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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, the
electromagnetic force, the weak force, and the strong force:
1. The strong force is responsible for holding quarks into larger particles, specifically hadrons and mesons. In
chemistry, 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 charge
of 0. The strong force is, as the name suggests, quite strong, but it is also VERY short ranged. The quantum of
the 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 a
long-range force that, although weaker than the strong force, is still relatively strong. The quantum of the
electromagnetic interaction or electromagnetic field, mediating the interaction between charged particles, is the
photon, 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 force
has the ability to change the identity of quarks, e.g., from up to down and down to up, which underlies many
nuclear decay processes. There are three quanta of the weak force, known as the W+ , W− , and Z0 bosons. Note
that 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 into
galaxies, galaxies into clusters, and clusters into superclusters. Gravitation is obviously also responsible for
keeping our feet planted firmly on the ground. Like the electromagnetic force, gravitation is long range. The
quantum of the gravitational fieldis a spin-2 boson known as the graviton. As of today, there is no definitive
theory on how to unify gravitation with the other three forces. The difficulty is that gravitation is described
by Einstein’s theory of general relativity, which describes space and time as being curved by massive objects
and whose mathematical structure is quite different from that of the other three forces. Devising a quantum
theory 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 experimental
conditions 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, while
the weak and gravitational forces are 10−4 and 10−41 time weaker. On the scale of protons and neutrons, the weak
and 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. Unlike
processes governed by the electromagnetic force, which conserves mass, the weak force does not obey mass conservation
and is not subject to this restriction. Thus, in nuclear binding or decay processes, mass changes. If we denote the
mass 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
p
E = p 2 c 2 + m2 c 4
2
This relation is consistent with fundamental condition that no particle can have a speed that exceeds c. When
pc ≪ mc2 , then we can expand the square root as follows:
r
p2
p2
p2
2
2
2
E = mc 1 + 2 2 ≈ mc 1 +
=
mc
+
m c
2m2 c2
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 so
that everything is measured relative to mc2 , and we do not need to worry about this. But at speeds high enough
that 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 proton
into a neutron. The former occurs spontaneously for free neutrons (we will soon show why this is), while the latter
does not, although it does occur in the environment of the nucleus. The two processes are
1
0n
1 +
1p
−→ 11 p+ +−10 e− + ν̄e
−→ 10 n +01 e+ + νe
Here, 01 e+ is a positron, which is the antiparticle of the electron (−10 e− ), and the particles νe and ν̄e are known as
the nuetrino and antineutrino, respectively. Neutrinos are uncharged, massless particles whose existence was posited
by Wolfgang Pauli. The prediction of the existence of neutrinos was made on the basis that without them, it is not
possible to conserve energy, momentum, and angular momentum in the above decay processes. Because neutrinos have
no charge or mass (or negligible mass), their detection is extremely tricky and usually requires indirect methods. It
is 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 ordered
that?” 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 each
other. 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 diagram
in 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 process
can 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-vertex
Feynman 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 energy
change in the reaction. If ∆E < 0, energy is lowered, and the process is spontaneous. In reality, we should look
at the change in the relevent thermodynamic energy, namely, the free energy, however, in nuclear processes, the
contribution 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, and
me = 9.1093819 × 10−31 kg. Plugging these numbers in, we find that ∆E = −1.394661 × 10−30 kg, hence, the process
is 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 corresponding
energy change is
∆E = −1.394661 × 10−30 kg
2
2.9979 × 108 m/s = −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 their
isotopes. We see that for Z less than approximately 40, it is possible for isotopes of these elements to be either
neutron rich Nn /Z > 1, where Nn is the number of neutrons, or neutron poor Nn /Z < 1. However, the plot veers
