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CHEM-UA 127: Advanced General Chemistry I

Notes for Lecture 22

I. THE FOUR FORCES OF NATURE AND NUCLEAR CHEMISTRY

Physicists have determined that all of nature is governed by four fundamental forces: The gravitational force, theelectromagnetic force, the weak force, and the strong force:

1. The strong force is responsible for holding quarks into larger particles, specifically hadrons and mesons. Inchemistry, the two hadrons of greatest interest are the proton and the neutron. The proton consists of two up(“u”) quarks and one down (“d)” quark. Quarks are spin-1/2 fermions with fractional charges (qup = +2/3,qdown = −1/3), giving it a charge of +1. The neutron consists of one up and two down quarks, giving it a chargeof 0. The strong force is, as the name suggests, quite strong, but it is also VERY short ranged. The quantum ofthe strong force that mediates interactions between quarks is a bosonic particle known as the “gluon”. In total,there are 8 possible gluon particles.

2. The electromagnetic force, as we have seen, holds electrons into atoms and atoms into molecules. Therefore,it operates at the scale of all ordinary matter and is, hence, responsible for all of ordinary chemistry. It is along-range force that, although weaker than the strong force, is still relatively strong. The quantum of theelectromagnetic interaction or electromagnetic field, mediating the interaction between charged particles, is thephoton, which is also a boson. There is only one type of photon.

3. The weak force is responsible for the nuclear decay processes we will discuss in this lecture. The weak forcehas the ability to change the identity of quarks, e.g., from up to down and down to up, which underlies manynuclear decay processes. There are three quanta of the weak force, known as the W+, W−, and Z0 bosons. Notethat W+ and W− are both charged particles.

Note that these three forces have been unified into a single theory known as the standard model.

4. The fourth force is gravitation, which is responsible for holding planets into solar systems, solar systems intogalaxies, galaxies into clusters, and clusters into superclusters. Gravitation is obviously also responsible forkeeping our feet planted firmly on the ground. Like the electromagnetic force, gravitation is long range. Thequantum of the gravitational fieldis a spin-2 boson known as the graviton. As of today, there is no definitivetheory on how to unify gravitation with the other three forces. The difficulty is that gravitation is describedby Einstein’s theory of general relativity, which describes space and time as being curved by massive objectsand whose mathematical structure is quite different from that of the other three forces. Devising a quantumtheory of space-time curvature has proved immensely challenging, and existing theories such as string theory(or M-theory) or loop quantum gravity cannot be tested because of the difficulty of creating the experimentalconditions under which quantum mechanics becomes important in gravitational interactions.

Relative to the electromagnetic force, on the scale of quarks, the strong force is roughly 60 times stronger, whilethe weak and gravitational forces are 10−4 and 10−41 time weaker. On the scale of protons and neutrons, the weakand gravitational forces are 10−7 and 10−36 times weaker.

In nuclear processes, the most important consideration is the change of one type of particle into another. Unlikeprocesses governed by the electromagnetic force, which conserves mass, the weak force does not obey mass conservationand is not subject to this restriction. Thus, in nuclear binding or decay processes, mass changes. If we denote themass change by ∆m, then there is an associated energy change ∆E given by Einstein’s well know formula

∆E = c2∆m

where c is the speed of light. Where does Einstein’s relation come from? According to his special theory of relativity,the kinetic energy of a free particle of mass m and momentum p is

E =√

p2c2 +m2c4

2

This relation is consistent with fundamental condition that no particle can have a speed that exceeds c. Whenpc ≪ mc2, then we can expand the square root as follows:

E = mc2√

1 +p2

m2c2≈ mc2

(

1 +p2

2m2c2

)

= mc2 +p2

2m

Note that the second term is just the usual non-relavistic kinetic energy. The first term, mc2 is known as the rest

energy, which is the energy of a mass m at rest (when p = 0). Normally, we can just shift the zero of energy sothat everything is measured relative to mc2, and we do not need to worry about this. But at speeds high enoughthat relativistic effects are important, the rest energy cannot be neglected. So, when p = 0, we have E = mc2, or∆E = c2∆m.

