Download CHAPTER 2 The nucleus and radioactive decay - Cin

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Hydrogen atom wikipedia , lookup

Standard Model wikipedia , lookup

Antimatter wikipedia , lookup

Nuclear fusion wikipedia , lookup

Conservation of energy wikipedia , lookup

Mass versus weight wikipedia , lookup

Lepton wikipedia , lookup

Mass wikipedia , lookup

Electromagnetic mass wikipedia , lookup

Anti-gravity wikipedia , lookup

Elementary particle wikipedia , lookup

History of subatomic physics wikipedia , lookup

Nuclear structure wikipedia , lookup

Nuclear transmutation wikipedia , lookup

Negative mass wikipedia , lookup

Chien-Shiung Wu wikipedia , lookup

Isotope wikipedia , lookup

Nuclear binding energy wikipedia , lookup

Beta decay wikipedia , lookup

Nuclear drip line wikipedia , lookup

P-nuclei wikipedia , lookup

Atomic theory wikipedia , lookup

Atomic nucleus wikipedia , lookup

Valley of stability wikipedia , lookup

Nuclear physics wikipedia , lookup

Transcript
1
CHEMISTRY OF THE EARTH
CHAPTER 2
The nucleus and radioactive decay
2.1 The atom and its nucleus
An atom is characterized by the total positive charge in its nucleus and the atom’s
mass. The positive charge in the nucleus is Ze, where Z is the total number of protons in
the nucleus and e is the charge of one proton. The number of protons in an atom, Z, is
known as the atomic number and dictates which element an atom represents. The
nucleus is also made up of N number of neutrally charged particles of similar mass as the
protons. These are called neutrons. The combined number of protons and neutrons,
Z+N, is called the atomic mass number A. A specific nuclear species, or nuclide, is
denoted by
A
2.1
ZΓ
where Γ represents the element’s symbol. The subscript Z is often dropped because it is
redundant if the element’s symbol is also used. We will soon learn that a mole of protons
and a mole of neutrons each have a mass of approximately 1 g, and therefore, the mass of
a mole of Z+N should be very close to an integer1. However, if we look at a periodic
table, we will notice that an element’s atomic weight, which is the mass of one mole of
its atoms, is rarely close to an integer. For example, Iridium’s atomic weight is 192.22
g/mole. The reason for this discrepancy is that an element’s neutron number N can vary.
Nuclides with different neutron numbers but the same atomic number Z are called
193
isotopes. Thus, Iridium has two isotopes, 191
77 Ir and 77 Ir , which are characterized
respectively by 114 neutrons and 116 neutrons; a mole of the former weighs 190.960580
g and the latter weighs 192.962917 g. In naturally occurring Iridium, 37.3% of the atoms
are 191Ir and the remaining atoms are 193Ir. The atomic weights reported in the periodic
table represent the average masses of each element.
2.2 The mass of an atom
How is the mass of an atom determined? One way to determine the mass of an atom
is to use a mass spectrometer. Figure 1 shows schematically how one type of mass
spectrometer works. The mass spectrometer begins with an ion source, which produces a
beam of ionized atoms or molecules (in Chapter 4, we will discuss the different types of
ion sources). This beam of ions passes through a velocity sector, through which only
ions with a particular velocity are sampled. The velocity sector makes use of an electric
(E) and a magnetic field (B). The E and B fields are oriented such that the electric and
magnetic forces on the ions are equal and opposite, that is,
qE = -qv×B
2.2
where q is the charge of the ion. Thus, only ions having a velocity equal to E/B are
permitted to continue on their flight paths as ions with any other velocity would be
deflected to the walls of the mass spectrometer. Once ions have been selected for their
velocities, they enter a magnetic sector, where the magnetic field is uniform and
perpendicular to the path of the ions. The magnetic field exerts a force qv×B acting
1
The mass of a proton is approximately 2000 times that of an electron, so electrons contribute only a small
part to the mass of an atom.
Cin-Ty Lee
2
CHEMISTRY OF THE EARTH
s
tor
ec
l
l
co
E
B
Ion source
Velocity Sector
B
Magnetic Sector
Modified from Krane 1988
Figure 1. One type of mass spectrometer used for separating isotopes.
perpendicular to the ion flight path. This force is balanced by the centripetal force mv2/r,
that is,
mv2/r = qv×B
2.3
where r is the radius of curvature of the ion flight path. Upon rearranging, Eq. 2.3 can be
expressed in terms of r
r = mv/qB
2.4
