Strand J. Atomic Structure Unit 2. Radioactivity Contents Page Unstable Nuclei 2 Alpha, Beta and Gamma Radiation 5 Balancing Equations for Radioactive Decay 10 Half Life 12
J.2.1. Unstable Nuclei. The binding energy of a nucleus is the energy required to hold the nucleus together. Repulsion between positively charged protons is constantly trying to tear the nucleus apart, whilst the strong nuclear force between is trying to hold everything together. Coulombic repulsion acts over an infinite range (recall the 1/r 2 relationship between Coulombic force and distance) whereas the strong nuclear acts over only a small distance, roughly the size of a small nucleus. This results in a difference in distribution of the two forces over the nucleus. As the number of protons in the nucleus increases, the radius of the nucleus increases and the number of neutrons required to keep the two forces balanced throughout becomes greater. Figure J.2.1.1 shows how neutron number increases with proton number for the first 90 elements in the periodic table. If the neutron number increased at the same rate as the proton number, the red trend line would be followed. Figure J.1.2.1 The green trend line that the proton and neutron number follows is known as the line of stability. When the atoms of an element are on the line of stability, the force of repulsion and the attractive strong nuclear force are perfectly balanced, and the binding energy holding the nucleus together is minimised. If the atom is above the line of stability then it is unstable, has more neutrons than it really needs, and the strong nuclear is dominates. In this case the nucleus has an excess of binding energy. If the atom is below the line of stability it is unstable, and has to few neutrons (or too many protons) resulting in a deficit of binding energy. Neutron Number N 140 120 100 80 60 40 20 0 Line of Stability 0 50 100 Proton Number Stable nuclei have the minimum binding energy required to hold the nucleus together, and exist on the line of stability. Unstable nuclei have an imbalance between the attractive and repulsive forces within the nucleus. Due to the excess or deficit in binding energy an unstable nuclei can loose or gain energy by either reshuffling protons and neutrons (a proton may spontaneously turn into a neutron or vice versa), ejecting subatomic particles, or releasing pure energy. The process of releasing energy from the nucleus to become more stable is known as radioactive decay and is a completely random process (there is no way of knowing when a particular unstable nucleus will decay). The energy or particles released when an unstable nucleus decays is radiation. 2
J.2.2. Alpha, Beta and Gamma Radiation. When an unstable nucleus undergoes radioactive decay, energy may be released in three ways (three forms of radiation); Alpha (α) particle emission the nucleus ejects two protons and two neutrons Beta (β) particle emission ejection of a high energy electron or positron Gamma emission (γ) the nucleus emits a high energy gamma photon α emmision 4 An alpha particle consists of 2 protons and 2 neutrons (a 2HHHH nucleus, or 4 2 αα). An alpha particle has a relative charge of +2 and a relative mass of 4u. When an unstable nuclei emits an alpha particle, its atomic number reduces by 4 and its proton number reduces by 2, and the unstable element becomes a different, 238 more stable element or isotope. For example, when uranium 92UU decays via alpha emission it becomes thorium 234 90TTh. Since alpha particles are relatively large with a charge of +2, they readily interact with other atoms, capturing electrons as they move through space due to Coulombic attraction. Alpha particles are therefore heavily ionising, and because each interaction reduces its kinetic energy, they only travel between a few mm, and 10 cm in air. In a distance of only 1mm however, an alpha particle can produce 10,000 ions in air. If the alpha particle captures 2 electrons it becomes a helium atom. Due to their highly interacting nature, alpha particles are not heavily penetrating, and can be stopped by paper or thin metal foil. As such alpha radiation cannot penetrate unbroken skin, but is extremely hazardous if ingested. β emission A beta particle is a high energy electron or positron (an anti electron). Electrons have a -1 charge and positrons +1 charge. Beta minus ( - β) decay occurs when a neutron in an unstable nucleus spontaneously turns into a proton and ejects an electron. Beta plus ( + β) decay occurs when a proton spontaneously turns into a neutron and ejects a positron. Since beta decay involves a nucleon turning into another nucleon, the nucleon number (or atomic mass number A) remains unchanged after the decay. 3
