CH 30 – Nuclear Physics
In the early part of the 1900’s Ernest Rutherford demonstrated that the atom consisted of
a small, positive nucleus that contained most of the mass surrounded by electrons. In the
previous chapter we discussed the configurations and energies of the electrons. In this
chapter we discuss the makeup of the nucleus.
The nucleus of an atom consists of protons and neutrons. The proton has a charge e =
1.602 x 10-19 C and the neutron has no net charge. The nucleus is characterized by
atomic number (Z), neutron number (N), and mass number (A = Z + N):



Atomic number Z: number of protons in nucleus
Neutron number N: number of neutrons in nucleus
Mass number A: number of nucleons (neutrons and protons) in nucleus
Symbolically, we describe a nucleus as ZA X , where X is the element (e.g., 11H , 24He , 37Li ,
etc.)
The Z number determines the element. Elements with different N and A numbers are
12 13 14
isotopes. For example, the isotopes of carbon are 11
6 C, 6C, 6C, 6C .
The mass number, A, which is given for the elements in the periodic table is generally not
an integer. The reason is that it represents an average of the naturally occurring isotopes
and it includes the mass equivalent of the binding energy of the nucleons.
Atomic mass unit (u)
The atomic mass unit (u) is defined as 1/12th the mass of 126 C . Since 12 grams of 126 C is
a mole, or Avogadro’s number of atoms, the atomic mass unit is
1u 
1  0.012kg / mole 

  1.6606 x10  27 kg
23
12  6.0221x10 / mole 
The rest mass energy of 1u is
E  mc 2  ( 1.6606 x10 27 kg )( 2.998 x108 m / s )2  1.492 x10 10 J
 ( 1.492 x10 10 J ) /( 1.602 x10 19 J / ev )  931.5Mev
From the equation E = mc2, we can write
1 u  931.5Mev / c 2
1
The masses of the proton and neutron are slightly larger than 1 u. This is because of the
binding energy of the nucleus, as will be discuss later. The proton, neutron and electron
masses are
Proton:
m = 1.6726 x 10-27 kg = 1.007276 u = 938.28 Mev/c2
Neutron:
m = 1.6750 x 10-27 kg = 1.008665 u = 938.57 Mev/c2
Electron:
m = 9.109 x 10-31 kg = 5.486 x 10-4 u = 0.511 Mev/c2
Size of Nucleus
The size of the nuclei can be determined by experiments in which charged particles such
as alpha particles (helium nuclei) are scattered from a material. The angular distribution
of the scattering depends on the distance of closest approach of the alpha particles to the
nuclei. The radii of the nuclei has been found to follow the relationship
r  r0 A1 / 3 ,
where r0 = 1.2 x 10-15 m and A is the atomic number. This equation suggests that the
densities of the nuclei of the different elements are all approximately the same.
The density of the nuclei can be estimated as follows:
V  4  r 3  4  r0 3 A ( nuclear volume )
3
3
m
A
( nuclear mass )
NA

