156

8738Sr ( 6 87

38Sr % ã .

Chapter 12

Nuclear Transformations

Introduction

Radioactivity occurs because some nuclei are unstable and spontaneously decay.

Important aspects of radioactivity:

Elements transform into other, different elements.

The energy released in radioactive decay comes from mass which is converted to energy.

Radioactivity is a quantum phenomenon. Radioactive decay is a statistical process.

12.1 Radioactive Decay

Be sure to read this section. I would expect you to associate the names Roentgen, Becquerel, andCurie with the discovery of radioactivity. I will not discuss the brief historical review in class.

There are five kinds of radioactive decay. Figure 12.3 shows them and gives the reasons for theiroccurrence. Understand figure 12.3.

Here are the five kinds of radioactive decay. We will go into more detail for each in later sectionsin this chapter.

(1) Gamma decay.

This occurs when a nucleus has excess energy.

A gamma ray (packet of energy) is emitted from the nucleus.

The parent and daughter nuclides are the same.

Example:

The * in the reaction denotes an excited nuclear state.

157

238 92 U 6 234

90 Th % 42He .

14 6 C 6 14

7 N % e & .

6429Cu % e & 6 64

28Ni .

(2) Alpha decay.

This occurs when the nucleus is too large.

An alpha particle is emitted, reducing the size of the nucleus.

The daughter nucleus has an atomic number 2 less and an atomic mass 4 less than theparent nucleus.

Example:

(3) Beta decay.

This occurs because the nucleus has too many neutrons relative to protons.

A neutron changes into a proton and emits an electron.

The daughter nucleus has an atomic number 1 more and an atomic mass the same asthe parent nucleus.

Example:

Later we will find there is something missing from this reaction.

(4) Electron capture.

This occurs because a nucleus has too many protons relative to neutrons.

A proton captures an electron and changes into a neutron.

The daughter nucleus has an atomic number 1 less and an atomic mass the same as theparent nucleus.

Example:

Again, we will find something is missing from this reaction.

158

6429Cu 6 64

28Ni % e % .

R ' &dNdt

.

(5) Positron emission.

As with electron capture, this occurs because a nucleus has too many protons relativeto neutrons.

A proton emits a positron and changes into a neutron.

The daughter nucleus has an atomic number 1 less and an atomic mass the same as theparent nucleus.

Example:

Again, we will find something is missing from this reaction.

Radioactive decay involves an unstable nucleus giving off a particle or ray, and in the processbecoming a more stable nucleus.

There are several ways to detect what the particle/ray is.

One way is to detect the radiation after it passes through a magnetic field. Positive andnegative charged particles will be deflected in different directions. Neutral particles or raysgo straight through.

One simple way to distinguish different types of radiation is to see what it penetrates. A pieceof paper can stop alpha rays. Beta particles can be stopped by a sheet of aluminum. Evenlead may not stop gamma rays.

This section defines activity.

The activity of a radioactive sample is the rate at which atoms decay.

If N(t) is the number of atoms present at a time t, then the activity R is

dN/dt is negative, so the activity is a positive quantity.

The SI unit of activity is the becquerel:1 becquerel = 1 Bq = 1 event/second.

Another unit of activity is the curie (Ci) defined by1 curie = 1 Ci = 3.70x1010 events/s = 37 GBq.

This section also contains interesting paragraphs about the relation between the geology of the earth

159

R ' R0 e &ët .

12

R0 ' R0 e&ëT1/2 .

eëT1/2 ' 2 ,

T1/2 '0.693

ë.

and radioactivity, and about radiation hazards. I won't lecture on them, but they are testable onquizzes or the final exam, so be sure to read them.

12.2 Half-Life

Experimental measurements show that the activities of radioactive samples fall off exponentially withtime.

The empirical law which describes this decay is

ë is called the decay constant.

Each radioactive nuclide has a different decay constant.

The half-life, T½, is the time it takes for the activity to drop by ½.