upward off the diagonal (where Nn = Z) considerably for Z > 40, and all isotopes are of the neutron rich type. Many
isotopes are not stable and ultimately into other stable nuclei. When Nn /Z > 1, this decay can occur via neutron
decay, 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
A Z+
zX
An example of β − decay is that of the
14
6 C
A
−→Z+1
Y(Z+1)+ +−10 e− + ν̄e
isotope. The process for nuclei is
14 6+
6 C
7+
−→14
+−10 e− + ν̄e
7 N
Note that I have written the explicit charge on each nucleus so that charge balance can be seen explicitly. We now
wish to determine if this process occurs spontaneously, and what the associated mass change is. Note that if we just
write down the mass change in terms of the reaction above as written, we obtain the mass change for the nuclei as
7+
6+
∆m = m 14
+ m −10 e− − m 14
6 C
7 N
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 and
associated 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 we
use this trick is to look at the left side of the equation, where we see a carbon nucleus with a +6 charge. In order to
create the neutral atom, we would need to add 6 electrons to the left side. However, whatever we do on the left, we
must also do on the right, so we add +6 electrons to both sides of the nuclear decay reaction equation, to give
14 6+
6 C
7+
+ 6 −10 e− −→14
+ 6 −10 e− +−10 e− + ν̄e
7 N
Now, on the left, we have a neutral carbon atom. On the right, if we take the 6 electrons we added together with the
electron that results from the β − decay, we have a total of 7 electrons, enough to make the nitrogen nucleus into a
neutral atom. Thus, in terms of neutral atoms, we can write the reaction as
14
6 C
−→14
7 N + ν̄e
We can now obtain the mass difference from that of the neutral atoms as
14
∆m = m 14
7 N −m 6C
which will give exactly the same value as that from the nuclei. How do we see this? The manipulation we did for
the reaction equation is equivalent to taking the equation for the mass difference in terms of nuclei and adding and
subtracting the mass of 6 electrons as follows:
7+
6+
∆m = m 14
+ 6m −10 e− + m −10 e− − m 14
− 6m −10 e−
7 N
6C
the mass of the nitrogen nucleus plus the mass of the 7 electrons we are adding to it, gives us the mass of the neutral
atom, while taking the mass of the carbon nucleus, which we are subtracting, and subtracting, in addition, the mass
of 6 electrons, we end up subtracting the mass of a neutral carbon atom, which leads to the correct mass difference
in 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 by
taking this mass difference and multiplying by c2 , and this gives ∆E = −2.5072207 × 10−14 J = -0.156 MeV.
β + decay
B.
When Nn /Z < 1, the isotope is neutron poor, and decay occurs via β + decay. A β + particle is just an energetic
positron (01 e+ ). The general reaction is
A Z+
ZX
An example of this is the decay of
11
6 C,
A
−→Z−1
Y(Z−1)+ +01 e+ + νe
which decays according to
11 6+
6 C
5+ 0 +
−→11
+ 1 e + νe
5 B
The mass change for the reaction as written is
∆m = m
11 5+
5 B
+m
0 +
1e
−m
11 6+
6 C
which is correcy for nuclei, but again, we need to transform this so that it’s appropriate for neutral atoms. Again, we
need to add 6 electrons to each side of the reaction to give
11 6+
6 C
5+
+ 6 −10 e− −→11
+ 6 −10 e− +01 e+ + νe
5 B
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 we
have a neutral boron atom. Thus, in terms of neutral atoms, the reaction becomes
11
6 C
−→11
5 B+
0 −
−1 e
+01 e+ + νe
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and the mass difference becomes
5+
6+
∆m = m 11
+ 6m −10 e− + m 01 e+ − m 11
− 6m
5 B
6 C
11
0 −
0 +
11
= m 5 B + m −1 e + m 1 e − m 6 C
0 −
= m 11
− m 11
5 B + 2m −1 e
6 C
0 −
−1 e
= (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 by
which a proton can be turned into a neutron according to
1 +
1p
+−10 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 for
electron capture is
A Z+
ZX
A
+−10 e− −→Z−1
Y(Z−1)+ + νe
An example of electron capture is the decay of Uranium-231 according to
91+
+−10 e− −→231
+ νe
91 Pa
231 92+
92 U
In terms of nuclei, the mass difference is
∆m = m
91+
231
91 Pa
−m
231 92+
92 U
−m
0 −
−1 e
In order to put this in terms of neutral atoms, we add 91 electrons to both sides:
231 92+
92 U
91+
+ 91 −10 e− +−10 e− −→231
+ 91 −10 e− + νe
91 Pa
Now on the left, if we take the electron already present from the electron capture together with the 91 we added, we
have a neutral U-231 atom, and on the right, we have 91 electrons to form neutral protactium:
231
92 U
−→231
91 Pa + νe
In terms of masses, we add and subtract 91 times the mass of the electron to give
91+
92+
∆m = m 231
+ 91m −10 e− − m 231
− m −10 e− − 91m
91 Pa
92 U
231
= m 231
91 Pa − m 92 U
0 −
−1 e
= (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 that
while this mass change is comparable to values we obtained for β-decay, relative to the mass of uranium, the mass
change is small.