A/ NA
m
1
1



3
3
26
4
4
4
V
( 6.02 x10 / kg )  ( 1.2 x10 15 m )3
 r0 A N A  r0
3
17
3
3
 2.3 x10 kg / m 3
Nuclear Stability
For small atomic numbers, the number of protons and neutrons in the nucleus is about the
35
same. For example, 24 He 126C , 17
Cl . But for Z greater than about 20, the number of
neutrons exceeds the number of protons (e.g., 198
79 Au ). The figure below is a plot of the
number of neutrons (N) versus the number of protons (Z). The solid points represent
stable nuclei while the shaded area represents radioactive nuclei. The stability of the
nucleus depends on competition between electrostatic repulsion between the protons and
the attractive nuclear force between nucleons. Without the nuclear force the nucleus
2
would fly apart due to the electrostatic repulsion. The nuclear force is a short-range force
that exists mainly between nearest-neighbor nucleons; whereas, the electrostatic force is
comparatively long range. If there were no neutrons in the nucleus, then the repulsive
electrostatic force would be too great for stability. In a sense, neutrons help to spread the
protons apart, reducing their repulsion. If there were no neutrons, then the electrostatic
force on the outer protons would increase in proportion to the nuclear radius. If Q is the
total nuclear charge and  the nuclear charge density, then the force on an outer proton
(e) would be
F k
Qe
r
2
k
 4  r 3e
3
r2
r
So, as the number of protons increases, there must be an increasing excess of neutrons to
reduce the nuclear charge density.
3
Binding Energy of Nuclei
The mass of a nucleus is less than the sum of the masses of the individual protons and
neutrons because of the binding energy of the nucleus. An amount of energy equal to the
binding energy must be supplied to the nucleus to break it apart into individual protons
and neutrons. The mass difference is related to the binding energy by the equation
Eb  m c 2
Example:
Calculate the binding energy of deuterium.
Solution:
The deuterium nucleus consists of a proton and a neutron. So we find the difference
between the sum of the masses of the protons and neutron and the mass of the deuterium
nucleus (deuteron),
m  ( m p  mn )  md
If we use the atomic mass for md, then this includes the mass of the nucleus and the mass
of the electron. If we also use the atomic mass of hydrogen in place of the mass of the
proton, then the mass of the electron will cancel out in the calculation. Then
m  ( 1.007825 u  1.008665 u )  2.014102 u  0.002388 u
 ( 0.002388 u )( 931.49 Mev / c 2 / u )  2.224 Mev / c 2
Eb  m c 2  2.224 Mev
Per nucleon, the binding energy is
Eb / A  2.224 Mev / 2  1.11 Mev
Example:
Calculate the binding energy of 126 C .
4
Solution:
m  ( 6m p  6mn )  mC 12
 6( 1.007825 u  1.008665 u )  12 u  0.09894 u
 ( 0.09894u )( 931.49 Mev / c 2 / u )  92.16 Mev / c 2
Eb  m c 2  92.16 Mev
Per nucleon,
Eb / A  92.16 Mev / 12  7.68 Mev
The binding energy per nucleon generally increases rapidly with increasing A up to about
A ~ 60 and then more slowly decreases with further increase in A. This means that the
nuclei with intermediate masses are the most stable compared with the light and heavy
nuclei. The binding energy per nucleon is shown below as a function of A. This figure
suggests that we can convert mass to energy by combining lighter nuclei to make nuclei
of intermediate size (fusion) or breaking apart heavy nuclei into nuclei of intermediate
size (fission).
5
Radioactive Decay
An unstable nucleus can transform to a more stable nucleus by emitting an alpha particle
or a beta particle (electron or positron). A nucleus in an excited state can also emit
gamma rays. The alpha particle is the nucleus of the helium-4 atom, 24 He . The positron
is the antiparticle of the electron. Its mass is the same as that of the electron but it has a
positive charge +e. A gamma ray is a high energy photon and has no mass or charge.
Alpha Decay
Alpha decay can be expresses symbolically as
A
A 4
4
Z X  Z  2Y  2 He
Note that both the total mass number A and the atomic number Z are the same before and
after the decay. An example of alpha-decay is the decay of radium-226 into radon-222:
226
222
4
88 Ra  86 Rn  2 He
Example:
Calculate the energy released in the alpha decay of radium.
Solution:
m  mRa  ( m Rn  mHe )
 226.025402 u  ( 222.017571u  4.002602 u )  0.005229 u
E  ( 0.005229 u )( 931.494 Mev / u )  4.97 Mev
Since the alpha particle is much lighter than the radon nucleus, most of the released
energy appears as the kinetic energy of the alpha particle.
Beta Decay
Beta decay can be expressed as
A
A
Z X  Z 1Y
 e
(negative beta decay)
A
A
Z X  Z 1Y
 e
(positive beta decay)
or
6
Note that the mass number doesn’t change, but the nuclear charge either increases (for
electron emission) or decreases (for positron emission).
The emission of an electron is equivalent to the decay inside the nucleus of a neutron into
a proton and an electron –
1
1

0 n1 p  e
Similarly, the emission of a positron is equivalent to the conversion of a proton in the
nucleus into a neutron and a positron.
A process related to beta decay is K-capture. In this process a nucleus ‘captures’ an
orbital electron in the K-shell which combines with a proton to form a neutron. This
process is expressed as
A
ZX
 e   Z A1Y
(K-capture)
Example:
Calculate the energy released in the beta decay of C-14 into N-14.
Solution:
The process is
14
14