It can be seen from Eq. 2.4, that a magnetic sector separates ions based on their
momentum mv. However, as q, B, and v are uniquely determined, each different mass
appears at a different radius r. If we know the magnitude of the electric and magnetic
fields precisely, then obtaining the mass of an atom is trivial.
To be useful, the precision of our mass determinations must be on the order of one
part in 106. Unfortunately, it is not possible to know the magnitude of the electric and
magnetic fields to this precision. We can, however, measure relative mass differences
very precisely by keeping the electric and magnetic fields constant and measuring the
difference in radii of curvature, which we can do more precisely. By convention, the 12C
atom is taken to be exactly 12.000000 units on the atomic mass scale, against which all
other masses are calibrated. The masses of other isotopes are determined using the
“doublet” approach, which is best explained through an example. If we set up the mass
spectrometer to collect the molecular ions, C9H20+ (nonane) and C10H8+ (naphthalene), the
difference ∆ in mass units on this scale is 0.09290032±0.00000012 units2
∆ = m(C9 H 20 ) − m(C10 H 8 ) = 12m(1H ) − m(12C )
2.5
where m(C9H20) is the mass of C9H20 and so forth. It follows that the mass of Hydrogen,
1
H, is given by
2
We neglect molecular binding energies, which are negligible for this treatment.
Cin-Ty Lee
3
CHEMISTRY OF THE EARTH
m(1H ) =
1
m(12 C ) + ∆
12
[
= 1.00000000 +
]
1
∆
12
2.6
= 1.00782503 u
Using this approach, we can work through the periodic table and obtain the mass of each
isotope of interest. We can also determine the masses of isotopes and elementary
particles by considering the energies of particles in nuclear reactions by assuming energy
and mass are interchangeable (E=mc2, where E is energy, m is mass, and c is the speed of
light). Regardless of the approach, all masses are calibrated against 12C. On such a scale
the proton is 1.00727647 u, the electron is 0.0005485803 u, and the neutron is
1.00866501 u. One mass unit is defined to be 1.660566×10-24 g, such that one mole or
6.022045×1023 atoms (Avogadro’s number) of 12C weighs exactly 12.000000 g. Nuclide
masses are often tabulated in terms of mass defects (∆=m - A), which denotes the
difference between the actual mass m and the atomic mass number A (Appendix Table 1).
Once nuclide weights are known, the atomic weight of an element can be
determined by measuring an element’s relative abundances. The atomic weight AW of an
element is simply the weighted average of each nuclide
AW = ∑ X i mi
2.7
where Xi is the atomic proportion of nuclide i and its mass mi. For most naturally
occurring elements, the relative nuclide abundances are constant and their numbers have
now been tabulated for convenience. As we will learn later, there are two situations
when isotopic abundances in nature vary. For example, isotopic abundances can be
fractionated by certain physical processes in nature or in the lab. Another situation
occurs when a nuclide is radioactive or is the product of the decay of a radioactive
nuclide. Both of these processes yield a variable atomic weight. For example, Pb has 4
naturally occurring isotopes 204Pb, 206Pb, 207Pb, and 208Pb. The latter three are decay
products of 238U, 235U, and 232Th, and therefore the relative abundances of the Pb isotopes
depends on how much radiogenic Pb has accumulated from U and Th. The nonconstancy of the atomic weight of lead was recognized very early on and was one line of
evidence suggesting that elements are made up of different isotopes.
2.3 Binding energies and mass defects
The astute reader might wonder now that if we know the masses of each elementary
particle (proton, neutron, and electron), then it seems trivial to calculate the masses of all
the nuclides as long as we know how many protons, neutrons and electrons are in a
particular nuclide. In fact, the mass of a nuclide determined by the “doublet method”
discussed above is slightly smaller than the combined masses of the constituent
elementary particles. We can use the case of 191Ir as an example. Iridium-191 is made up
of 77 protons, 77 electrons and 114 neutrons. Therefore, the combined mass of the
elementary particles is given by
Cin-Ty Lee
CHEMISTRY OF THE EARTH
M = Zm p + Zme + Nmn
= 77(1.00727647) + 77(0.0005485803) + 114(1.00866501)
2.8
= 192.590340 u
(where mp, me, and mn refer to the masses of a proton, electron and neutron). The true
mass of 191Ir (e.g., m( A Γ) ) is actually 190.960584 u, which is 1.62976 u lower than the