A number of nuclear decay processes are governed by the transformation of a neutron into a proton or of a protoninto a neutron. The former occurs spontaneously for free neutrons (we will soon show why this is), while the latterdoes not, although it does occur in the environment of the nucleus. The two processes are

10n −→

11p

+ + 0−1 e− + ν̄e

11p

+−→

10n+0

1 e+ + νe

Here, 01e

+ is a positron, which is the antiparticle of the electron ( 0−1e

−), and the particles νe and ν̄e are known asthe nuetrino and antineutrino, respectively. Neutrinos are uncharged, massless particles whose existence was positedby Wolfgang Pauli. The prediction of the existence of neutrinos was made on the basis that without them, it is notpossible to conserve energy, momentum, and angular momentum in the above decay processes. Because neutrinos haveno charge or mass (or negligible mass), their detection is extremely tricky and usually requires indirect methods. Itis interesting to note that the Nobelaureate physicist Isaac Isidor Rabi (known for his discovery and theory of NMR),when he first heard about that neutrinos had been postulated, was said to have quipped, “Neutrino? Who orderedthat?” Since the mass of a neutrino is 0, its energy is determined solely from its momentum and is Eνe,ν̄e = pc.

The electron and positron, being antiparticles of each other, undergo an annihilation process if they encounter eachother. This process can be represented in a diagram called a “primitive Feynman vertex” or a one-vertex Feynman

diagram, shown in Fig. 1. Mathematically, this processs has the same form as an electron scattering process in which

FIG. 1. Electron-positron annihilation one-vertex Feynman diagram.

an electron radiates a photon and, in so doing, changes its direction of travel (recall that this is a form of acceleration,and accelerating charged particles emit electromagnetic radiation). This is shown as a one-vertex Feynman diagramin Fig. 2 below.

FIG. 2. Electron scattering one-vertex Feynman diagram.

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In order for a neutron to decay into a proton, one of its down quarks needs to change to an up quark. This processcan also be represented as a one-vertex Feynman diagram, as shown in Fig. 3. In the process of changing from down

FIG. 3. One-vertex Feynman diagram for tranformation from a down quark to an up quark via the weak interaction.

to up, a quantum of the weak force W− is emitted, which is needed in order to carry away one unit of negative charge,and change the charge state from −1/3 to 2/3. We can also express this process as

d −→ u +W−

The charge on the left is −1/3, while that on the right is +2/3− 1 = −1/3, so charge is balanced.

In proton decay, an up quark must transform to a down quark. This process is

u −→ d +W+

Note that, in this case, the W+ boson must be emitted in order to balance charge. The corresponding one-vertexFeynman diagram is shown in Fig. 4.

FIG. 4. One-vertex Feynman diagram for tranformation from an up quark to a down quark via the weak interaction.

If we wish to determine whether the neutron or proton decay process is spontaneous, we should examine the energychange in the reaction. If ∆E < 0, energy is lowered, and the process is spontaneous. In reality, we should lookat the change in the relevent thermodynamic energy, namely, the free energy, however, in nuclear processes, thecontribution from entropy changes is comparatively negligible, and the two will be essentially identical. However,since ∆E = c2∆m, we can also just look at the change in mass. If ∆m < 0, the process is spontaneous.

For neutron decay, the mass change is

∆m = mp +me− −mn

Here, the masses of the proton, neutron, and electron are mp = 1.6726216×10−27 kg, mn = 1.6749272×10−27 kg, andme = 9.1093819× 10−31 kg. Plugging these numbers in, we find that ∆E = −1.394661× 10−30 kg, hence, the processis spontaneous. Note that for the corresponding proton decay process, we would find ∆E = 1.394661× 10−30 kg > 0,hence, it is not spontaneous for a free proton, as previously noted. In the neutron decay process, the correspondingenergy change is