6 C 7 N  e
We can neglect the mass of the electron on the right side of the expression if we use the
atomic masses, since atomic N has one more electron than atomic C. Then
m  mC 14  m N 14
 14.003242 u  14.003074 u  0.000168 u
E  ( 0.000168 u )( 931.494 Mev / u )  0.156 Mev
If the electron were the only particle emitted in this decay process, then essentially all of
this energy would be imparted to the electron since its mass is small compared to the
mass of the nitrogen nucleus. However, it is observed that most of the time the electron’s
kinetic energy is significantly less than this amount. The observed distribution of kinetic
energies is as shown in the figure on the next page.
This discrepancy led to the prediction by Pauli that a third particle must be involved to
carry away the unaccounted for energy. In addition, the third particle was required to
have a spin angular quantum number of ½ to account for conservation of angular
7
momentum. This particle was expected to have no charge and a very small mass and was
called the ‘neutrino’. It was eventually discovered in 1956. The complete decay process
would be given as
14
14

6 C 7 N  e 
 is an anti-neutrino. In + beta decay
processes, a neutrino ( ) is emitted. It has
recently been shown that there are several types
of neutrinos and that they have a finite, but
small, mass. Because of their small mass they
travel very near the speed of light. Neutrinos
are extremely abundant (the sun produces a
large flux of neutrinos); however, they are
extremely difficult to detect.
Gamma Decay
Quite often in alpha or beta decay or in nuclear reactions a nucleus is left in an excited
stated, somewhat like an excited electronic state in an atom. The nucleus then decays to
its ground state by emitting one or more high energy photons (gamma rays). This process
is described as
A *
A
Z X Z X

Below is an example of gamma decay:
60
60 *

27 Co 28 Ni  e  
60 *
60
28 Ni  28 Ni  
Penetrating power of radiation
The penetrating power of radiation depends on the type and energy. Gamma rays are
generally the most penetrating since they have no charge to interact with the electrons
and nuclear charges in matter. Several centimeters of lead would be required to provide
safe shielding from a high energy gamma emitter source. Alpha particles are the least
penetrating both because of their charge and their mass. They easily lose energy in air
and can be shielded by thick paper. Beta particles have intermediate penetrating capacity.
8
Radioactive Decay Rates
The number of nuclei N that will decay in a time interval t is proportional to the
number of radioactive nuclei N and the time interval,
N  Nt
By using calculus it can be shown that this leads to an exponential dependence of N and
the decay rate N/t on time –
N  N 0 e  t
R
N
 N 0 e   t  N
t
 is the decay constant and has units of inverse time. The unit for decay rate, R, is the
curie.
1 curie  1Ci  3.7 x1010 decay / s
The half-life T1/2 is the time for one-half of the
nuclei to decay. The relationship between the halflife and the decay constant can be obtained as
follows.
1
2
N 0  N 0 e   T1 / 2
e  T1 / 2  2
ln( e  T1 / 2 )  T1 / 2  ln( 2 )
Or,
T1 / 2 
ln( 2 )

The exponential decay formula is somewhat like that for the discharge of a capacitor
through a resistor, where  is analogous to the RC time constant.
9
Example:
Tritium (hydrogen-3) decays into helium-3 by beta decay with a half-life of 12.3 yrs:
3
3

1 H 2 He  e  
What is the activity (decay rate) of 1 mg of tritium?
Solution:
number moles  n 
0.001g
 3.33x10  4
3g
N  nN A  ( 3.33x10  4 mole )( 6.02 x10 23 / mole )  2.01x1019

ln( 2 )
0.693

 1.79 x10  9 / s
T1 / 2 ( 12.3 yr )( 3.15 x10 7 s / yr )
R  N  ( 1.79 x10  9 / s )( 2.01x1019 )  3.60 x1010 / s
 ( 3.60 x1010 / s )( 1 Ci / 3.7 x1010 / s )  0.97 Ci
Example:
A curie is a very large and dangerous amount of radioactivity. How long would one have
to wait for the tritium activity to reduce to 1 mCi?
Solution:
R  N 0 e  t  R0 e  t
0.001Ci  0.97Cie   t
e t 
0.97
 970
0.001
ln( 970 )
ln( 970 )
t

 9.9 T1 / 2  ( 9.9 )( 12.3 yr )  122 yr

ln( 2 ) / T1 / 2
Carbon Dating
CO2 in the earth’s atmosphere has a small amount of radioactive C-14. The C-14 is
produced from N-14 in the upper atmosphere by neutrons which are generated in cosmic
ray interactions :
1 14
14
1
0 n  7 N  6 C 1 H
10
C-14 is radioactive and decays back into N-14 through beta decay:
14
14