combined mass of the elementary particles. The reason for this decrease in mass is that
the fusion of these elementary particles into the nucleus is energetically favorable, that is,
energy is released when protons and neutrons come together. As energy is equivalent to
mass according to the well-known equation
E = mc2
2.9
where E is the energy, m is the mass, and c is the speed of light, the decrease in energy is
equivalent to a decrease in mass. The unit conversion between mass and energy is
c2=931.50 MeV/u.
The energy released by combining N neutrons and Z protons to make a nucleus ZA Γ
is called the binding energy (Appendix Table 1):
2.10
E B = Zm p + Nm n − (m ( A Γ ) − Zm e ) c 2
The binding energy for electrons is small compared to the nuclear binding energies,
which is why in Eq. 2.10, we subtract out the mass energy of the electrons. Eq. 2.10 can
be simplified by recognizing that the most elementary nucleus is the Hydrogen nucleus,
1
H:
[
Average Binding Energy Per Nucleon (MeV)
4
]
9
Fe
8
C
7
O Ne
Mg Si
B
Be
6
Li
5
4
3
2
He
0
Fusion
4 8 12 16 20 24
Fission
30
60
90
120
150
180
210
240
Mass Number A
Figure 2. Binding energy per nucleon.
Cin-Ty Lee
5
CHEMISTRY OF THE EARTH
[
]
E B = Z (1H ) + Nmn − m( A Γ) c 2
2.11
191
For Ir, we find that its binding energy is 1478.77 MeV.
Figure 2 shows the binding energy per nucleon (for 191Ir this is 1478.77 MeV/191
= 7.7423 MeV) as a function of mass number, e.g. EB/A. Except for mass numbers below
~20 u, it can be seen that EB/A is remarkably constant, lying between 7 and 10 MeV per
nucleon. This feature indicates that the binding energy increases approximately linearly
with mass number. Another feature of the EB/A curve is that there is maximum at 56Fe.
This means that fusion of masses lower than mass number 56 or fission of masses greater
than 56 will release energy. It is this property that we take advantage of when generating
nuclear energy by fusion or fission. We will see in Section 2.4 that an understanding of
binding energy is fundamental to understanding why certain elements undergo
radioactive decay. A nuclide will energetically tend towards decay by a particular mode
(α emission, β emission, or spontaneous fission) if its atomic mass is greater than the sum
of the masses of the products formed by that decay mode. This means that nuclei above
A>100 are unstable towards spontaneous fission into two nuclei of approximately the
same mass and that nuclei above A>140 will tend to decay by α emission.
A physical explanation for the binding energy
Figure 3 shows a plot of Z versus N for all stable nuclei. One of the most
important features is that at Z<20, the number of protons and neutrons is approximately
equal. For nuclides having Z>20, there are more neutrons than protons as shown by the
deviation of the mass curve towards the neutron-rich side of the Z=N line. The stable
nuclide array represents the most energetically favorable nuclides, that is, the
combination of neutrons and protons that yields the highest binding energy. This array is
often called the “valley of stability”. The excess of neutrons at high mass numbers can be
qualitatively explained by the fact that the more protons one packs together in the nucleus,
the higher the Coulombic repulsion (which lowers the binding energy) and therefore the
more neutrons needed to physically increase the separation distance of the protons so that
their Coulombic repulsion is decreased.
In more detail, physicists have come up with a semi-empirical equation to predict
the binding energy of a nuclide. The binding energy can be expressed in terms of A and
Z as follows
( A − 2Z ) 2
E B ( Z , A) = av A − a s A 2 / 3 − ac Z ( Z − 1) A −1 / 3 − a sym
+δ
2.12
A
We now discuss each term in sequence. The first term is an empirical relationship that
describes the fact that the binding energy is approximately a linear function of the atomic
mass number A. That EB ~ avA indicates that each nucleon interacts with only its nearest
neighbors and that each nucleon contributes roughly the same amount to the binding
energy. The constant av must be on the order of 8 MeV per nucleon. The second term
accounts for the fact that nucleons on the surface of the nucleus interact less with other
nucleons. As the surface area is proportional to R2 and R ∝ A1/3, the surface term is
proportional to A2/3 (this assumes that the nucleons are rather incompressible as is implied
by the fact that EB ~ avA). The third term accounts for the Coulombic repulsion of the