∆E =(

−1.394661× 10−30 kg) (

2.9979× 108 m/s)2

= −1.2535× 10−13 J = −0.782 MeV

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II. NUCLEAR DECAY PROCESSES

A. β− decay

Fig. 5 shows a plot of the number of neutrons vs. the number of protons in all of the known elements and theirisotopes. We see that for Z less than approximately 40, it is possible for isotopes of these elements to be eitherneutron rich Nn/Z > 1, where Nn is the number of neutrons, or neutron poor Nn/Z < 1. However, the plot veersupward off the diagonal (where Nn = Z) considerably for Z > 40, and all isotopes are of the neutron rich type. Manyisotopes are not stable and ultimately into other stable nuclei. When Nn/Z > 1, this decay can occur via neutrondecay, which releases an energetic electron known as a β− particle.

FIG. 5. Number of neutrons versus number of protons for the elements and their isotopes.

The general reaction for β− decay is

Az X

Z+−→

AZ+1 Y(Z+1)+ + 0

−1 e− + ν̄e

An example of β− decay is that of the 146 C isotope. The process for nuclei is

146 C6+

−→147 N7+ + 0

−1 e− + ν̄e

Note that I have written the explicit charge on each nucleus so that charge balance can be seen explicitly. We nowwish to determine if this process occurs spontaneously, and what the associated mass change is. Note that if we justwrite down the mass change in terms of the reaction above as written, we obtain the mass change for the nuclei as

∆m = m(

147 N7+

)

+m(

0−1e

−

)

−m(

146 C6+

)

There is nothing incorrect about the way this is written and, if we had the masses of the nuclei readily available,we could easily calculate ∆m this way. However, the data we have for masses comes from mass spectrometry and,

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therefore, is for neutral atoms. Consequently, if we wish to use this data, we need to express the reaction andassociated mass change as if they were occurring for neutral atoms (although they really do occur for nuclei). Thus,we are employing a trick that will allow us to obtain a correct mass difference due to the nuclear decay. The way weuse this trick is to look at the left side of the equation, where we see a carbon nucleus with a +6 charge. In order tocreate the neutral atom, we would need to add 6 electrons to the left side. However, whatever we do on the left, wemust also do on the right, so we add +6 electrons to both sides of the nuclear decay reaction equation, to give

146 C6+ + 6 0

−1e−

−→147 N7+ + 6 0

−1e− + 0

−1 e− + ν̄e

Now, on the left, we have a neutral carbon atom. On the right, if we take the 6 electrons we added together with theelectron that results from the β− decay, we have a total of 7 electrons, enough to make the nitrogen nucleus into aneutral atom. Thus, in terms of neutral atoms, we can write the reaction as

146 C −→

147 N+ ν̄e

We can now obtain the mass difference from that of the neutral atoms as

∆m = m(

147 N

)

−m(

146 C

)

which will give exactly the same value as that from the nuclei. How do we see this? The manipulation we did forthe reaction equation is equivalent to taking the equation for the mass difference in terms of nuclei and adding andsubtracting the mass of 6 electrons as follows:

∆m = m(

147 N7+

)

+ 6m(

0−1e

−

)

+m(

0−1e

−

)

−m(

146 C

6+)

− 6m(

0−1e

−

)

the mass of the nitrogen nucleus plus the mass of the 7 electrons we are adding to it, gives us the mass of the neutralatom, while taking the mass of the carbon nucleus, which we are subtracting, and subtracting, in addition, the massof 6 electrons, we end up subtracting the mass of a neutral carbon atom, which leads to the correct mass differencein terms of neutral atoms. Plugging in the numbers for the masses, which can be looked up in atomic units, we find

∆m = (14.003074 amu− 14.003242 amu)(

1.6605387× 10−27 kg/amu)

= −2.789705× 10−31 kg

from which we see that ∆m < 0, so the process is spontaneous. The corresponding energy change is obtained bytaking this mass difference and multiplying by c2, and this gives ∆E = −2.5072207× 10−14 J = -0.156 MeV.