The half-life and decay constant are related, and the expression for the half-life is found bysubstitution R0/2 for the activity and T½ for the time it takes for the activity to become R0/2.

The result is

or

Here's a plot of the activity of a radionuclide.

The initial activity was chosen to be 1000 for this plot.

The half-life is 10 (in whatever time units we areusing).

All decay curves look like this; only the numbers onthe axes will differ, depending on the radionuclide(which determines the half-life) and the amount of

160

dN ' &Në dt .

N ' N0 e &ë t ,

T̄ ' 1.44T1/2 .

radioactive material (which determines the initial activity).

The empirical activity law can be derived if we assume that ë is the probability per unit time for thedecay of a nucleus.

Then ëdt is the probability that the nucleus will undergo decay in a time dt.

If a sample contains N undecayed nuclei, then the number dN that will decay in the time dtis just N times the probability of decay,

This equation can be integrated to give

which you should recognize as looking like the activity law with N's instead of R's. In fact,the activity R of a sample of N radioactive nuclei is just R=ëN.

What's the difference between equations 12.2 and 12.5? One talks about rates and the otherabout numbers. More important, equation 12.2 is empirical, and you should always besuspicious of empirical equations. On the other hand, equation 12.5 is derived under theassumption that ë is the decay probability per unit time. Equation 12.5 is part of a testabletheory, while an empirical equation may or may not have any physical meaning.

Example: radon (nasty stuff) has a half-life of 3.8 days. If you start with 1 mg of radon, after 3.8days you will have 0.5 mg of radon. After 7.6 days, you will have 0.25 mg of radon. After 11.4 days,you will have 0.125 mg of radon.

Notice that we are halving the amount of radioactive radon with every half-life that passes,but after 3 half-lives we still have a significant amount of radon left.

The mean lifetime of a nucleus is different than its half-life. It turns out that

See Beiser for more details.

For another example, see the Mathcad document RADON1.MCD. I will put it in the problemsubdirectory, so you can download it with the chapter 12 problems.

161

t '1ë

lnR0

R.

Radiometric Dating

Carbon-14 dating is the best-known example. Carbon-14 is formed in the atmosphere by the reaction

714N + 0

1n 6 6 14C + 1

1H.

This reaction is continually taking place, and the carbon-14 atoms are continually betadecaying to nitrogen-14, with a half-life of 5760 years.

Because carbon-14 is continually being created and decaying, we eventually reach a steady-state condition, where there is a constant amount of carbon-14 in the atmosphere.

Living things take up carbon-14 as long as they are alive, and have the same ratio of carbon-14 to carbon-12 as does the atmosphere.

When living things die, they stop taking up carbon-14, and the radioactive carbon-14 decays.If we compare the carbon-14 to carbon-12 ratio in a dead organism with a living one, we cantell how long the carbon-14 has been decaying without replenishing, and therefore how longthe organism has been dead.

What are the assumptions behind carbon-14 dating?

The carbon-14 to carbon-12 ratio in the atmosphere is the same now as it was when theorganism died.

Living organisms now are essentially the same in their carbon content as were similarorganisms long ago.

Carbon-14 dating takes us back a relatively short time, and both assumptions seem to bevalid.

The equation for radiocarbon dating is like the one in example 12.2, except that it uses rates insteadof numbers of decays.

Remember, rates and numbers are just related by the multiplicative factor ë, so the ratio R/R0

is equal to N/N0.

The formula for radiocarbon dating, derived from R=R0e-ët, is

So, you see, we need to know the activity R0 of the organism at death, which is the reasonfor the second assumption above.

162

t '1ë

lnR0

R'

5760 years0.693

ln1613

' 1726 years .

Radiocarbon dating is good for a few half-lives of carbon-14, or 50,000 or so years.

A similar approach can be taken with radioactive potassium, rubidium, or uranium, to go backmuch further in time.