protons, which decreases the binding energy. The Coulombic repulsion is directly
proportional to Z(Z-1) and inversely proportional to the separation distance R, or in other
words, proportional to A-1/3. The A-1/3 proportionality allows for the Coulombic repulsion
2.3.1
Cin-Ty Lee
6
CHEMISTRY OF THE EARTH
N
Z=
Z
Lawrence Berkeley National Laboratory
N
Figure 3. Chart of Nuclides in a plot of Z versus N.
energy to decrease as more neutrons are added and has its greatest effect for high mass
numbers. The fourth term accounts for the fact that at low masses (Z < ~20), there is a
tendency for the number of protons and neutrons to be symmetric, e.g. A = 2Z. The (A2Z)2 proportionality is to account for the fact that as A becomes greater than 2Z, the
binding energy is reduced. The inverse proportionality with respect to A is to account for
the decreasing importance of symmetry and the increasing importance of the Coulombic
term as the mass number increases. Finally, the last term accounts for the fact that
nuclides with even numbers of Z or N are energetically more stable as is exemplified by
the observation that there are 167 stable nuclei in nature with even N and even Z, but only
four stable nuclei in nature with both odd N and odd Z (2H, 6Li, 10B, 14N). This term thus
accounts for the pairing force, which accounts for the tendency of like nucleons to pair up.
For even Z and N, the pairing energy is usually expressed as +apA-3/4, for odd Z and N the
term is -apA-3/4, and for odd A the pairing term is zero. The constants av, as, ac, asym, and
ap must be chosen to best fit the data in Figure X. One choice of constants is av=15.5
MeV, as = 16.8 MeV, ac = 0.72 MeV, asym = 23 MeV, and ap = 34 MeV.
Using Eq. 2.12 for EB, we can rearrange Eq. 2.12 to yield the semi-empirical
mass formula
m( ZA Γ) = Zm(1H ) + Nmn − E B ( Z , A) / c 2
Eq. 2.13
For constant A, Eq. 2.13 represents a parabola of m versus Z. On Figure X, nuclides of
constant A are said to lie on an isobar, which represent lines having a slope of negative
one. A cross-section at constant A through the nuclide mass chart will therefore look like
Cin-Ty Lee
7
CHEMISTRY OF THE EARTH
a valley, hence the trend of naturally occurring nuclei is called the “valley of stability”.
For constant A, we can calculate the minimum mass or the most energetically favorable
nuclide configuration by letting (∂m( ZA Γ) / ∂Z ) A = 0 :
mn − m(1H ) + a c A −1 / 3 + 4a sym
Z min =
2.14
2a c A −1 / 3 + 8a sym A −1
Since ac = 0.72 MeV and asym = 23 MeV, the first two terms in the numerator are
negligible and thus,
A
1
Z min ≈
2.15
2/3
2 1 + A a c /(4a sym )
It can be seen from Eq. 2.15 that for small A, Zmin is approximately equal to A/2. For
heavy nuclides, Z/A ~ 0.4 as expected from observations of the heavy stable nuclei.
[
]
2.4 Radioactive decay
If we look at the chart of nuclides, which is a plot of naturally occurring nuclides by
atomic number (Z) versus neutron number (N), we see the resulting array defines the
region of greatest stability, often referred to as the “valley of stability”. For nuclides of
low atomic mass, greatest stability is achieved when Z ~ N. However, as atomic mass
increases, the stable Z/N ratio increases to ~1.5, that is, a larger number of neutrons are
needed in order to maintain stability. This can be explained by the semi-empirical mass
formula (Eq. 2.12). In words, we can say that as the number of protons increases in the
nucleus, extra neutrons (e.g., N>Z) are needed to offset the increase in Coulombic
A
Z+1
N-1
β−
A
Z
N
Z
α
y
ce a
d
β+
E.C.
A
Z-1
N+1
A-4
Z-2
N-2
N
Figure 4. Z versus N diagram showing how Z, A and N change during
different decay schemes.
Cin-Ty Lee
8
CHEMISTRY OF THE EARTH
repulsion caused by the larger number of protons. It follows that nuclides, which lie
away from the “valley of stability”, will be unstable, and have a tendency to decay
towards the “valley of stability”. In other words, a nuclide will energetically tend
towards decay by a particular mode (e.g., α emission, β emission, or spontaneous fission;
Fig. 4) if its atomic mass is greater than the sum of the masses of the products formed by
that decay mode. It turns out that nuclei above A>100 are unstable towards spontaneous
fission into two nuclei of approximately the same mass and that nuclei above A>140 will
tend to decay by α emission. Below, we describe various types of radioactive decays.
Alpha Decay
At masses above ~A>140, many nuclides may decay by alpha emission because
of the Coulombic repulsion between the increased number of protons being forced into