However, the proton number Z must go up or down by 1. Since proton number changes, a new element (or isotope of a new element) is formed. For example, when the carbon-14 isotope ( 14 6 CC) decays through - β decay, its atomic number remains the same and its proton number increases by 1, forming nitrogen ( 14 ). Electrons and positrons have a mass 10,000 times smaller than a proton or neutron, and a charge of +1 or -1. They therefore have a greater range in air than an alpha particle (on the order of 1m). Beta particles may have less charge and therefore be less ionizing than alpha radiation, but beta radiation is still dangerously ionizing, producing approximately 100 ions per mm in air. Beta radiation is more penetrating and can pass through paper and thin metal foil, but is stopped by approximately 5mm of aluminium. NN 7 γ emmision Unlike α and β radiation, γ rays are not particles but pure energy in the form of photons (electromagnetic wave packets). These photons have no mass and no charge so are weakly ionizing, but have very small wavelengths (10-11 m) and very high energies. Since the γ ray is a photon it has infinite range, and can be stopped by several centimeters of lead or several feet of concrete. It is worth noting here however that due to quantum effects, no barrier completely stops γ radiation, not even several meters of lead (see quantum tunneling for further reading). Some radioactive materials (sources) will only emit one of the three types of radiation whist others will emit a combination or all three types. To experimentally investigate an unknown radioactive source, and in addition to investigate the penetrating capabilities of each type of radiation, the following experimental set up may be used. Radioactive Source GM Tube Absorber Figure J.2.2.1 To Counter A Geiger counter consists of a Geiger Muller (GM) tube and a digital counter. When radiation enters the tube and causes an ionization event, a charged ion is created. This ion is then attracted to a conducting rod within the tube that registers the ion s a tiny current. The counter counts the pulses of current, often registering each count an audible click. GM tubes can detect all three types of radiation by detecting the ionised atoms in the air. 4
An unknown source of radiation is placed in front of the GM tube and the distance between the tube and the source is varied. Alternatively, strips of absorber such as paper, aluminium or lead may be placed between the source and the detector. If the counter stops registering at distances greater than 5cm, or paper reduces the count rate to zero, only α radiation is present. If it reduces but not to zero, α and another type of radiation must be present. If the count rate remains unchanged, no α is present. If the count rate reduces or falls to zero at a distance of approximately 1m or with a few mm of lead between the source and detector, this indicates the presence of β radiation. If the count rate does not fall to zero at distances greater than 1m, but is greatly reduced by placing lead between the source and the detector, γ radiation must be present. α β γ α β γ Paper Aluminium Figure J.2.2.2 Lead Exercise J.2.2 1. A student tests two unknown radioactive sources. He places source A next to a GM tube and records the count rate. He moves the GM tube to a distance of 8cm from the source and the count rate reduces. He continues to move the GM tube away from the source to a distance of 1.2m and the count rate reduces to expected background levels. He repeats this procedure for source B, but this time he finds only a very gradual reduction in count rate as he moves the GM tube and still records radiation from the source at the maximum experimental distance of 2m. For each source, state the type of radiation emitted and explain your reasoning. 2. The student then places paper in front of unknown source C whilst the count rate from the source is recorded. The paper is found to have no effect. Placing 7mm of aluminium between the source and the detector however reduces the count rate to background levels. Which type of emitter is source C? Explain your reasoning. 298 3. The unstable element 99 XX decays via alpha emission. State the chemical symbol of the resulting new element Y. 5
4. Element B is unstable. It has 80 neutrons and 62 protons. If element B decays 4 into element C, firstly by emitting a 2HHHH nucleus, and then by emitting an electron, what is the proton number Z and atomic mass number A of element C? Challenge Question 5. When the student in question 1 set up his experimental equipment, he noted that the GM tube and counter recorded a small number of radioactive events over time even though there were no sources present. His teacher informed him that the GM tube was picking up background radiation and that he should average this count and during the experiment adjust his zero count level accordingly. Suggest some possible sources of this background radiation level. J.2.3. Balancing Equations for Radioactive Decay. The decay of an unstable nuclide (species of nucleus) through the emission of an alpha or a beta particle can be represented in equation form (in the case of gamma photon