There is a difference. With geological dating, we measure the numbers of radioactive andstable daughter nuclei in a given sample.

If we assume the daughter nuclei came only from the original radioactive nuclei, we cancalculate the original number, and then calculate the decay time.

We have to find parent-daughter decay schemes that give us unique daughter nuclei; i.e., theycould have only come from decay of the parent. That's why so few methods are available.

We also may be unsure what event caused the clock to start "ticking;" i.e., what event frozeinto the sample the particular number of parent atoms.

Example on page 427. A piece of wood has 13 disintegrations per minute per gram of carbon. Theactivity of living wood is 16 dpm per gram. How long ago did the tree die?

12.3 Radioactive Series

Most radionuclides belong to one of four radioactive series. The masses of the nuclides in these fourseries are given by

A=4nA=4n+1A=4n+2A=4n+3

Table 12.3 in Beiser summarizes this further:

Mass Numbers Series Parent Stable End Product

4n Thorium 90232Th 82

208Pb

4n+1 Neptunium 93237Np 83

209Bi

163

4n+2 Uranium 92238U 82

206Pb

4n+3 Actinium 92235U 82

207Pb

The fact that there are only four decay series may seem surprising at first, but it's not, if you thinkabout it. The kinds of radioactive decay we mentioned above involve either a change in mass numberA of 4 (alpha decay) or of 0 (all of the others).

164

140

130

80 84 88 92Z

90Th232

88Ra228

89Ac228

90Th228

88Ra224

86Rn220

84Po21682Pb212

83Bi212

48Po21281Tl206

82Pb208

á

â

165

Thus, if we start with a nuclide of mass number A0, it can only decay into nuclides of massnumber A0-4, A0-8, etc.

Let's look at one of these series in more detail.

Discuss the thorium decay series.

Things to note: alpha decay, beta decay; the branch at the decay of 83212Bi.

It is interesting to calculate the number of daughter nuclei present as a function of time.

Suppose A is a parent and B is a daughter. Initially, there is all A and no B.

A decays into B, building up the concentration of B. But if B is radioactive, B also decaysaway.

166

dNB

dt' ëA NA & ëB NB

dNB

dt' ëA N0 e

& ëA t& ëB NB .

NB 'N0 ëA

ëB & ëA

e&ëAt

& e&ëB t

.

NB ' N0

ëA

ëB

1 & e&ëBt

,

Does the concentration of B build up and remain at a peak, or build up but then decreasebecause B decays.

We can do the calculation easily.

The rate of change of B nuclei is equal to the rate at which A decays to B, minus the rate atwhich B decays away (number of B's = B's in minus B's out).

Mathematically:

If we do some calculus on the above equation (integrate and use the fact that NB=0 at t=0 tofind the constant of integration), we get:

This equation is used to plot NB and NA as afunction of time in the figure to the right.

This isn't quite the whole story, because in allof the radioactive series mentioned above, theparent nuclide is much longer lived than any ofthe daughters.

In that case, ëA<<ëB, and we obtain (by ignoring ëA in the denominator of the fraction, andexpanding the exponentials in a power series)

Finally, after a time long relative to the half-life of B but short relative to the half-life of A,

167

NA ëA ' NB ëB .

NA ëA ' NB ëB ' NC ëC . . .

Q ' ( mi & mf & má )c 2 .

which shows that the ratio NB/NA is constant.

The above analysis can be extended to all the members of the radioactive series as long as theparent is longer-lived than any of the daughters. The result is

12.4 Alpha Decay

Alpha decay happens when a nucleus is too large for the short-range forces between nucleons to holdit together.

There are several questions we ought to ask about alpha decay.

Why does an alpha particle escape from an unstable nucleus, instead of individual neutronsand protons.

The nucleus is a deep potential well. Does the alpha particle have enough energy to escape?

Where, in fact, does the alpha particle's energy come from?

