 min. At the start of an experiment 32.0 mg of the radioactive material is present. Determine how much will be left after 18.0 min.

39An unstable isotope X decays into a stable isotope Y. What is the ratio of the number of nuclei Y to the number of nuclei X after 3 half-lives? No nuclei of Y were present initially.

40An unstable isotope X decays into a stable isotope Y. After 12 minutes the ratio of the number of nuclei Y to the number of nuclei X is 3. No nuclei of Y were present initially. What is the half-life of X?

41The half-life of an isotope is 5 minutes. It is observed that a nucleus in a sample of the isotope has not decayed after 5 minutes. A student claims that it is now more likely that this nucleus will decay in the next 5 minutes. Comment on this statement.

42The count rate for a radioactive sample is 10 000 counts per minute. After 36 days the count rate is reduced to 625 counts per minute. What is the half-life?

43At a particular location, the background count rate is 20 counts per minute. An experiment at this location shows a count rate of 310 counts per minute for a radioactive sample. The half-life is four days. What will be the count rate 4 days from the start of the experiment?

44At a particular location, the background count rate is 25 counts per minute. An experiment at this location shows a count rate of 565 counts per minute for a radioactive sample. The count rate becomes 160 counts per minute after 12 days. What is the half-life?

45In a study of the intensity of gamma rays from a radioactive source it is suspected that the counter rate C at a distance d from the source behaves as C∝(1 d+d 0)2 where d 0 is an unknown constant. A set of data for C and d is given. Outline how the data must be plotted in order to get a straight line.

46The isotope K 19 40 of potassium is unstable, with a half-life of 1.25 × 109 years. It decays into the stable isotope Ar 18 40. moon rocks were found to contain a ratio of potassium to argon atoms of 1 : 7. Find the age of the moon rocks. (You may assume that no argon was present in the rocks when they were formed.)

47Figure 23.7 shows the variation with time of the activity of a radioactive sample.

Figure 23.7: For question 47.

aState what is meant by activity.

bUse the graph to estimate the half-life of the sample.

cOn a copy of the graph, extend the curve to show the variation of the activity for a time up to 12 minutes.

dThe sample contains a radioactive element X that decays in to a stable element Y. At t = 0, no atoms of element Y are present in the sample. Determine the time after which the ratio of Y atoms to X atoms is 7.