D.
α-decay
An α-particle is an energetic helium-4 nucleus, 42 He2+ . The general form of α-decay, which causes a change in both
A and Z, is
A Z+
ZX
(Z−2)+
+42 He2+
−→A−4
Z−2 Y
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An example of this is the decay of U-238 according to
238 92+
92 U
90+ 4
−→234
+2 He2+
90 Th
The change in mass is
∆m = m
90+
234
90 Th
+m
2+
4
2 He
−m
238 92+
92 U
In order to express this in terms of neutral atoms, we need to add 92 electrons to both sides of the reaction:
238 92+
92 U
90+
+ 92−10 e− −→234
+ 92−10 e− +42 He2+
90 Th
On the left, we now have a neutral U-238 atom, and on the right, if we divide the 92 electrons between the Thorium
and the helium, 90 for the former, 2 for the latter, we end up with neutral Thorium and Helium atoms:
238
92 U
4
−→234
90 Th +2 He
In terms of the mass change, we obtain
90+
∆m = m 234
+ 92m −10 e− + m
90 Th
4
238
= m 234
90 Th + m 2 He − m 92 U
2+
4
2 He
−m
238 92+
92 U
− 92m
0 −
−1 e
= (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 =
−4.27M eV .
E.
γ decay
Just as electrons bound into atoms have a set of discrete quantum mechanical energy levels, so do the nucleons that
are bound into atomic nuclei. When excited into an excited state, these particles decay back down to the ground state
and emit electromagnetic radiation in the process. This radiation tends to be very high frequency, typically lying in
the gamma (γ) region of the electromagnetic spectrum. When the photons emitted are specifically in the gamma part
of the spectrum, the decay process is known as γ decay. There is no change in the number of neutrons or protons, so
the nucleus does not change its elemental form. Thus, we do not consider this type of decay in the same category as
those 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, this
condition says nothing about the rate at which the reaction will occur. The study of the kinetics of a process seeks to
answer this question. The key to this answering this question is to determine the rate constant k and the corresponding
populations of different species in a reaction as a function of time. You will learn much more about this next semester
when you study chemical kinetics. In brief, rate constants tell us how many decays (or chemical events) occur, on
average, 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 probability
of occurring consistent with the rate constant for the decay process. Under this condition, the decay rate −dN/dt
will be proportional to the current population N (t). The fewer nuclei there are, the fewer decays there can be, so
that the slower the change in population with respect to time will be. This means that
−
dN
∝ N (t)
dt
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Note that the minus sign in the decay rate indicates that we are, indeed, expressing a loss of population rather than
a gain. The rate constant is the constant of proportionality, i.e.,
−
dN
= kN (t)
dt
Solving this little first-order differential equation, we obtain
N (t) = N (0)e−kt
The decay rate, itself,
−
dN
= kN (t) = kN (0)e−kt
dt
is denoted A(t), which is called the activity. Although the units of A(t) are also inverse time units, in order to
distinguish 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 the
time at which N (t1/2 ) = N (0)/2. From the decay law, we see that
N (t1/2 ) = N (0)e−kt1/2 =
1
N (0)
2
ln 2
k
Half lives are often used to characterize nuclear energy waste products and come into play significantly in the technique
of carbon dating, which is used to determine ages of archaeological artifacts.
t1/2 =
A.