emission the unstable nuclide does not change type since no change in proton number occurs). This representation of nuclear decay can be useful, especially when many decay events occur in the decay chain (the route of decay an unstable atom takes toward stability). Recall that a nuclide AA zz XX; Contains Z protons and A Z neutrons. Has a charge +Z times the charge of an electron (since the charge of a proton and electron is of the same magnitude) Has a mass = A in atomic mass units α decay During alpha emission, a particle consisting of 2 protons and 2 neutrons is ejected from the nuclide. This alpha particle may therefore be represented in 4 symbol form as 2 αα since it has 2p + 2n = 4 nucleons (A) and 2 protons. The new nuclide formed after the emission of the α particle must therefore have two less protons and 4 less nucleons, i.e. may be represented as AA 4 zz 2YY. Hence the alpha emission by the nucleus may be represented as; The original unstable nuclide AA AA 4 4 ZZXX zz 2YY + 2 αα The particle emitted The new nuclide formed after the decay 6
Note that the nuclear equation representing the decay process does not use an equals sign but instead uses an arrow. The arrow may be thought of as goes to or turns into. However, just like an equals sign, the equation must balance on either side of the arrow. Hence the number of both A and Z on the left of the arrow must equal the total number of A and the total number of Z to the right of the arrow. Worked Example Uranium-238 has 146 neutrons and 92 protons. When Uranium decays it is likely to emit an alpha particle and become thorium (Th) and then emit another alpha particle to become radon (Ra). Represent this decay process using a nuclear equation. Answer Uranium- 238 has A = 238 and Z = 92. Since the uranium nuclide ejects an alpha particle the thorium nuclide must have A = 238 4 = 234 and Z = 92 2 = 90. Hence 238 TTh + 234 4 92UU 90 2 αα The thorium then decays into radon by alpha emission. The radon nuclide therefore has A = 234 4 = 230 and Z = 90 2 = 88. Hence 234 90TTh 88RRRR + 230 4 2 αα The entire decay process may therefore be represented as 238 234 4 TTh + αα RRRR + 92UU 90 2 230 4 88 2 αα β - decay When a nuclide decays via beta minus emission (the ejection of an electron), a neutron inside the nucleus turns into a proton. In this case the nucleon number stays the same, but the proton number goes up by one. The beta particle ejected is represented as. In addition to the ejection of a beta minus particle, the nucleus also ejects another sub atomic particle, the electron anti-neutrino (νν ee ). This particle has no charge and negligible mass (for the interested, see lepton conservation). The beta minus decay process may therefore be represented as; 1 0 ββ AA ZZXX ZZ+1 AA YY + 1 0 ββ + νν ee 7
β + decay When a nuclide decays via beta plus emission (the ejection of an anti-electron (positron)), a proton inside the nucleus turns into a neutron. Hence the nucleon number stays the same, but the proton number goes down by one. The beta particle ejected is represented as +1 0 ββ since it has positive charge. Again, the nucleus also ejects another sub atomic particle, this time the electron neutrino (νν ee ). The beta plus decay process may therefore be represented as; AA ZZXX ZZ 1 AA YY + +1 0 YY + νν ee Worked Example 190 An unstable nuclide 76XX decays into a Y nuclide via beta minus decay. This nuclide then decays into Z via alpha decay. Represent the decay chain for this process using a nuclear equation. Answer. 190 76XX has 76 protons and 190-76 = 114 neutrons. In β - decay an electron and an anti electron neutrino is ejected when a neutron turns into a proton. Thus the nucleon number A stays the same but the proton number goes up by 1. 190 190 YY + ββ + νν ee 76XX 77 1 0 190 The 77YY then decays via α emission, ejecting 2 protons and 2 neutrons. Thus the nucleon number A goes down by 4 and the proton number by 2. Adding this to the nuclear equation above represents the entire decay chain; 190 190 186 4 76XX 77YY + 1 0 ββ + νν ee 75ZZ + 2 αα Exercise J.2.3. 1. Complete the following nuclear equations representing α decay; 236 92XX YY + 235 92UU TTh + 2. Complete the following nuclear equations representing β - decay; 218 84PPPP RRRR + ββ + νν ee 8
