Nuclear Physics Part 2: Radioactive Decay Last modified: 17/10/2017
Part A: Decay Reactions What is a Decay? Alpha Decay Definition Q-value Example Not Every Alpha Decay is Possible Beta Decay β rays are electrons Anti-particles Neutrinos Basic Reactions β Decay β + Decay Not All Beta Decays Are Possible Electron Capture Example Branching Ratios Decays & the Line of Stability Gamma Decay Decay Chains Part B: Decay Statistics
What is a Decay? A radioactive decay process is a special case of a nuclear reaction, where there is only one nucleus at the start. A decay will occur spontaneously, without external influences. A Z X A1 Z 1 Y 1 + A2 Z 2 Y 2 +... The nuclei A1 Z 1 Y 1, A2 Z 2 Y 2... are known as decay products or the daughter nuclei of the parent nucleus A ZX. Many different decay processes are possible, but three are very much more common than the others. We will now take a closer look at these common decay mechanisms, all first observed in the late 1800 s, and very unimaginatively named: α decay β decay γ decay
Alpha Decay Alpha decay is the emission of an α particle (a 4 2He nucleus) from a larger nucleus. We saw in a previous lecture that 4 2He is a particularly tightly bound nuclide, so overall binding energy can often be reduced by such a decay. Because the α is relatively large, it is mostly heavier nuclei which will decay this way. Nearly all of the energy released (typically 5 MeV) is seen as kinetic energy of the α particle. At these energies, the α is non-relativistic. The general equation for the alpha decay of nuclide A ZX is straightforward: A Z X A 4 Z 2Y + 4 2He
Some examples of α decay: 238 92 U 234 90Th + 4 2He 212 84 Po 208 82Pb + 4 2He Because the α is relatively large and low energy, it cannot travel any great distance - only a few centimetres in air - and is blocked completely by a sheet of paper, or human skin. Alpha radiation from an external source is not normally a health risk. Many domestic smoke alarms take advantage of the easily stopped alpha particles. A small amount of the artificially produced radioisotope americium-241 is used. 241 95Am decays via alpha decay. The flow of the emitted alpha particles across a small air gap is measured as a current. If the density of the air suddenly increases (i.e. with smoke particles), alphas will be blocked and the reduced current triggers the alarm.
Q-value Example Write down the reaction and calculate Q for the α decay of 241 Am The reaction must be : 241 95 Am A ZX + 4 2He Simple arithmetic gives the unknown Z = 93 and A = 237. After checking the periodic table, we can complete the reaction: and calculate the Q-value: 241 95 Am 237 93Np + 4 2He Q = [M Am (M Np + M He )] 931.5 = [241.0568291 237.0481734 4.002602] 931.5 = 5.64 MeV
Not Every Alpha Decay is Possible It is easy to write down an alpha decay equation for any nuclide larger than 4 2He. For example 12 6C: The Q-value of this reaction is: 12 6 C 8 4Be + 4 2He Q = [M C (M Be + M He )] 931.5 = [12 8.0053051 4.002602] 931.5 = 7.37 MeV This negative value indicates that 12 6C is stable against α decay (i.e. the reaction does not occur). Since there is only a single initial particle, it is not possible to add kinetic energy as we have seen with other endothermic reactions. Just because we can write down the equation for a decay, it doesn t automatically follow that the reaction will occur in nature.
Beta Decay Beta radiation was determined to be the emission of a low-mass, charged particle. The mass of this emitted particle is always the same, but the charge can be positive (a β + decay) or negative (β decay). β Decay β + Decay + Unlike α decays, which are mostly seen in heavier nuclides, nuclides with a wide range of masses are observed to β decay - from very light (e.g. 3 1H) to mid-size (e.g. 60 27Co) to very heavy (e.g. 234 90Th). Further investigations identified the negatively charged particle to be an electron (e ) and the one with positive charge, an anti-electron (e + ) also known as a positron.
Anti-Particles Every particle has a corresponding anti-particle, which has the same mass as the particle, but opposite values of all other properties, including charge. Except for the positron e +, antiparticles are represented by placing a bar over the particle symbol. For example the anti-proton is p, and the anti-neutron n. When a particle and its anti-particle meet, they will annihilate into two photons. e.g. For an electron and positron: e + e + γ + γ Fortunately for us, the universe is made up predominantly of matter with only very small amounts of anti-matter. The Big Bang is believed to have produced equal amounts of matter and anti-matter, before a currently not understood process caused the balance to change.