B. β+ decay

When Nn/Z < 1, the isotope is neutron poor, and decay occurs via β+ decay. A β+ particle is just an energeticpositron (01e

+). The general reaction is

AZX

Z+−→

AZ−1 Y(Z−1)+ +0

1 e+ + νe

An example of this is the decay of 116 C, which decays according to

116 C6+

−→115 B5+ +0

1 e+ + νe

The mass change for the reaction as written is

∆m = m(

115 B5+

)

+m(

01e

+)

−m(

116 C6+

)

which is correcy for nuclei, but again, we need to transform this so that it’s appropriate for neutral atoms. Again, weneed to add 6 electrons to each side of the reaction to give

116 C6+ + 6 0

−1e−

−→115 B5+ + 6 0

−1e− +0

1 e+ + νe

On the left, we have a neutral atom of carbon-11, and on the right if we take 5 of the 6 electrons we added, then wehave a neutral boron atom. Thus, in terms of neutral atoms, the reaction becomes

116 C −→

115 B+ 0

−1e− +0

1 e+ + νe

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and the mass difference becomes

∆m = m(

115 B5+

)

+ 6m(

0−1e

−

)

+m(

01e

+)

−m(

116 C6+

)

− 6m(

0−1e

−

)

= m(

115 B

)

+m(

0−1e

−

)

+m(

01e

+)

−m(

116 C

)

= m(

115 B

)

+ 2m(

0−1e

−

)

−m(

116 C

)

= (11.0093055 amu + 2 ∗ 0.0005485799 amu− 11.011433 amu)(

1.6605387× 10−27 kg/amu)

= −1.7109198× 10−31 kg

which is negative and corresponds to an energy change of -0.96 MeV.

C. Electron capture

If the mass change in a reaction is too small, β decay might not be a possible nuclear decay channel. In such cases,nuclear decay is still possible through a process called electron capture. Electron capture is another mechanism bywhich a proton can be turned into a neutron according to

11p

+ + 0−1 e− −→

10 n+ νe

which clearly involves the change of an up quark to a down quark via emission of a W+ boson. The general form forelectron capture is

AZX

Z+ + 0−1 e− −→

AZ−1 Y(Z−1)+ + νe

An example of electron capture is the decay of Uranium-231 according to

23192 U92+ + 0

−1 e− −→23191 Pa91+ + νe

In terms of nuclei, the mass difference is

∆m = m(

23191 Pa91+

)

−m(

23192 U92+

)

−m(

0−1e

−

)

In order to put this in terms of neutral atoms, we add 91 electrons to both sides:

23192 U92+ + 91 0

−1e− + 0

−1 e− −→23191 Pa91+ + 91 0

−1e− + νe

Now on the left, if we take the electron already present from the electron capture together with the 91 we added, wehave a neutral U-231 atom, and on the right, we have 91 electrons to form neutral protactium:

23192 U −→

23191 Pa + νe

In terms of masses, we add and subtract 91 times the mass of the electron to give

∆m = m(

23191 Pa91+

)

+ 91m(

0−1e

−

)

−m(

23192 U92+

)

−m(

0−1e

−

)

− 91m(

0−1e

−

)

= m(

23191 Pa

)

−m(

23192 U

)

= (231.0385879 amu− 231.036289 amu)(

1.6605387× 10−27 kg/amu)

= −6.8082087× 10−31 kg

which is less than zero, hence the process is spontaneous. The corresponding energy change is -0.382 MeV. Note thatwhile this mass change is comparable to values we obtained for β-decay, relative to the mass of uranium, the masschange is small.

