8 Radioactive Decay Processes PDF

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8 Radioactive Decay Processes...

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8.0 Radioactive Decay Processes 8.1 Radioactive Decay Most naturally-occurring isotopes are stable, but many are not. Unstable nuclei emit radiation in a process known as radioactivity, and the unstable nuclei are referred to as radioactive nuclei. Radioactivity is a random process: it is not possible to predict exactly when a given nucleus will decay, it is only possible to estimate the probability that a nucleus will decay within a given period of time. The rate of radioactive decay is called the activity (A), and it is measured in terms of the average number of decays per second (the actual number of decays per second will fluctuate about the mean). If the activity is high, then in percentage terms the fluctuations will be small, but if the activity is low, then the fluctuations will be significant, hence the use of the mean number of decays per second. The rapidity with which radioactive decay takes place is characterised by the half-life (t½)of the isotope: this is the time taken for half of the initial number of the unstable nuclei to decay, and can be less than 1ms to more than 1000 yr. Naturally, the more unstable a nucleus is, the shorter its half-life. Radioactivity involves parent (unstable) nuclei, P, decaying to form daughter nuclei, D, accompanied by the emission of radiation and an amount of energy, Q (sometimes called the disintegration energy). Unstable nuclei that undergo radioactive decay are known as radionuclides, or radioisotopes. Parent Nucleus (P)  Daughter Nucleus (D) + Emitted Radiation + Energy (Q) Q is manifested in the kinetic energies of the decay products. During radioactive decay, there must be conservation of charge, conservation of the total number of nucleons (A), and conservation of relativistic mass & energy (E = m c2). This conservation of relativistic mass & energy leads to a means of calculating Q: Q = (rest mass energy of parent nucleus) – (total rest mass energy of daughter nucleus & emitted radiation) If Q is positive, then the decay is allowed, but if Q is negative, the decay is forbidden (i.e. it cannot occur spontaneously). Example: Is the following decay scheme feasible, given the accompanying data ? 222 221 1 88 Ra  87 Fr 1 H  Q rest-mass of Ra = 222.015353 u rest mass of Fr = 221.014183 u rest-mass of H = 1.007825 u Answer: Q = (m(Ra) - [m(Fr) + m(H)] ) c2 = (222.015353 u - [221.014183 u + 1.007825 u]) c2 = - 0.006655 u c2 Since Q is negative the decay not possible. There are several mechanisms of radioactive decay, including:  alpha (α) decay;  beta (β) decay;  gamma (γ) decay; 1



electron capture;

These mechanisms will be considered in greater detail in the following sections.

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Practice Question - Section 8.1 Try to answer this question yourself, before checking your work against the model answer provided at the end of this document. 1. For each of the following decays, decide whether or not the decay is forbidden. Give a reason for your decision in each case. 235 4 U  230 90Th  2 He  p a) 92 239 4 Pu  235 92 U  2 He b) 94 Atomic masses: 235 230 239 4 92 U 235.043925 u ; 90Th 230.033131u ; 2 He 4.002603 u ; 94 Pu 239.052158 u m p 1.67262 10 27 kg 1 u = 1.66054 x 10-27 kg.

8.2 Alpha Decay An α-particle is a helium nucleus (Z=2, N=2, A=4; charge = +2e). The general decay scheme for α-decay is: A Z

P 

A-4 Z-2

D 2-  42 He 2+  Q 

A-4 Z-2

D  42 He  Q

So with α-decay the daughter nucleus has a value of A that is 4 less than the parent nucleus, and values of Z and N that are 2 less than the parent nucleus. With reference to the stability plot above, this means that α-decay involves a transition parallel to the N=Z line, which potentially could take an unstable parent nuclide (in the light shaded region above the dark points) and turn it into a stable daughter nuclide (one of the dark points). Sometimes α-decay can result in a daughter nucleus which either still has the wrong neutron to proton ratio for stability, or which is in an excited state, in which case it will decay again, for example by beta-decay or gamma decay, respectively (see next sections). Although the daughter nucleus is produced as doubly negative ion, the extra two electrons quickly break away to leave a neutral atom; likewise, the fast-moving doubly positive -particle quickly loses kinetic energy by ionising and exciting 3

atoms along its path, soon acquiring 2 electrons to become a neutral helium atom. When calculating the Q for the process it is a good approximation to deal with the masses of the neutral final products, rather than the masses of the charged ions initially formed. But why are α-particles emitted, rather than the individual nucleons or other small nuclei ? The answer lies in the high binding energy of the α-particle, which in turn provides the kinetic energy necessary for the particle to escape the parent nucleus. One way of viewing α-decay is to imagine the αparticle as being held within the parent nucleus by an electrostatic (Coulomb) barrier. According to quantum theory it is possible for the α-particle to tunnel through the Coulomb barrier, and the probability that this will occur increases as the height and thickness of the barrier decrease. Indeed, the probability of the α-particle escaping can be calculated from Schrödinger's equation and the results agree well with experimental data.