Let's answer the last question first. Consider a parent nucleus, which emits an alpha particle, and adaughter nucleus plus the alpha particle.

The initial mass is just the mass of the parent.

The final mass is the mass of the daughter+alpha. In the case of alpha decay, this final massmust be less than the parent mass.

The energy released is the mass difference times c2. This is the energy available to get thealpha particle out of the nucleus.

Why does an alpha particle escape, rather than its constituents.

An alpha particle's mass is considerably smaller than the sum of the masses of its constituents.

168

Ká 'A & 4

AQ .

(This is another way of saying the alpha particle has a high binding energy.) We need this bigmass difference to provide the energy to get the alpha out.

According to Beiser, if you do the calculation for Q for neutrons or protons, you don't endup with enough available energy to get them out of the nucleus. In fact, you need to putenergy in.

Now, we have seen that an alpha particle can escape, and mass is converted to energy in the process.

This is "like" an explosion. Both the alpha particle and the daughter nucleus will recoil, inopposite directions.

The kinetic energy carried away by the alpha particle is approximately

This equation comes from the conservation of energy and momentum (an easy Physics 23calculation).

Since most alpha emitters are very massive compared to the alpha particle, the alpha particle'skinetic energy is nearly equal to the total energy of the disintegration.

For 86222Rn, Q=5.587 MeV and Ká=5.486 MeV.

Tunnel Theory of Alpha Decay

The potential well of the nucleus is about 25 MeV deep. How does the alpha particle getout?

We already know the answer (I hope). The alpha particle tunnels out.

The probability of tunnelling may be extremely small. See Beiser. An alpha may "try" totunnel out 1021 times per second, but take 1010 years to escape. That's something on the orderof 1038 attempts to escape before the alpha particle gets out.

I won't test you on the details of the subsection of the tunnel theory of alpha decay.

12.5 Beta Decay

Beta decay occurs when a nucleus has too many neutrons relative to protons. Emission of an electronby a neutron changes it to a proton.

169

Energy

energy equivalent of mass

lost by decaying nucleus

The electron immediately leaves the nucleus upon creation.

Again, there are some questions.

The energy spectrum of emitted electronstypically looks like this:

It is logical to claim that electrons are createdwith the maximum energy, and lose some of itto the nucleus on the way out. But this is notverified experimentally. So, what happened tothe missing energy?

Also, beta decay involves n 6 p + e-, but spinsof all three particles are 1/2. There is no way (in the above reaction) for the net spin to be thesame before and after; i.e., angular momentum appears not to be conserved.

Also, conservation of linear momentum would require electrons and nuclei to recoil inopposite directions, but this is also not observed experimentally.

The solution to these problems is simple in hindsight; another particle must be involved in beta decay.

The "extra" particle is the neutrino. It is not charged and has no detectable mass (if it doeshave mass, the mass is very small).

Lacking charge and mass, the neutrino interacts very weakly with matter, which is why it tookso long to discover.

Beta decay is, then,n 6 p + e- + í

-.

The neutrino in beta decay is actually an antineutrino. (That's what the bar over the neutrinosymbol means.)

Positron emission is like beta decay, except a +e particle is emitted. In positron emission, a protonis converted into a neutron, a positron, and, of course, a neutrino:

p 6 n + e+ + í .

Competitive with positron emission is electron capture:p + e- 6 n + í .

The absorbed electron usually comes from the K shell, and later on an electron from an outershell drops into the empty state, emitting a gamma ray.

Electron capture occurs more often than positron emission in heavy elements because the

170

inner electrons are close to the nucleus.

There is another nuclear reaction by which a proton can change into a neutron. It is called inversebeta decay:

p + í-

6 n + e+.

Note that the absorption of an antineutrino is equivalent to the emission of a neutrino, so thisreaction involves the same physical processes as beta decay.

Inverse beta decay was used to confirm the existence of neutrinos. See Beiser.

Another kind of inverse beta decay isn + í 6 p + e-.