D. α-decay

An α-particle is an energetic helium-4 nucleus, 42He

2+. The general form of α-decay, which causes a change in bothA and Z, is

AZX

Z+−→

A−4Z−2 Y(Z−2)+ +4

2 He2+

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An example of this is the decay of U-238 according to

23892 U92+

−→23490 Th90+ +4

2 He2+

The change in mass is

∆m = m(

23490 Th90+

)

+m(

42He

2+)

−m(

23892 U92+

)

In order to express this in terms of neutral atoms, we need to add 92 electrons to both sides of the reaction:

23892 U92+ + 92 0

−1e−

−→23490 Th90+ + 92 0

−1e− +4

2 He2+

On the left, we now have a neutral U-238 atom, and on the right, if we divide the 92 electrons between the Thoriumand the helium, 90 for the former, 2 for the latter, we end up with neutral Thorium and Helium atoms:

23892 U −→

23490 Th +4

2 He

In terms of the mass change, we obtain

∆m = m(

23490 Th90+

)

+ 92m(

0−1e

−

)

+m(

42He

2+)

−m(

23892 U92+

)

− 92m(

0−1e

−

)

= m(

23490 Th

)

+m(

42He

)

−m(

23892 U

)

= (234.043595 amu + 4.002603 amu− 238.050783 amu)(

1.6605387× 10−27 kg/amu)

= −7.61357× 10−30 kg

which is less than zero, indicating that the process is spontaneous. The corresponding energy change is ∆E =−4.27MeV .

E. γ decay

Just as electrons bound into atoms have a set of discrete quantum mechanical energy levels, so do the nucleons thatare bound into atomic nuclei. When excited into an excited state, these particles decay back down to the ground stateand emit electromagnetic radiation in the process. This radiation tends to be very high frequency, typically lying inthe gamma (γ) region of the electromagnetic spectrum. When the photons emitted are specifically in the gamma partof the spectrum, the decay process is known as γ decay. There is no change in the number of neutrons or protons, sothe nucleus does not change its elemental form. Thus, we do not consider this type of decay in the same category asthose previously discussed, which do change the identity of the nucleus, however, it is important to mention it here,as we will discuss its biological effects a bit later.

III. KINETICS OF NUCLEAR DECAY

The term kinetics refers to dynamics, specifically, the rates of processes. Even if ∆E < 0 for a reaction, thiscondition says nothing about the rate at which the reaction will occur. The study of the kinetics of a process seeks toanswer this question. The key to this answering this question is to determine the rate constant k and the correspondingpopulations of different species in a reaction as a function of time. You will learn much more about this next semesterwhen you study chemical kinetics. In brief, rate constants tell us how many decays (or chemical events) occur, onaverage, in a given unit of time.

Given a collection of radioactive nuclei such that at time t = 0, we have N(0) of them. What is N(t) at a time t later,assuming that the decay of a particular nucleus at a particular time is essentially a random event with a probabilityof occurring consistent with the rate constant for the decay process. Under this condition, the decay rate −dN/dtwill be proportional to the current population N(t). The fewer nuclei there are, the fewer decays there can be, sothat the slower the change in population with respect to time will be. This means that

−dN

dt∝ N(t)

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Note that the minus sign in the decay rate indicates that we are, indeed, expressing a loss of population rather thana gain. The rate constant is the constant of proportionality, i.e.,

−dN

dt= kN(t)

Solving this little first-order differential equation, we obtain

N(t) = N(0)e−kt

The decay rate, itself,

−dN

dt= kN(t) = kN(0)e−kt

is denoted A(t), which is called the activity. Although the units of A(t) are also inverse time units, in order todistinguish activities from rate constants, we give A(t) different units called Becquerels, denoted Bq. Since A(0) =kN(0), we see that A(t) satisfies the same decay law as N(t):

A(t) = A(0)e−kt

In nuclear decay, an important concept is that of the half life of a decaying species. The half life t1/2 is defined as thetime at which N(t1/2) = N(0)/2. From the decay law, we see that

N(t1/2) = N(0)e−kt1/2 =1

2N(0)

t1/2 =ln 2

k

Half lives are often used to characterize nuclear energy waste products and come into play significantly in the techniqueof carbon dating, which is used to determine ages of archaeological artifacts.