6 C 7 N  e 
The lifetime of C-14 is 5730 yr, and the fraction of C-14 in the atmosphere is about 1 part
in 1012. Living organisms ingest C-14 from the atmosphere and from ingesting other
organisms and contain this same fraction of C-14. The C-14 in a living organism
produces a decay rate of 15 decays/min·g.
When a living organism dies, it no longer ingests C-14 and its fraction of C-14 decreases
with time. By comparing the decay rate of a dead organism with that of a live organism,
the time of death can be determined.
Example:
20 g of carbon is extracted from a fossil and the C-14 activity is measured to be 100
decays/min. What is the age of the fossil?
Solution:
The activity of 20-g of ‘live’ carbon would be
R0  ( 15 decays / min g )( 20 g )  300 decays / min
So,
R  R0 e  t
100  300e   t
ln( 3 ) ln( 3 )
ln( 3 )
t

T1 / 2 
( 5730 yr )  9082 yr

ln( 2 )
ln( 2 )
Radioactive Series
All elements for which Z > 83 are radioactive. These elements decay through alpha or
beta decay into lower Z elements which may also be radioactive. Eventually, they decay
into an element which is stable. The various radioactive elements that are produced in
this process can be classified into four series depending on the starting and ending nuclei.
These four radioactive series are given listed as follows:
11
Series
Uranium
Actinium
Thorium
Neptunium
Starting isotope
Half-life (yrs)
Stable end product
238
92 U
235
92 U
232
90Th
237
93 Np
4.47x109
206
82 Pb
207
82 Pb
208
82 Pb
209
83 Bi
7.04x108
1.41x1010
2.14x106
The figure below shows the Thorium Series.
Nuclear Reactions
In a nuclear reaction a target nucleus is struck by an energetic particle such as a neutron,
alpha particle or another nucleus, and the particles are transformed into one or more other
particles.
Examples:
4
21
24
1
2 He 10 Ne12 Mg  0 n
12
1
4
2
3
0 n 2 He1 H 1H
Nuclear reactions can be either exothermic or endothermic, depending on whether the
final kinetic energy is greater or smaller than the initial kinetic energy. The ‘Q-value’ of
the reaction is the energy released (or absorbed). For exothermic reactions Q > 0; for
endothermic Q < 0. The Q-value can be obtained from the difference in mass of the
initial and final products.
Q  mc 2  ( mi  m f )c 2
A reaction with a negative Q requires that the incident particle have a minimum amount
of kinetic energy in order for the reaction to occur. If the target particle is at rest, then the
threshold kinetic energy for the reaction must exceed |Q| in order to conserve momentum.
For example, if KEi = |Q|, then KEf = 0 and the final momentum is zero, so momentum is
not conserved. By requiring that both energy and momentum be conserved, it can be
shown that the threshold kinetic energy of the incident particle for an endothermic
reaction to occur is
m

KEmin  1 
|Q|
M

where m is the mass of the incident particle and M is the mass of the target particle.
Example:
Calculate the Q for the following reaction, and calculate the minimum energy of the
incident neutron for the reaction to occur.
1 14
11
4
0 n  7 N  5 B  2 He
Solution:
Q  (mn  m N14 )  (m B11  m He 4 )
 (1.008665  14.003074)  (11.009306  4.002603)  0.00017 u
 (0.00017 u )(931.494 Mev / u )  0.158 Mev