Not all α-decays predicted as being allowed by positive Q values will actually take place. For example, consider the following decay scheme: 209 83 Bi



205 81Tl

 42 He  Q

Q = +3.11 MeV, so decay is allowed; however, it is not observed ! If the Coulomb potential energy barrier is calculated, it comes to approximately 28 MeV. Clearly this is much greater than the energy available to the α-particle, so classically the decay is actually forbidden, and even quantum mechanically the probability of decay is vanishingly small because of the height and width of the barrier. On the other hand, for the decay scheme: 222 88 Ra



218 86 Rn

 42 He  Q

the Q value is +6.68 MeV and the decay does indeed take place, with a half-life of only 38 s. In α-decay the Q of the process is divided between the daughter nucleus and the α-particle and becomes their kinetic energies (KX’ and Kα, respectively).  Q = KX’ + Kα But KX’ = ½ ( mX’ vX’2) = ½ (pX’2 / mX’)

and Kα = ½ ( mα vα2) = ½ (pα2 / mα)

 KX’ / Kα = (mα pX’2) / (mX’pα2) Conservation of momentum applies:  0 = pX’ + pα

 pX’ = - pα

 KX’ / Kα = (mα pα2) / (mX’pα2) = mα / mX’ = 4 / (A – 4) where A is the mass number of the original nucleus. But Q = Kα + KX’ 4



Q = Kα + (Kα KX’ / Kα) = Kα (1 + [KX’ / Kα]) = Kα (1 + [4 / (A – 4)]) = Kα (A / [A – 4])



Kα = Q (A – 4) / A

This means that the α-particle has a fixed kinetic energy for a given α-decay scheme. Also, since most αemitters have A > 210, i.e. A >> 4, most of the Q of the decay appears as kinetic energy of the α-particle. Practice Question - Section 8.2 Try to answer this question yourself, before checking your work against the model answer provided at the end of this document. 2. For the decay

210 84

Po 

206 82

Pb  42 He , calculate the Q-value in MeV and estimate the kinetic energy of

both the α-particle and the lead (Pb) nucleus. Assume that the polonium nucleus decays from rest. 210 84

Po 209.982876 u

206 82

Pb 205.974455 u

4 2

He 4.002603 u 1 u = 1.66054 x 10-27 kg c = 2.9979 x 108 ms-1

8.3 Beta Decay There are two types of beta () decay: negative (-) and positive (+). - decay: In - decay a neutron in the parent nucleus is spontaneously converted into a proton and an electron: the electron is emitted immediately after its creation and is the observed -particle, whilst the proton stays inside the nucleus. This decay mechanism therefore decreases the proportion of neutrons in the nucleus, as the daughter nucleus has one more proton and one less neutron than the parent. The energies of the emitted electrons are found to vary continuously from zero to a maximum (Kmax) characteristic of the nuclide, as shown in the graph opposite. In every case, the maximum total energy carried off by the electron (i.e. the rest-mass energy of the electron, me c2, plus Kmax) is equal to the Q of the decay; however, only a small fraction of the electrons have this maximum energy. Calorimetric measurements have shown that for the remaining electrons the “missing” energy cannot be accounted for by collisions with the atomic electrons surrounding the nucleus, which would seem to contradict the law of conservation of energy.