Both kinds of inverse beta decay have extremely low probabilities, which is why we aren'tbothered by the billions of neutrinos passing through us right now.

This section concludes with the problem of the missing neutrinos from the sun, which is one of theoutstanding problems of physics today.

12.6 Gamma Decay

We already mentioned how gamma decay is a means by which a nucleus can get rid of excess energy.

An alternative to gamma decay is internal conversion, in which the nucleus gives up its excessenergy to an orbital electron.

The orbital electron then exits the atom with its excess kinetic energy.

Finally, most gamma decay occurs in a very short time.

It is very unusual for nuclei to remain in excited states for a few hours, although some do.A long-lived excited nucleus is called an isomer.

12.7 Cross Sections

171

only these

particles interact

dNN

'nAó dx

A' nó dx ,

&dNN

' nó dx .

In the rest of this chapter, we will discuss ways in which nuclei can react, and some of theconsequences of these reactions.

In the first part of chapter 12, we considered radioactive nuclei which can spontaneouslydecay.

In the following sections, we consider reactions which take place when nuclei (or otherparticles) collide with high energies.

A cross section is basically just the probability that a projectile will interact in some specified waywith a target.

A geometrical cross section is like a targetarea. If the projectile's path takes it throughthis area, an interaction takes place. Ageometrical cross section is based on somephysical dimension, which might be the atomicsize (for x-rays) or the nuclear size (forneutrons).

It may be that the effective cross section forsome interaction is different than thegeometrical cross section. This effective crosssection is called the interaction cross section.

The cross section for a projectile incident on target atoms is defined by

where N is the number of incident particles, dN is the number of interacting particles, n is the numberof target atoms per volume, ó is the cross section for the particle to interact with the atoms, A is thetarget area, and dx its thickness.

The unit of cross section (not an SI unit) is the barn:

1 barn = 1 b = 10-28 m2 = 100 fm2.

If the projectiles are scattered away or somehow lost upon interaction, we must write

254

N ' N0 e &n ó x .

ë '1

nó.

This equation is easy to integrate, and the number of projectile particles which remainunscattered as a function of thickness x is

The cross section typically is a function of energy (and other things).

Figure 12.14 shows the cross section for absorption of a neutron by cadmium, as a functionof neutron energy.

Things to note in this figure: (1) the log scales (can be misleading if you are used to linearscales--small features on plots with log scales are usually extremely large features on linearscales); (2) the resonance in the cross section.

The next page in these notes is the document CH12-1.MCD.

The mean free path is the average distance a projectile travels before interacting. We have alreadyseen mean free paths in chapter 10.

In this chapter, the mean free path can be written in terms of a cross section:

Reaction Rate

Skip this subsection of chapter 12.7. You will not be tested on it.

12.8 Nuclear Reactions

Here's how to think of nuclear reactions:

An energetic projectile strikes a target.

If the projectile energy is large enough, it can reach and stick to the target.

The result is a new, compound nucleus, which is probably unstable.

After a relatively short period of time, the compound nucleus decays, leaving a new nucleus.Energy and radiation are given off.

255

ÄE Ät $ h&

2.

ô 'h&

Ã.

23592 U% 10n 6 236

92 U ( 6 14054 Xe% 94

38Sr% 10n% 10n .

Example:

Figure 12.15 gives a number of reactions which can produce the compound nucleus 714N*.

The resulting 714N* nucleus cay decay in a number of ways; see page 448.

A peak in the cross section curve, like the one for neutrons on cadmium, represents a resonance, oran excited state nucleus.

The energy width of the resonance can be used to define the lifetime of the resonant state.

Heisenberg's uncertainty principle says

If we consider the width of the resonance to be ÄE and the lifetime ô to be the uncertainty intime Ät, then

Center of Mass Coordinate System -- skip this subsection. You will not be tested on it.