232 94XX YY + 3. Complete the following nuclear equations representing β + decay; 228 88RRRR AAAA + ββ + 232 94XX YY + 228 236 4. For elements 87AAAA and 90YY write down the number of protons and the number of neutrons in each nucleus. Then, state a possible route of decay that element Y could take to become Ac. Challenge Question 5. An unstable isotope X which has 94 protons and 126 neutrons decays by emitting a β - particle to form an unstable isotope Y which then decays to form Z via α emission. Write down a nuclear equation that describes the entire decay chain. J.2.4. Radioactive Decay and Half Life. When an unstable element decays by emitting α or β - radiation, it becomes a new element. This is because the proton number changes. If we had a lump of unstable element X, after a time the lump would no longer be element X but a combination of element X and Y. Eventually it would be made up of more of element Y than X. As the initial number N of the original nuclides X decreases, the mass of the original unstable element or isotope X decreases. The rate of conversion of the original element must get slower over time because at a later time t there are less of the X nuclides to decay into Y nuclides. This rate of decay is called an exponential decay curve and is shown in Figure J.2.4.1. The curve is exponential because the number of nuclides that will decay in a certain time is proportional to the number remaining. nuclides of element X remaining 100 80 60 40 20 t1/2 2t1/2 Figure J.2.4.1 time 9
The half life (t1/2) of a radioactive element is the time it takes for the mass of the original element to reduce to half its initial value. This would be the same as the time taken for the original nuclides to reduce to half the initial number. We call the initial mass m0 and the initial number N0. The half life t1/2 of a radioactive isotope is the time taken for the mass or number of nuclides to reduce to half the initial value Worked Example 131 A radioactive isotope of iodine 53II has a half life of 8 days. The initial mass of the iodine isotope is 1 10-3 kg. If 1 atomic mass unit = 1.67 10-27 kg, calculate the initial number of nuclides in the sample, and the number remaining after 24 days. Answer Initially we have 1 10-3 kg of the iodine isotope and each nuclide has a relative atomic mass given by the nucleon number, therefore the mass of one nuclide in kilograms is; 131 1.67 10-27 = 2.19 10-25 kg and the number of nuclides in the original sample is 1 10 3 kkkk 2. 19 10 25 = 4.57 1021 kkkk The half life of the isotope is 8 days. Therefore after 8 days (1 half life), 4.57 10 21 /2 = 2.285 10 21 remain after 16 days (2 half lives), 2.285 10 21 / 2 = 1.1425 10 21 remain after 24 days (3 half lives), 1.1425 10 21 / 2 = 5.71 10 20 remain leaving 5.71 10 20 2.19 10-25 kg = 1.25 10-4 kg of the iodine isotope remaining. NOTE: It is important to remember that there is not 1.25 10-4 kg of the substance remaining, since the substance now comprises of the 1.25 10-4 kg iodine nuclides AND the mass of the new isotope resulting from the decay. Naturally occurring radioactive decay is a random process. The time at which a particular nuclide will decay cannot be predicted. However, we can predict how many nuclides will decay in a certain time. The random nature of radioactive 10
decay can be modeled by throwing dice. If you threw 1 die, you could not predict the number the die will show. But if you threw 1000 dice, you could expect 1000/6 to show a 1. The activity of a radioactive isotope is the number of nuclides that disintegrate per second. Activity has the unit Becquerel (Bq), where 1Bq = 1 decay per second. Like half life, the activity is different for different isotopes and in addition, the activity of the isotope is also proportional to the mass of the isotope. This is because the greater the mass, the greater the number of unstable nuclides, the greater the rate of decay (activity). Thus as the number of nuclides in a radioactive sample decreases exponentially with time, so does the activity of the sample. Worked example. A safety monitor at a nuclear power plant measures the count rate from a sample of radioactive waste. The Geiger counter measures 1200 events over a 40s interval. Calculate the activity of the sample. The worker returns 54 hours later and measures the activity of the sample once more. If the half life of the sample is 36 hours, what activity does the worker record? Answer The Geiger counter measures 1200 clicks in 40s. The activity of the sample is the number of decays per second, hence the activity a = 1200/40s = 30Bq. The half life of the sample is 36 hours. Thus in 36+18=54 hours the sample goes through exactly 2 half lives and the activity reduces to 30/4 = 7.5Bq. Exercise J.2.4. 212 has a half life of 24 hours. If the initial mass of the 212 after 4 days? 1. A radioactive sample of 94XX sample is 20g, what would be the mass of the remaining 212 2. Calculate the number of 94XX nuclides left in the sample from question 1 after 14 days. 3. A sample of a radioactive isotope contains 360 10 9 nuclides. How many atoms of the isotope are present after; A. 3 half lives B. 7 half lives C. 15 half lives XX 94 11
4. Consider the graph of count rate vs. time for an unknown radio isotope Y. Calculate the half life of the radio isotope. State the number of decay events occurring per second within the sample at a time t = 450s. Activity (Bq) 100 80 60 40 20 Challenge Question 100 200 300 400 500 time/s 131 5. A radioactive isotope of iodine 53II has a half life of 8 days. After 32 days 1.1425 10 21 131 53II remain. Calculate the initial mass of the sample. 12