In a beta decay, the number of protons in a nucleus changes. The basic reactions appear to be: n p + e and p n + e + Because the electron/positron is so much lighter than the nucleus, energy and momentum conservation predict that the kinetic energy of the emitted beta particle should be very close to the Q-value of the decay. Experimentally however, this is not what is observed. Betas are seen with a range of kinetic energies, including zero! Plotting the numbers of emitted electrons/positrons observed in a β ± decay against their kinetic energy will look something like the graph shown at right. The exact curve will vary, depending on the particular decay, but the general shape is always the same. 0 0.2 0.4 0.6 0.8 1 KE electron /Q
Neutrinos The explanation for this is that another particle - the neutrino - is also emitted, and the decay energy is shared between the beta and neutrino. The neutrino (ν) is a particle with zero charge and a very small mass (less than one-millionth of the electron mass). It interacts extremely weakly with other matter - billions of neutrinos produced by the Sun are passing through our bodies every second without us noticing. Because of this the neutrino is very difficult to observe experimentally. After its existence was first proposed in 1930 to solve the beta decay energy problem, it took until 1956 before it was directly observed in an experiment. It wasn t until about 2000 before it was conclusively shown that, though tiny, the neutrino mass is NOT zero.
Including the neutrino, the full reactions for β and β + decay, along with a third variation of the same process known as electron capture are: β : β + : electron capture : n p + e + ν p n + e + + ν p + e n + ν Where the ν is an anti-neutrino. The mass of a neutron is a little larger than the mass of the proton plus the mass of an electron, so the first of these reactions is exothermic - a free neutron is unstable and will decay into a proton and an electron (taking, on average, about 15 minutes). The second reaction is endothermic, so the proton is stable. Beta decays occur when a nucleon inside a nucleus undergoes one of the above processes.
β Decay In a β decay, one of the neutrons in a nucleus decays to a proton and an electron: A Z X Z+1Y A + 1β 0 + ν (The emitted electron is often written as β to emphasize that it is an energetic decay product, not one of the regular orbital electrons. Setting Z = 1 on the β looks a little odd, but helps with book-keeping.) The distinction between atomic and nuclear masses is especially important in calculating the Q-value of a β decay. Q = (m X (m Y + m e )) c 2 = ((M X Zm e ) (M Y (Z + 1)m e + m e )) c 2 (nuclear masses) = (M X M Y ) c 2 (atomic masses)
Write down the reaction, and calculate the Q-value of the β decay of tritium ( 3 1H) The reaction is: 3 1H 3 2He + 1β 0 + ν And, using atomic masses, the Q-value is: Q = [ M ( 3 H) M ( 3 He)] 931.5 = [3.0160492779 3.0160293191] 931.5 = 18.6 kev The maximum emitted electron energy is 18.6 kev ( 3% electron rest mass energy). Most beta decays have larger Q-values (in the MeV range) which means that Special Relativity is usually necessary to describe the energy and momentum of the emitted electrons.
β + Decay A single free proton cannot beta decay, but sometimes a proton inside a nucleus will undergo this process to give a β + decay: A Z X For example, 18 9F can β + decay: Z 1Y A + +1β 0 + + ν 18 9 F 18 9O + 0 1β + + ν Calculating the Q-value again requires care with mass definitions: Q = [m X (m Y + m e )] c 2 (nuclear masses) = [(M X Zm e ) (M Y (Z 1)m e + m e )] c 2 = [M X M Y 2m e ] c 2 (atomic masses)
Example Q-value Calculate the Q-value for the β + decay of 18 9F. Using the equation from the previous page: Q = [M F M O 2m e ] 931.5 = [18.000938 17.9991610 2(0.00054858)] 931.5 = 0.63 MeV 18 9 F is often used medically in PET ( Positron-Emission Tomography) scans. Measurements of the gamma rays produced from positron-electron annihilation can be used to constuct images of cancerous tumours etc.
Not All Beta Decays Are Possible Again, we can write down hypothetical β-decay reactions for any nuclide: e.g 12 6 C 12 7N + 0 1β + ν and 12 6 C 12 5B + 0 1β + + ν but these reactions will only actually be possible if Q > 0. Q β = (M C M N ) 931.5 use atomic masses! = (12 12.0186132) 931.5 = 17.33 MeV and Q β + = (M C M B 2m e ) 931.5 atomic masses, again = (12 12.0143521 2 0.0005486) 931.5 = 14.4 MeV These negative Q-values tell us that 12 6C is stable against β ± decay. (As well as α decay, shown previously!).