the nucleus. Alpha particles are made up of 2 protons and 2 neutrons (Z=2, N=2) and
therefore have the nuclear make-up of the 4He. Alpha decay can be expressed as follows:
A
A− 4
4
Eq. 2.16
Z Γ → Z −2 Γ + 2 He + Q
where Q is the energy released by the decay process. The question arises as to why a
radionuclide prefers to decay by alpha emission rather than by emission of a larger
particle. Assuming the parent radionuclide ZA Γ is at rest, conservation of energy requires
that the energy released by alpha decay is
Q = (mA − m A−4 − mα )c 2
Eq. 2.17
It turns out that the alpha particle is one of the most tightly bound nuclides, having a low
mass defect (or low binding energy). It can be seen from Eq. 2.17 that the smaller the
mass defect of the emitted particles, the greater the release of energy Q and the more
favorable the decay reaction is. The alpha particle, 4He, has a mass defect of +0.002603,
which is smaller than the mass defects of other low mass nuclides, such as 1H, 2H, 3H and
3
He, which means that emission of an alpha particle is an energetically more favorable
process than emission by any of these other nuclides. We leave you to verify this in the
Problem Set.
In general, the greater Q of alpha decay is for a given nuclide, the more likely it is
to decay by alpha decay. Thus, it turns out that if we look at the half-lives of alpha decay
for various radionuclides, the half-life decreases with increasing Q. There also appears to
be a difference between atomic masses with even numbers of neutrons and protons from
atomic masses with odd numbers of neutrons and/or protons. For a given Z and Q, the
alpha decay half-lives of the latter are much longer than those of even-even nuclides.
Geologically relevant alpha decay systems include the decay chain of 238U to
206
235
Pb, U to 207Pb, and 147Sm to 143Nd.
2.4.1
Alpha recoil energies
We now consider the distribution of energy caused by alpha decay. When a
radionuclide decays to generate an alpha particle and a product nuclide, energy Q is
released. This energy is taken up as kinetic energy due to the recoil of the product
particles. Assuming that the radionuclide was at rest, the following relationship must
hold
1
1
2
2
Q = mP vP + mα vα
2.18
2
2
Cin-Ty Lee
9
CHEMISTRY OF THE EARTH
where the two terms on the right hand side represent the kinetic energies of the product
nuclide and the alpha particle (where v represents velocity). Because linear momentum
must also be conserved, the following relationship also holds
mP vP = mα vα
2.19
Combining Eqs. 2.19 and 2.18, we arrive at
1
m 
2
Q = mα vα 1 + α 
2.20
2
 mP 
which shows that the velocity of the alpha particle is uniquely determined by the release
energy Q, its own mass and the mass of the product nuclide. Because the mass of the
alpha particle is very small, Eq. 2.20 also shows that most of the decay energy is taken up
by the alpha particle. Typically, the alpha particle accounts for ~98% of the decay energy.
The recoil distance of an alpha particle is therefore much larger than the product nuclide.
In many instances, an alpha particle has enough energy to rip apart the lattices of its host
material, leaving a short path of destruction, called an “alpha recoil track”. Recoil tracks
can be seen in geologic materials that have high concentrations of radionuclides, such as
238
U and 235U, and that have had sufficient time to accumulate noticeable amounts of
alpha decay products. Very old zircons are often so full of alpha recoil tracks that they
serve as transport pathways for the diffusive loss of U or its decay products.
2.4.2
Beta decays and electron capture
In addition to alpha decay, nuclides can undergo either negative beta decay (β−),
positive beta decay (β+), or orbital electron capture (E.C.). These decay processes are as
follows:
Negative beta decay
n → p + e−
Positive beta decay
p → n + e+
Electron capture
p + e− → n
In words, negative beta decay results in the conversion of a neutron to a proton and an
electron (the negative beta particle), positive beta decay results in the conversion of a
proton to a neutron and a positron (positive beta particle), and electron capture results
from the capture of an orbital electron by a proton in the nucleus to form a neutron. One
of the dilemmas facing nuclear physicists during the early stages of beta decay research
was that the measured kinetic energy of the beta particle and the daughter nuclide did not
add up to the energy that should be released by the decay. The beta particles were in fact