Carbon dating
Carbon dating is based on the decay of one the isotopes of carbon, specifically, carbon-14. In the stratosphere, high
energy neutrons are created by cosmic rays which collide with nitrogen to produce this carbon isotope according to
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7 N
1 +
+10 n −→14
6 C +1 p
The carbon-14 isotope enters the carbon cycle of the earth continuously and has a half life, t1/2 , of 5730 years. When
a plant or animal dies, carbon exchange with the environment ceases, causing carbon-14 to decay, and we can tell the
age 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
= 1.21 × 10−4 yr−1
5730 yr
The problem is stated in terms of activities, however, the decay law is the same.
A(t) = A(0)e−kt
1
t = − ln
k
=−
A(t)
A(0)
1
ln
1.21 × 10−4
= 2217 yr
0.195
0.255
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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 tissue
of living organisms through the process of ionization. When these high energy particles collide with the molecules
in these tissues, they knock electrons out of stable orbits causing them to become essentially free electrons, leaving
behind 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 living
organisms are ripped apart, OH· radicals are left behind, and these, being highly reactive, degrade proteins and
other important molecules on which these organisms rely. The different types of radioactive decay processes we have
discussed have different biological implications.
α particles are extremely high energy and can do considerable damage. However, their high mass and relatively large
size 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 example
of a substance that decays via α decay, which is thought to have been what caused the illness and ultimate death of
Marie 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 or
roughly 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 is
essentially infinite, and their high energy gives them the potential to cause ionization over the entire range of their
trajectory.
Noting that radiation is caused by the transfer of energy from nuclear decay products to the environment, we can
define a measure of radiation dosing. 1 rad is deinfed as the amount of radiation needed to deliver 0.01 J of energy
to living tissue. However, the rad does not distinguish between different types of radiation, some of which is more
damaging than others, as our discussion above makes clear. Thus, a unit known as the rem is often used to account
for both the energy delivered and the toxicity of the radiation. 1 rem = 1 rad for β and γ radiation. However, because
α-radiation is more energetic, 1 rad of α radiation is equivalent to a dose of 10 rems. The SI unit of radiation is the
Sievert, 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 to
about 0.1 mSv. A 7-hour domestic flight exposes you to about 0.02 mSv, caused by increased exposure to cosmic
radiation at the heights planes typically fly. A chest CT scan, which is a high dose of X-rays, is around 7 mSv, which
is quite substantial. The average person is exposed to roughly 1 mSv annually. Small amounts of radiation can cause
damage to our DNA, however, when this happens, naturally occurring repair enzymes in our cells can repair the
damage and restore the DNA to a healthy state. Too much radiation exposure causing more extensive damage cannot
generally 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 living
tissue. By damaging the DNA of the cells that make up the tumor beyond the reach of the aforementioned repair
mechanisms, such therapies ensure that the tumor cells cannot replicate, and the tumor is made to shrink in size over
a series of such treatments. Unfortunately, exposure to photons of the frequency used, typically X-rays or gamma
rays, in cancer therapies, causes considerable damage to surrounding tissue as well, hence radiation therapy has very
serious side effects. This fact has prompted medical researchers to search for alternative therapies that can target
tumor 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 is
that 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 the
Bethe-Bloch equation. For a particle of charge z and mass m traveling with velocity v ≪ c through a medium of
electron number density n, the energy delivered dE over a distance dx is
dE
4πnz 2
−
≈
dx
me v 2
e2
4πǫ0
ln
2me v 2
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 Bragg
peak. Comparing the three curves, we see that protons and carbon ions have a sharp, narrow peak in their Bragg
curves. The implication of this peak is that these particles deliver the majority of their energy over a very narrow
range at a distance given by the location of the Bragg peak. It should be noted that the location of the Bragg peak
will, 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 to
photons for use in radiation therapies targeting certain types of cancers. In particular, if the distance between a
patient and a source of energetic protons is set to coincide with the location of the Bragg peak, then a typical tumor
will lie within the width of the Bragg peak, and the proton beam can be used to destroy the cells in the tumor while