Electron Capture Electron capture is a variation on the β + decay reaction: p + e n + ν This process can occur within a nucleus, where the electron is one of the atom s orbital electrons, captured by the nucleus. A Z X + e A Z 1Y + ν For example: 7 4Be + 1e 0 7 3Li + ν Yet again, we need to be careful when calculating the Q-value of this reaction: Q = (m X + m e m Y ) c 2 = [(M X Zm e + m e ) (M Y (Z 1)m e )] c 2 (nuclear masses) = (M X M Y ) c 2 (atomic masses)
Example Q-value As well as β + decay, 18 9F can also decay via electron capture. Write down the full reaction for this decay, and calculate the Q-value. Writing the reaction is straightforward: And the Q-value is: 18 9 F + 0 1e 18 8O + ν Q = (M F M O ) 931.5 = (18.0009380 17.9991610) 931.5 = 1.66 MeV Most of this energy will be kinetic energy of the neutrino.
Branching Ratios We have just seen that 18 9F can decay in two different ways. This is not unusual for radioisotopes, particularly for those (like 18 9F) that can decay via both β + and electron capture. Some nuclides can decay in three or more possible ways. Remember that nuclei are quantum systems, so we can never predict exactly what a single nucleus will do. We can only express our knowledge in terms of probabilities. In the case of 18 9F, the probability of a β + decay is 97%, and for electron capture, 3%. These percentages are known as the branching ratios of the decays. 18 9 F is typical, in that one of the decays is much more common than the others, but this is not always true. 212 83Bi, for example, has branching ratios of 64% for α decay and 36% for β.
Decays & the Line of Stability In the last lecture, we saw the line of stability plot shown again at right. N 160 Isotope Lifetimes It is possible to use a nuclide s position on this plot to predict which type of decay is most likely to occur. Remember that a longer lifetime indicates a more stable nucleus. 140 120 100 80 stable We expect that the daughter nucleus of a decay should be a darker colour on this plot than the parent, and more generally, that a decay will 60 40 20 move towards the line of stability (in black). 20 40 60 80 100 longer life Z
An alpha decay will decrease both Z and N, so the daughter nucleus will be closer to the origin of the N-Z plot than the parent. N N 160 140 120 α decay likely N = 2 100 Z = 2 Z 80 In which part of the plot does this move us closer to the line of stability? The top right. 60 40 20 We predict that nuclides in this region will predominantly α decay. 20 40 60 80 100 Z
decays also involve a diagonal movement on the plot: β ± N 160 α decay likely N N = 1 β N = 1 N Z = 1 β + 140 120 Z Z = 1 We expect whichever of these diagonal movements gets closer to the line of stability will be the likely decay, so: Z 100 80 60 β likely β + likely A β decay is likely for a nuclide above the line. β + decay (or possibly electron capture) will be likely for nuclides below the line. 40 20 20 40 60 80 100 Z
N 160 The plot at right is shaded to show the decay mode with the highest experimentally measured branching ratio for the unstable nuclides. We can see that it is a little more complicated than our simple predictions, but the broad trends are correct. 140 120 100 80 60 40 20 α decay β decay β + decay electron capture 20 40 60 80 100 Z
Gamma Decay A gamma ray is a high-energy photon, usually emitted by a nucleus in an excited state dropping to a lower energy (usually the ground) state: A Z X A ZX + γ The * indicates that the nucleus is in an excited state. Gamma decay usually occurs after an α or β decay where the daughter nucleus is often not in the ground state. For example, some of you will have done a lab experiment involving the β decay: 137 55 Cs 137 56Ba + β + ν followed by the γ decay: 137 56 Ba 137 56Ba + γ
The gamma rays emitted in the decay of an excited state nucleon typically have energies in the range of a few MeV. Slightly lower energy gamma rays are also seen after a β + decay as positrons annihilate with electrons. Unlike α and β radiation which are readily blocked by small thicknesses of material, gamma rays can be very penetrating. Because of their high energy, these photons can cause significant damage to cells if they are absorbed in the body and can cause cancer and other illnesses. Blocking gamma rays requires a significant thickness of dense material. A few centimetres of lead is often used in smaller equipment in labs. In large facilities such as nuclear power plants, concrete walls several metres thick are generally used to block gamma radiation.