A. Carbon dating

Carbon dating is based on the decay of one the isotopes of carbon, specifically, carbon-14. In the stratosphere, highenergy neutrons are created by cosmic rays which collide with nitrogen to produce this carbon isotope according to

147 N+1

0 n −→146 C+1

1 p+

The carbon-14 isotope enters the carbon cycle of the earth continuously and has a half life, t1/2, of 5730 years. Whena plant or animal dies, carbon exchange with the environment ceases, causing carbon-14 to decay, and we can tell theage of the object based on carbon-14 amounts, as the following example illustrates:

Example: Living organisms contain a relatively constant amount of carbon-14 with an activity of 0.255 Bq/g.Suppose a fossilized plant sample is observed to have an activity of 0.195 Bq/g. What is the age of the fossil?

Solution: The half-life of carbon-14 is 5730 years, so the rate constant is

k =ln 2

5730 yr= 1.21× 10−4 yr−1

The problem is stated in terms of activities, however, the decay law is the same.

A(t) = A(0)e−kt

t = −1

kln

(

A(t)

A(0)

)

= −1

1.21× 10−4ln

(

0.195

0.255

)

= 2217 yr

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IV. RADIATION IN BIOLOGY

The high energy of the particles that are the products of radioactive decay can cause significant damage to the tissueof living organisms through the process of ionization. When these high energy particles collide with the moleculesin these tissues, they knock electrons out of stable orbits causing them to become essentially free electrons, leavingbehind radical and/or charged species that are chemically modified from their original form and, often, highly reactive.Damage of this type done to DNA, for example, can lead to unfavorable mutations, or if water molecules in livingorganisms are ripped apart, OH· radicals are left behind, and these, being highly reactive, degrade proteins andother important molecules on which these organisms rely. The different types of radioactive decay processes we havediscussed have different biological implications.

α particles are extremely high energy and can do considerable damage. However, their high mass and relatively largesize means that they dissipate their energy quickly and have a range of only about 0.5 mm in water and living tissue,roughly that of the outer skin layer. The greatest potential for suffering the negative effects of α radiation, therefore,is through ingestion of α-radioactive species, either through respiration or in water or food. Radium is an exampleof a substance that decays via α decay, which is thought to have been what caused the illness and ultimate death ofMarie Curie, who is credited with having discovered radium and identifying its radioactivity.

β− particles, being energetic electrons, are smaller and lighter than α-particles and, therefore, have a longer range orroughly 1-3 cm in water and living tissue. This gives them greater penetration potential, however, like α-particles,they do the greatest damage when they are ingested.

γ radiation tends to do the most immediate damage to living organisms. Since γ rays are photons, their range isessentially infinite, and their high energy gives them the potential to cause ionization over the entire range of theirtrajectory.

Noting that radiation is caused by the transfer of energy from nuclear decay products to the environment, we candefine a measure of radiation dosing. 1 rad is deinfed as the amount of radiation needed to deliver 0.01 J of energyto living tissue. However, the rad does not distinguish between different types of radiation, some of which is moredamaging than others, as our discussion above makes clear. Thus, a unit known as the rem is often used to accountfor both the energy delivered and the toxicity of the radiation. 1 rem = 1 rad for β and γ radiation. However, becauseα-radiation is more energetic, 1 rad of α radiation is equivalent to a dose of 10 rems. The SI unit of radiation is theSievert, where 1 Sv = 100 rems.

In order to give you an idea of radiation amounts from everyday activities, a dental or chest X-ray is equivalent toabout 0.1 mSv. A 7-hour domestic flight exposes you to about 0.02 mSv, caused by increased exposure to cosmicradiation at the heights planes typically fly. A chest CT scan, which is a high dose of X-rays, is around 7 mSv, whichis quite substantial. The average person is exposed to roughly 1 mSv annually. Small amounts of radiation can causedamage to our DNA, however, when this happens, naturally occurring repair enzymes in our cells can repair thedamage and restore the DNA to a healthy state. Too much radiation exposure causing more extensive damage cannotgenerally be fixed by such enzymatic mechanisms; for example, an exposure of 0.25 Sv can cause radiation sickness,and a dose of 5 Sv is generally fatal.