13.5 Nuclear Fission

Here's the "classic" reaction:

Several hundred MeV of energy, mostly in the form of kinetic energy of the fission fragments,is also released.

Other reactions are possible. See Beiser figure 12.20. Other numbers of neutrons arereleased in other reactions.

256

Here are the important things:

Lots of energy is released by a single fission.

Each fission produces, on the average, 2.5 neutrons.

If, on the average, one of those 2.5 neutrons can be used to produce another fission, you havea self-sustaining chain reaction.

How do you sustain a chain reaction?

What stops a chain reaction? Neutrons escape without causing fission (on the average, morethan 1.5 per fission escaping will stop a chain reaction).

To sustain a chain reaction, you "trap" the neutrons (e.g., surround the fissioning materialwith a neutron reflector, such as beryllium).

Or you make the fissioning mass physically bigger in extent, so that neutrons have morematter to pass through before they escape.

12.10 Nuclear Reactors

Most of this material you can read yourselves.

I won't test you on technical details, like what's the water pressure and temperature in a PWR(pressurized-water reactor). I will expect you to understand the main concepts.

Problems to overcome in a nuclear reactor:

92238U is abundant, but it absorbs neutrons without fissioning.

92235U is not abundant, but it absorbs neutrons and fissions.

Fast (energetic) neutrons are produced by fission. These are the neutrons that 92238U absorbs.

Slow (not very energetic) neutrons are the ones that cause 92235U to fission, but they are not

produced by fission.

To overcome these problems:

Enrich the uranium (find a way to get more of the mass-235 isotope). This has some of thesame problems as the next option.

257

Use another fuel. None others occur naturally, but uranium-238 can be used to produceplutonium, which is extremely dangerous and does undergo fission. Thorium-232 also can beused to create the fissionable uranium-233.

Problems: dangerous materials. Atomic bomb components.

Slow down the fast neutrons, so that they don't react with uranium-238 but they do react withuranium-235. That way you could use a fuel of natural uranium.

Slowing down (moderating) neutrons.

Remember, protons would work best. Water has lots of protons. Use water.

Problem--the hydrogen in water can also capture neutrons to form deuterium.!The solution -- light water reactors require fuel enriched in uranium-235.

Another solution: use heavy water. The deuterium in heavy water doesn't react with theneutrons. It doesn't slow the neutrons down as well, but it works. Then you can use naturaluranium fuel.

!A problem with this solution -- obtaining heavy water.

Another solution: use graphite. Graphite is carbon. It is about the next-lightest elementcompared to water that is feasible to use as a moderator of neutrons.

!There are lots of benefits in properly designed reactors.!A major problem is that graphite burns, as we all know.

Breeder Reactors

The material in the discussion in the notes above is all I expect you to know.

A Nuclear World?

Very interesting. No time to cover it. Read but I won't test you on it.

12.11 Nuclear Fusion in Stars

Very interesting. No time to cover it. Read but I won't test you on it.

258

21H% 21H 6 31H% 11H% 4.0 MeV

21H% 21H 6 32He% 10n% 3.3 MeV

31H% 21H 6 42H% 10n% 17.6 MeV

12.12 Fusion Reactors

Fusion is the combining of light nuclei to form a single, heavier nucleus, but one with less mass thanthe two light ones.

Here are some of the reactions which might be feasible:

The last reaction is particularly attractive because it liberates so much energy.

Advantages of fusion:

Abundant materials. Deuterium is readily available and tritium is easy to produce.

Clean. "No" nasty radioactive fission products. Only hydrogen, helium, neutrons, energy.(Tritium, however, is radioactive.)

Disadvantages of fusion:

Extremely high energies needed to begin a reaction.

How do you "hold" the "burning" fuel.Answer--you don't. Not with a physical container like we are used to thinking of.

How do you manage to bring a material object close enough to the burning fuel to extractenergy without vaporizing the material?

This is an engineering problem that can be solved. See the subsection on confinementmethods. I won't test you on this subsection.