Decay Chains Often the daughter nuclide of an α or β decay will also be radioactive. Heavy nuclides in particular will usually undergo a chain of decays to a series of other radionuclides, before reaching a stable isotope. There may also be multiple pathways for this decay chain to occur. For instance the (naturally occurring) nuclide 238 92U can decay to the stable nuclide 206 92Pb by a series of fourteen different decays, in one of sixteen different paths (Not all are equally likely). Showing this in a diagram (next page) is usually the best way to understand what is happening. Analyzing data from such decay chains becomes quite complicated!
238 92 U α 234 92 U β 234 91 Pa β 234 90 Th α 230 90 Th A Decay Chain Z α 226 88 Ra 238 92 U decaying to 204 82Pb α 222 86 Rn 218 86 Rn α β 218 85 At α Decays indicated by a dashed arrow have a branching ratio < 1%. β 218 84 Po α α 214 84 Po β 214 83 Bi α 210 84 Po β 210 83 Bi α β 214 82 Pb α β 210 82 Pb β 210 81 Tl α α 206 82 Pb β 206 81 Tl β 206 80 Hg
Part A: Decay Reactions Part B: Decay Statistics Decay Formula Decay Constant & Mean-Life Example Half-Life Example Half-Life & the Decay Equation Example Example Activity Example Example Application: Radiocarbon Dating
Decay Formula Because it is a quantum process, we can only ever determine the probability that a particular nucleus will decay. Fortunately, we are usually dealing with very large numbers of nuclei, so we can make useful calculations of the numbers of decays occurring in a sample of material. Logically, the rate of decay of any sample of a nuclide must be proportional to the number of nuclei present. If N (t) is the number of nuclei present at time t, then we expect that the rate of change in N will satisfy: dn (t) dt N (t) Since N is decreasing, the constant of proportionality must be negative. We therefore have (with λ being a positive constant called the decay constant): dn (t) = λ N (t) dt
This is another simple example of a differential equation. To solve this equation, we can rewrite it as: dn N then integrate with respect to time: t t=0 dn N = λ dt = λ t t=0 dt [ln N ] t t=0 = λt ( ) N (t) ln N (t) ln N (0) = ln = λt N (0) Taking the exponential of both sides gives our desired solution: N (t) = N (0)e λt N 0 e λt
Decay Constant The decay constant λ is a property of the particular nuclide. Every nucleus of this type will have the same value of the decay constant. From the decay equation we see that: larger λ faster decay smaller λ slower decay, and if λ = 0, then there is no change in N i.e. the nuclide is stable. Again looking at the decay equation, we should realize that λt must be dimensionless, so the dimensions of λ are (time) 1. The SI unit of λ is s 1 The decay constant is connected to the probability of a nucleus decaying. τ = 1 λ is the average lifetime (or mean-life) of an individual nucleus of this nuclide.
Example An object is analyzed and found to contain 10 21 nuclei of a nuclide X. 3 days later, the same object is measured again and found to now have 10 20 X nuclei. What is the decay constant of X? The first measurement is taken to be at time t = 0, so N 0 = 10 21. At t = 3 days = 3 24 60 60 = 259, 200 seconds, N (t) = 10 20. The decay formula gives: N (t) = N 0 e λt 10 20 = 10 21 e 259,200 λ ln 1 = 259, 200 λ 10 λ = (ln 10)/(259, 200) = 8.88 10 6 s 1
Half-Life The half-life T 1/2, of a nuclide is defined to be the time taken for 50% of a sample of that nuclide to decay. In other words: N (T 1/2 ) = 1 2 N 0 Using the decay equation: and rearranging gives: 1 2 N 0 = N (T 1/2 ) = N 0 e λt 1/2 1 2 = e λt 1/2 ln 2 = λt 1/2 T 1/2 = ln 2 λ = (ln 2)τ
Example What are (a) the mean-life and (b) the half-life of the nuclide in the previous example? We previously calculated the decay constant λ = 8.88 10 6 s 1 The mean-life is: and the half-life: τ = 1 λ = 1 = 112, 600 s 31.3 hours 8.88 10 6 T 1/2 = ln 2 λ = ln 2 = 78, 100 s 21.7 hours 8.88 10 6
Half-Life & the Decay Equation Half-life is an alternate way of giving the same information as the decay constant. Given a value for either half-life or decay constant, it is a simple calculation to find the other. Substituting λ = ln 2 T 1/2 into the decay equation gives: N (t) = N 0 e λt = N 0 e ln 2 t T 1/2 = N 0 ( 1 2 ) t T 1/2 So after n half-lives: N (t = nt 1/2 ) = ( ) n 1 N 0 2
Example 13 7 N has a half-life of 10 minutes. If the initial number of nuclei in a sample is 1.024 10 20, how many will be left after 50 minutes? Quick Method: If we are clever, we realize 50 minutes = exactly FIVE half-lives. N (50 mins) = ( 1 2 )5 N 0 = 1 32 1.024 1020 = 3.2 10 18 Using Formula: N (50 min) = N 0 e t=50 mins ln 2( T 1/2 =10 mins ) = (1.024 10 20 5 ln 2 )e = 3.2 10 18 Notice that because this equation contains a ratio of two times, it is not necessary to convert minutes to seconds.