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In addition, neutrons, protons and electrons all have spin ½, so the decay would seem to involve one spin ½ particle being converted into two spin ½ particles, which would contravene the law of conservation of angular momentum. Furthermore, the directions of movement of the emitted electrons and the recoiling daughter nuclei are never exactly opposite, which would contravene the law of conservation of linear momentum. All three conservation laws can be satisfied if the decay involves the emission of another particle, also of spin ½, but with an extremely small rest-mass, that can carry off the balance of the energy (i.e. the difference between Kmax and the observed kinetic energies of the electrons). In the case of - decay this particle is called an antineutrino (  ). The full - decay scheme is therefore: A Z

P



A Z+1

D+ + e-    Q

in which the inherent change is 10 n



1 + 1

p + e-  

Notice that because the daughter atom has one more proton in its nucleus than the parent does, it lacks one orbital electron for electrical neutrality, therefore is initially formed as a positive ion. Although the daughter quickly acquires an extra orbital electron from the surrounding medium, the fact that the daughter initially is a singly charged positive ion needs to be factored into the energetics when calculating Q values (the mass of the daughter ion can be approximated to the mass of the daughter atom minus the mass of one electron). Beta decay is caused by the 4th fundamental force: the weak interaction (the others being gravitational, electromagnetic, and the strong interaction). Antineutrinos ( ) are examples of antimatter: they are the antiparticle to neutrinos (  ). Both neutrinos and antineutrinos have negligible mass, no charge, are unaffected by strong interactions, are unaffected by electromagnetic forces, interact with matter only through the weak interaction, and cannot ionize matter: this makes them extremely difficult to detect, e.g. they can pass through the Earth unaffected ! The difference between neutrinos and antineutrinos is that they have opposite spins. Another example of a matter/antimatter pair are electrons (e-) and positrons (e+): these have the same mass but opposite charges. When particles and antiparticles collide they annihilate each other, producing a large amount of energy, e.g. electrons and positrons annihilate each other, producing two high energy photons (gamma rays, ).

+ decay: In + decay a proton in the parent nucleus is spontaneously converted into a neutron and a positron: the positron is emitted and is the observed -particle, whilst the neutron stays inside the nucleus. This decay mechanism therefore increases the proportion of neutrons in the nucleus, as the daughter nucleus has one more neutron and one less proton than the parent. The full + decay scheme is therefore: A Z

P



A Z-1

D - + e+    Q

1 + in which the inherent change is 1 p



1 0

n + e + 

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As with - decay, the spectrum of positron kinetic energies (opposite) again exhibits a continuous variation from zero to a maximum (Kmax = Q) characteristic of the nuclide.

As the daughter atom has one less proton in its nucleus than the parent does, it has one too many orbital electrons for electrical neutrality, therefore is initially formed as a negative ion. Although the daughter quickly releases this extra orbital electron to the surrounding medium, the fact that the daughter initially is a singly charged negative ion needs to be factored into the energetics when calculating Q values (the mass of the daughter ion can be approximated to the mass of the daughter atom plus the mass of one electron). Another complication is that sometimes the daughter nucleus is initially left in an excited state, in which case this excitation energy also enters into the Q calculation. Competing with + decay is the process of electron capture. This is where an electron from the innermost energy level is captured by the nucleus, where it combines with a proton to form a neutron, with the emission of a neutrino: A Z

P



A Z-1

D   Q

1 + in which the inherent change is 1 p + e 

1 0

n 

Since the electron came from an inner shell, the daughter nucleus is initially in an excited state, and subsequently relaxes by having an electron from a higher energy level fall down to fill the vacancy, thereby releasing a photon whose wavelength is characteristic of the daughter nucleus. Electron capture is actually much more common than positron emission; however, the two processes are very similar because the absorption of an electron is effectively equivalent to the emission of a positron.

Example: Find the maximum kinetic energy of the -particle in each of the following decays: 40 40 + (a) 19 K  18 Ar + e   (b)

23 10 40 19 40 18 23 10 23 11

Ne 

23 11

Na + + e -  

K 39.963999 u Ar 39.962383u Ne 22.994466 u

Na 22.989770u me = 9.11 x 10-31 kg 1 u = 1.66054 x 10-27 kg c = 2.9979 x 108 ms-1

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Answer: In both cases the maximum kinetic energy of the -particle will be the Q of the decay. (a) Q = (m(K) - [m(Ar -) + m(e+)] ) c2 = (m(K) - [m(Ar) + 2me] ) c2  Q = (39.963999 u - [39.962383 u + {2 x 9.11 x 10-31 / 1.66054 x 10-27} u] ) c2 = + 0.0005188 u c2 = + 0.483 MeV (b) Q = (m(Ne) - [m(Na +) + m(e-)] ) c2 = (m(Ne) - [m(Na)] ) c2  Q = (22.994466 u - 22.989770 u) c2 = + 0.004696 u c2 = + 4.37 MeV