doing minimal damage to the surrounding tissue. Because the location of the Bragg peak is energy dependent, it is
important that the proton beam be energy filtered so that only those protons with an energy that places the Bragg
peak 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 dissipation
curves between photons and protons. It can be seen that the photon curve is very broad and decays slowly, indicating
that photons deliver considerable energy over a very wide distance range. This is the cause of the ancillary damage
they do and the significant side effects of radiation therapy. By looking at the iron Bragg curve in Fig. 6, we see
why iron ions would not be suitable for cancer therapies: Although there is a sharp, narrow Bragg peak, there is
also a broad region in the Bragg curve before the peak at which iron ions can deliver considerable energy in much
the same way that photons do. The search for alternative cancer therapies via charged particles relies heavily on the
characteristics 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 to
develop an atomic bomb and suggesting that he encourage the develop of chain fission reactions in the US. What he
was referring to is the fact that is uranium-235 is bombarded with energetic neutrons, the nucleus can break apart in
a process called fission into lighter isotopes via reactions of the form
235
95 U
162
1
+10 n −→ 72
30 Zn +62 Sm + 20 n
235
95 U
153
1
+10 n −→ 80
38 Sr +54 Xe + 30 n
235
95 U
1
139
+10 n −→ 94
36 Kr +56 Ba + 30 n
In fact, there are close to 400 possible fission pathways for U-235 occurring with different likelihoods. Criteria such
as 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 these
neutrons 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 right
conditions, such reactions can actually be realized. There is a partial differential equation that describes how chain
reactions work. In particular, let u(r, t) be the density of neutrons at position r in space at time t. The evolution of
this density follows a diffusion equation with an extra term that serves as a control of the process:
∂
u(r, t) = a2 ∇2 u(r, t) + v(r, t)u(r, t)
∂t
Here, a is a constant, and v(r, t) is a control function determined by the conditions under which the reactions are
carried out. How this function is engineered determines whether the reactions are controlled for use in the production
of nuclear power or whether the reactions are allowed to run away in order to create the type of massive release of
energy 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% and
0.72%, respectively. Thus, in naturally occurring uranium, isolating enough U-235 for use in nuclear power requires
separating the two isotopes from each other in a process known as uranium enrichment. Techniques that can be used
include centrifugation and other diffusion techniques, sieving, thermal processing,.... While nuclear power has many
advantages such as no air pollution, and large power production at low operating costs, many disadvantages also
exist. These include the problem of disposing with the radioactive waste products (which have long half lives), and
the possibility of overheating causing meltdowns at nuclear power plants (think Chernobyl, Fukushima, or Three-Mile
Island). Disposal of nuclear waste products is one of the uses of a material that appeared on a previous problem
set: the aluminosilicates. In particular, borosilicates and boroaluminosilicates have been engineered for this purpose
because 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 nuclear
fission power.
Nuclear fusion is a process that does not suffer from these downsides and is a potential source of clean, manageable
nuclear 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 are
shown below:
2
1H
+21 H −→ 32 He +10 n
2
1H
+21 H −→ 42 He + γ
2
1H
+21 H −→ 31 H +11 p+
The problem with fusion reactions is the tremendous Coulomb barrier that must be surmounted to get two light nuclei
to approach each other. Generally, such reactions can be produced by subjecting a plasma to a very high magnetic
field (recall that charged particles in magnetic fields execute circular motion, and if the fields are large enough, high
velocities can be achieved). Currently, however, the energy needed to create fusion reactions is larger than the energy
that 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 accumulate
large 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:
4
2 He
+42 He ⇀
↽
8
4 Be
+42 He −→ 12
6 C
8
4 Be
(t1/2 = 7 × 10−17 s)
The first reaction is shown as an equilibrium because of the short half-life of the Be isotope, which can immediately
revert back to helium nuclei. However, occasionally, the Be-8 nucleus can fuse with another helium-4 nucleus to form
carbon-12, one of the key atoms in the molecules of life. The above fusion processes occur when the star starts to
contract under the action of its own gravity, thereby allowing nuclei to overcome the fusion barrier, occurring at
temperatures at 108 K.
In fact, other heavy element synthesis occurs via additional fusion reaction in a process known as the CNO (carbonnitrogen-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.