A. Radiation therapy in cancer treatments

High energy photons are often used in the treatment of cancers because of their very ability to damage livingtissue. By damaging the DNA of the cells that make up the tumor beyond the reach of the aforementioned repairmechanisms, such therapies ensure that the tumor cells cannot replicate, and the tumor is made to shrink in size overa series of such treatments. Unfortunately, exposure to photons of the frequency used, typically X-rays or gammarays, in cancer therapies, causes considerable damage to surrounding tissue as well, hence radiation therapy has veryserious side effects. This fact has prompted medical researchers to search for alternative therapies that can targettumor cells while causing significantly less damage to surrounding tissue.

An important concept in the study of the ionization capabilities of charged particles or electromagnetic radiaion isthat of the Bragg curve. The Bragg curve measures the energy transfer to the medium of a charged particle or photon

10

as a function of the distance traveled through the medium. Mathematically, it is given by a formula known as theBethe-Bloch equation. For a particle of charge z and mass m traveling with velocity v ≪ c through a medium ofelectron number density n, the energy delivered dE over a distance dx is

−dE

dx≈

4πnz2

mev2

(

e2

4πǫ0

)

ln

(

2mev2

I(x)

)

where I(x) is the ionization potential of the medium at position x.

Bragg curves for protons, carbon ions, and iron ions are shown in Fig. 6. The peak in the curve is known as the Braggpeak. Comparing the three curves, we see that protons and carbon ions have a sharp, narrow peak in their Braggcurves. The implication of this peak is that these particles deliver the majority of their energy over a very narrowrange at a distance given by the location of the Bragg peak. It should be noted that the location of the Bragg peakwill, as the Bethe-Bloch equation implies, depend on the velocity (hence the energy) of the particle.

The sharpness of the Bragg peak suggests that charged particles like protons and carbon ions might be superior tophotons for use in radiation therapies targeting certain types of cancers. In particular, if the distance between apatient and a source of energetic protons is set to coincide with the location of the Bragg peak, then a typical tumorwill lie within the width of the Bragg peak, and the proton beam can be used to destroy the cells in the tumor whiledoing minimal damage to the surrounding tissue. Because the location of the Bragg peak is energy dependent, it isimportant that the proton beam be energy filtered so that only those protons with an energy that places the Braggpeak exactly at the distance between the source and the tumor are allowed through the filter.

In order to see the advantage of proton therapy over photon therapy, look at Fig. 7, which compares energy dissipationcurves between photons and protons. It can be seen that the photon curve is very broad and decays slowly, indicatingthat photons deliver considerable energy over a very wide distance range. This is the cause of the ancillary damagethey do and the significant side effects of radiation therapy. By looking at the iron Bragg curve in Fig. 6, we seewhy iron ions would not be suitable for cancer therapies: Although there is a sharp, narrow Bragg peak, there isalso a broad region in the Bragg curve before the peak at which iron ions can deliver considerable energy in muchthe same way that photons do. The search for alternative cancer therapies via charged particles relies heavily on thecharacteristics of their Bragg curves.

FIG. 6. Bragg curves of protons, carbon ions, and iron ions.

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FIG. 7. Comparison of Bragg curves for protons and photons.

V. NUCLEAR FISSION AND FUSION

In 1939, Albert Einstein wrote a letter to President Roosevelt warning him of the intention of the Germans todevelop an atomic bomb and suggesting that he encourage the develop of chain fission reactions in the US. What hewas referring to is the fact that is uranium-235 is bombarded with energetic neutrons, the nucleus can break apart ina process called fission into lighter isotopes via reactions of the form

23595 U+1

0 n −→7230Zn +162

62 Sm + 210n

23595 U+1

0 n −→8038Sr +

15354 Xe + 310n

23595 U+1

0 n −→9436Kr +139

56 Ba + 310n

In fact, there are close to 400 possible fission pathways for U-235 occurring with different likelihoods. Criteria suchas the magnitude of ∆m for these reactions give us clues as to which of the myriad pathways are more favorable.