Example If the initial number of nuclei in a sample of 13 7N is 1.024 10 20, how many will be left after 53 minutes? There is no quick method in this case. We have to use the decay equation: N (53 min) = N 0 e t=53 mins ln 2( T 1/2 =10 mins) = (1.024 10 20 5.3 ln 2 )e = 2.56 10 18
Half-lives can vary enormously in size - from the very short to the very long, as the following examples demonstrate: Nuclide decay Half-Life 3 1H β 12.3 years 8 4Be α 6.7 10 17 sec 14 6 C β 5730 years 13 7 N β 9.96 mins 131 53 I α 8.04 days 204 82 Pb β 1.4 10 17 years 226 88 Ra α 1600 years 235 92 U α 7.04 10 8 years 238 92 U α 4.47 10 9 years Note that there are no really clear patterns here: There is no connection with the type of decay. Half-lives for both α and β decays can be very long or very short. There is no connection with mass - light or heavy nuclei can have long or short half-lives.
Activity The activity A(t) (sometimes called instead the rate of decay R(t)) of a sample at time t is the number of decays occurring at that time. A(t) is proportional to the number of nuclei present: A(t) = dn dt = λ N (t) Since A(t) is proportional to N (t), the connection between A(t) and A 0 (the activity at time t = 0) is the same as for N : A(t) = A 0 e λt The SI unit of activity is the becquerel (Bq). 1 Bq = 1 decay/s The becquerel is a rather small unit, so it is common to see for example MBq = 10 6 Bq. An often used non-si unit is the curie (Ci), where 1 Ci = 3.7 10 10 Bq.
Example 228 88 Ra has a half-life of 1.6 10 3 years. What is the activity of a sample containing 3 10 18 nuclei? A = λ N = ln 2 T 1/2 N = ln 2 1600 365.25 24 60 60 3 1018 = 41 MBq Notice that when we use this formula, we must convert times to seconds.
Example What will be the activity of this 228 88Ra sample in 1000 years time? Alternatively: 1000 ln 2 A(t = 1000 years) = A 0 e 1600 = 41 10 6 e 0.433 = 26.6 MBq A(t = 1000 years) = λ N (t = 1000 years) 1000 ln 2 = λ N 0 e 1600 = ln 2 1600 365.25 24 60 60 3 1018 e 0.433 = 26.6 MBq
Radiocarbon Dating Measurements of radioactivity have a number of applications. One well-known example is radiocarbon dating (or just carbon dating ). Carbon appears in the atmosphere in the form of CO 2. There are three different isotopes of carbon present: 12 6 C (stable) 99% 13 6 C (stable) 1% 14 6 C (radioactive, T 1/2 5730 years) 0.0000000001% The amount of 14 6C in the atmosphere is constant because it is constantly being created by the impact of cosmic rays on the atmosphere at the same rate as it is decaying. All organic material obtains carbon from the atmosphere plants directly via photosynthesis, and everything else by eating plants, or by eating something that eats plants so has the same percentages of carbon isotopes.
However after death, an organism will no longer be obtaining carbon from the atmosphere and so the ratio of 14 6C to 12 6C will decrease over time. By measuring this ratio (perhaps using a mass spectrometer), and using the known value of the half-life, the time since death can be calculated. This can be useful for archaeologists who can date ancient bones or charcoal from an excavation. For example, an archaeologist finds some chicken bones. Do these give us useful information about the diet of ancient people, or did one of the locals recently have KFC? After sending the bones off to the mass spectrometer lab, it is found that the ratio of 14 C to 12 C is one quarter of the atmospheric ratio. Therefore the amount of 14 C in the bones is one quarter of the original amount when the chicken was alive. So we know the age of the bones must be 2 half-lives of 14 C = 2 5730 = 11460 years. Probably not KFC.