Practice Question - Section 8.3 Try to answer this question yourself, before checking your work against the model answer provided at the end of this document. 3. Calculate the values for the following decays: 14 14 + (a) β- decay: 6C  7N + e   13 13 + (b) β+ decay: 7 N  6C + e  given that: 13 6 C = 13.003355u 14 6 C = 14.003242u 13 7

N = 13.005739u 14 7 N = 14.003074u 1 u ≡ 931.5MeV

8.4 Gamma Decay Just as an atom can have excited electronic states, a nucleus can also exist in excited states (bound states whose energies are greater that that of the ground state). Excited nuclei return to their ground state by emitting photons of the appropriate energy for the transitions involved: typically this can be of the order of MeV. Such high energy photons are termed gamma-rays (-rays), and they are more energetic than Xrays. γ-rays have wavelengths less than about 0.05 nm, whereas X-rays have wavelengths between 0.03nm and 3nm. Where -decay is found it is effectively a by-product of -decay or -decay, where they have produced a daughter nucleus in an excited state.

For example,

27 12

Mg -decays to

27 13

Al .

27 12

Mg

The decay may produce the 27 13 Al in either of two excited states, which may then undergo one or two -decays to reach the ground state (see energy level diagram opposite), giving a total of three different -ray energies. The -rays produced thus have very welldefined energies (or wavelengths), hence -decay produces a line spectrum.

3

1.015 MeV 0.834 MeV

2

1 0 MeV 27 13

Al 8

A In order to indicate that a nucleus of Z X is in an excited state, the symbol is given an asterisk: hence the -decay scheme can be written as:

A Z

X*



A Z

A Z

X* ;

X +γ

The half-lives for the -decay of most nuclei are very short, but just a few nuclei have half-lives of several hours. A long-lived excited nucleus is called an isomer of the ground state nucleus. There is an alternative to -decay, in which the excited nucleus relaxes by giving up its excitation energy directly to one of the orbital electrons. This process is known as internal conversion.

8.5 Summary & Comparison of Decay Mechanisms -decay:

A Z

A-4 Z-2

P 

D 2-  42 He 2+  Q 

A-4 Z-2

D  42 He  Q

- compared to the parent, daughter nucleus has 2 less protons & 2 less neutrons; -particle emitted.

- decay:

A Z

P

A Z+1



D + + e -   Q

- compared to the parent, daughter nucleus has 1 more proton & 1 less neutron; electron emitted (- particle). + decay: AZ P  Z-1A D - + e +   Q - compared to the parent, daughter nucleus has 1 less proton & 1 more neutron; positron emitted (+ particle). electron capture:

A Z

P

A Z-1



D   Q

-decay:

Proton Number, Z

- compared to the parent, daughter nucleus has 1 less proton & 1 more neutron. X*



A Z

X +γ

- p ent & daughter are the same nuclide.  -decay: ZZ-2 NN-2

stability curve

+ decay: ZZ-1 NN+1

- decay: ZZ+1 NN-1

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The above graph illustrates how the various decay mechanisms can bring about increased nuclear stability through changes to the numbers of neutrons and protons in the nucleus.

Practice Question - Section 8.5 Try to answer this question yourself, before checking your work against the model answer provided at the end of this document. 4. Complete the following decay equations: 27 (a) 27 14 Si  13 Al  (b)

74 33

(c) (d)

208 84 131 53

(e)

242 94

As 

74 34

Se 

Po   

I

131 54

Xe 

Pu   

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8.6 Decay Chains & Radioactive Series Many unstable nuclei do not achieve stability in a single decay: the daughter nucleus may itself be unstable, so that further decays take place. This leads to the generation of a decay chain.

Most of the naturally occurring radioactive elements are members of one of four radioactive series. Each series consists of a succession of daughter products, all ultimately derived from a single parent nuclide. There are four such series because -decay reduces the mass number, A, by one. These series are summarised in the table below, in which the “n” in the A column is an integer. A 4n

Parent Nuclide Stable End Product 232 208 90Th 82 Pb

4n + 1

237 93

4n + 2

238 92

4n + 3

235 92

Np

209 83

Bi

U

206 82

Pb

U

207 82

Pb

As an example, the 4n + 3 series is shown opposite. Not all elements in all of the series are found today - the age of the Universe is such that some nuclides have completely decayed away, although they can ...