found to have a spectrum of kinetic energies, ranging from nearly zero to a maximum
energy that was equal to the energy difference between the initial and final states. This
“missing” energy appears to violate conservation of energy.
Physicists have
hypothesized that this “missing” energy is taken up by a second particle, which has a
neutral charge. For negative beta decay, this second particle is called the antineutrino ν .
For positive beta decay, the second particle is called the neutrino ν . Therefore, the
complete decay process for negative beta decay is
A
A
−
2.21
Z Γ → Z +1 Γ + β + ν + Qβ −
where Qβ − is the mass difference between the initial and final states
Qβ − = [m( ZA Γ) − m( Z +A1 Γ)]c 2
2.22
Cin-Ty Lee
10
CHEMISTRY OF THE EARTH
Qβ − represents the kinetic energies shared by the negative beta particle and the neutrino.
Negative beta decay results in an increase in proton number at the expense of a neutron.
Some geologically relevant negative beta decay systems include the decays of 87Rb to
87
Sr, 187Re to 187Os, and 176Lu to 176Hf.
The decay equation for positron emission is as follows
A
A
+
2.23
Z Γ→ Z −1 Γ + β + ν + Qβ +
In this case, the proton number decreases by one and the neutron number increases by
one. An example of positron emission is the decay of 26Al to 26Mg, the former a very
short-lived radionuclide found only as a cosmogenic nuclide or as a nucleosynthetic
product of supernovas.
In the electron capture decay process, an electron reacts with a proton in the
nucleus to generate a neutron. Thus, the neutron number increases at the expense of a
proton, and for this reason, electron capture decay looks similar to positive beta decay
A
A
2.24
Z Γ → Z −1 Γ
The electron that is captured into the nucleus usually comes from the lowest energy levels
because such electrons are closest to the nucleus and hence have the highest probability
of being captured. Once such an electron is captured, an electron vacancy appears which
is then filled by an electron from a higher energy level. The transferring of a higher
energy electron to a lower energy level results in the release of characteristic X-rays. The
decay of 40K to 40Ar is an example of an electron capture decay relevant to geologists.
In all of the above decays, the nucleus can be left in an excited state. This excited
nuclide can then decay to a lower energy level. This process results in the release of
characteristic gamma rays (or photons).
2.4.3
Spontaneous Fission
Recall from our discussion of the binding energy (Fig. 2) that 56Fe has the highest
binding energy. Nuclides with mass numbers above 56 have lower binding energies and
therefore it is energetically favorable for very high mass particles to separate into two
nuclides with subequal masses. The release in energy caused by fission decay has been
harnessed in nuclear reactors. Some nuclides, such as 235U can be induced to fission if
they are bombarded by neutrons. The fission products are not exactly equal in mass and
are not always the same. There appears to be a spectrum of fission product masses. A
possible neutron-induced fission reaction is
235
U + n = 93Rb + 141Cs + 2n
where the product nuclides invariably fall on the neutron-rich side of the valley of
stability and therefore undergo negative beta decay.
Cin-Ty Lee
11
CHEMISTRY OF THE EARTH
Useful units
Speed of light
Charge of electron
Avogadro’s Number
c
e
NA
2.99792458 x 108 m/s
1.602189 x 10-19 C
6.022045 x 1023 mole-1
Mass units (u)
5.485803 x 10-4
1.00727647
1.00866501
4.00150618
MeV/c2
0.511003
938.28
939.573
3727.409
Particle rest masses
Electron
Proton
Neutron
Alpha
Conversion Factors
1 eV = 1.602189 x 10-19 J
1 u = 931.502 MeV/c2 = 1.660566 x 10-27 kg
1 barn = 10-28 m2 = 10-24 cm2
1 Ci = 3.7 x 1010 decays/s
References and useful websites
Faure, G., 1986, Principles of Isotope Geology, John Wiley and Sons, NY, 589 pp. Classic text on the
application of isotope chemistry to geology.
Krane, K. S., 1988, Introductory Nuclear Physics, John Wiley and Sons, NY, 845 pp. Comprehensive
treatment of nuclear chemistry and physics at the moderately advanced level.
Table of nuclides (http://www2.bnl.gov/ton/) – Lawrence Livermore National Laboratory website.
Table of isotopes (http://ie.lbl.gov/toi.html) – Lawrence Livermore National Laboratory website.
Web of Elements (http://www.webelements.com/) – comprehensive source of elemental properties,
electron configurations, ionization potentials, naturally occurring isotopes, binding energies, etc.
Cin-Ty Lee