As these reactions indicate, more neutrons tend to be produced than are consumed by the reactions, and theseneutrons are of very high energy. Consequently, these neutrons can be used to cause additional fission reactions,which, in turn, create more energetic neutrons, and the result is a self-propagating chain reaction. Under the rightconditions, such reactions can actually be realized. There is a partial differential equation that describes how chainreactions work. In particular, let u(r, t) be the density of neutrons at position r in space at time t. The evolution ofthis density follows a diffusion equation with an extra term that serves as a control of the process:

∂

∂tu(r, t) = a2∇2u(r, t) + v(r, t)u(r, t)

Here, a is a constant, and v(r, t) is a control function determined by the conditions under which the reactions arecarried out. How this function is engineered determines whether the reactions are controlled for use in the productionof nuclear power or whether the reactions are allowed to run away in order to create the type of massive release ofenergy characteristic of a nuclear bomb.

While U − 238 and U − 235 are the stable isotopes of uranium, the relative abundances of these are 99.27% and0.72%, respectively. Thus, in naturally occurring uranium, isolating enough U-235 for use in nuclear power requiresseparating the two isotopes from each other in a process known as uranium enrichment. Techniques that can be usedinclude centrifugation and other diffusion techniques, sieving, thermal processing,.... While nuclear power has manyadvantages such as no air pollution, and large power production at low operating costs, many disadvantages alsoexist. These include the problem of disposing with the radioactive waste products (which have long half lives), andthe possibility of overheating causing meltdowns at nuclear power plants (think Chernobyl, Fukushima, or Three-MileIsland). Disposal of nuclear waste products is one of the uses of a material that appeared on a previous problemset: the aluminosilicates. In particular, borosilicates and boroaluminosilicates have been engineered for this purposebecause of their high stability and resistance to degradation from water when buried deep in the ground. Control

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issues that lead to meltdowns are much more difficult to anticipate. While quite rare, when these events do occur,they are extremely serious because of the potential for wide-spread radiation to escape the nuclear power facility.Waste disposal and safety concerns are among the reasons for the resistance to more comprehensive use of nuclearfission power.

Nuclear fusion is a process that does not suffer from these downsides and is a potential source of clean, manageablenuclear power. Nuclear fusion employs reactions in which light nuclei are fused together to create heavier nuclei,releasing a tremendous amount of energy in the process. Some examples of fusion reactions of deuterium nuclei areshown below:

21H+2

1 H −→32He +

10 n

21H+2

1 H −→42He + γ

21H+2

1 H −→31H+1

1 p+

The problem with fusion reactions is the tremendous Coulomb barrier that must be surmounted to get two light nucleito approach each other. Generally, such reactions can be produced by subjecting a plasma to a very high magneticfield (recall that charged particles in magnetic fields execute circular motion, and if the fields are large enough, highvelocities can be achieved). Currently, however, the energy needed to create fusion reactions is larger than the energythat is gained, and the break-even point has yet to be reached.

Another interesting example of fusion is the nuclesynthesis of heavy elements in stars. As stars age, the accumulatelarge amounts of helium, which is a starter light element that can fuse into heavier elements. Consider, for example,the two-step fusion process leading to carbon-12:

42He +

42 He ⇀↽ 8

4Be (t1/2 = 7× 10−17 s)

84Be +

42 He −→

126 C

The first reaction is shown as an equilibrium because of the short half-life of the Be isotope, which can immediatelyrevert back to helium nuclei. However, occasionally, the Be-8 nucleus can fuse with another helium-4 nucleus to formcarbon-12, one of the key atoms in the molecules of life. The above fusion processes occur when the star starts tocontract under the action of its own gravity, thereby allowing nuclei to overcome the fusion barrier, occurring attemperatures at 108 K.

In fact, other heavy element synthesis occurs via additional fusion reaction in a process known as the CNO (carbon-nitrogen-oxygen) cycle. Several such cycles have been suggested, one of which is shown below: The cycle shows how

FIG. 8. A CNO cycle.

it is possible, through a series of fusion reactions and β decay processes, to cycle through various isotopes of carbon,nitrogen, and oxygen.