mn
KEmin  1 
 m N14

1
 | Q | 1  (0.158Mev )  0.17 Mev
 14 

13
Earlier in this chapter we discussed the binding energy of a nucleus. The mass of a
nucleus is less than the mass of its constituent protons and neutrons and this mass
difference gives the binding energy through the relationship Eb = mc2. When we
calculate the binding energy per nucleon, then we find that the nuclei with intermediate
mass have greater binding energy per nucleon than those light or very heavy nuclei. A
graph of binding energy per nucleon was shown earlier.
By combining light nuclei to form heavier nuclei or by splitting apart heavy nuclei to
form lighter nuclei, we can take advantage of the change in binding energy to convert
mass to energy. The former process is fusion and the later process is fission.
Nuclear Fission
239
In nuclear fission a heavy nucleus such as 235
92 U or 94 Pu split into two smaller nuclei
along with the production of one or more neutrons. The fission is prompted by the
capture of a slow neutron which makes the nucleus unstable. In the fission of uranium235, the capture of a neutron forms uranium-236 in an excited state which subsequently
breaks apart as shown below.
1 235
236 *
0 n 92 U  92 U  X  Y  neutrons
The product nuclei produced in nuclear fission are typically radioactive.
There are many possible fission products. Some specific examples for uranium-235 are
1 235
141
92
1
0 n 92 U  56 Ba  36 Kr 30 n
1 235
140
94
1
0 n 92 U  54 Xe 38 Sr  2 0 n
An estimate of the energy released in a typical fission event can be determined as follows.
From the above graph, we see that the binding energy per nucleon for U-238 is about 7.5
Mev and the binding energy per nucleon for a nucleus with A = 119 (= 238/2) is about
8.5 Mev. So, the energy released per nucleon is about 1 Mev, and the total energy
released is about 238 Mev. A more precise value is Q  208 Mev. This lower value is in
part because the binding energy per nuclei of the products is smaller than 8.5 Mev since
they are unstable and not typical of the values shown on the graph.
Example:
Calculate the total energy released in the fission of 1 g of U-235.
Solution:
We calculate the total number of atoms in 1 g of U-235 and multiply this by 208 Mev.
14
N  nN A 
1g
6.02 x10 23 / mole  2.56 x10 21
235 g / mole
E  NQ  ( 2.56 x10 21 )( 208Mev )  ( 5.33x10 29 ev )( 1.6 x10 19 J / ev )
 8.53x1010 J
Example:
A typical US household uses 1000 W of electricity. How long would the energy released
in the fission of 1 g of U-235 power a household if all the energy could be converted in to
electricity?
Answer:
E 8.53x1010 J
t 
 8.53x107 s
P
1000 J / s
Converting to years –
t  ( 8.53x10 7 s )(
1yr
)  2.7 yr
24 x3600 x365s
Nuclear Reactors
A nuclear reactor is designed to allow the sustained fission through a chain reaction. The
fission of a U-235 nucleus produces neutrons which can, in turn, be absorbed by other
nearby U-235 nuclei. On average each fission event produces about 2.5 neutrons. If an
average of one of these neutrons prompts another fission event, then this will go on until
all the U-235 nuclei have been depleted. The chain reaction is said to be critical. If more
than one neutron from each fission event prompts another fission event, then the number
of nuclei undergoing fission will rapidly increase in time and the chain reaction is super
critical. The rate at which energy is released will rapidly increase and can lead to a
nuclear explosion of a melt-down of the reactor. On the other hand, if an average of less
than one neutron produced in a fission event leads to another fission event, then the chain
reaction is not self-sustaining and is subcritical. The figure below illustrates a super
critical chain reaction.
15
The neutrons released in a fission event have a
kinetic energy of about 2 Mev. This energy is
too large to be absorbed by another U-235
nucleus and initiate another fission event. A
reactor contains a moderator material that slows
down the neutrons. A typical moderator material
is “heavy water”, D2O, where D is deuterium. An
illustration of a reactor core is shown to the right.
Reactors are designed to operate near critical
conditions. To prevent a reactor from going
super critical control rods are inserted between
fuel elements. The control rods are made of a
material such as cadmium that is a good absorber
of neutrons. The control rods can be moved up
and down as needed to keep the reactor operating near the critical condition.
The energy released in nuclear reactors is in the form of heat. A heat exchanger is used
to generate steam which drives a turbine and produces electricity.
Nuclear Fusion
In nuclear fusion lighter nuclei combine to form heavier nuclei. For A less than 60 or so
the heavy nuclei have a greater binding energy per nuclei than the light nuclei, so fusion
results in a decrease in mass with the mass difference appearing as energy released in the
process. Nuclear fusion requires overcoming the electrostatic repulsion between the
16
nuclei. This requires that the nuclei just collide at very high speeds, which means very
high temperatures.
Nuclear Fusion in the Sun
Nuclear fusion is the source of the radiation emitted by the sun and the stars. Stars are
formed by the gravitational collapse of dust and gas. At the gas collapses inward the
molecules gain speed and the temperature rapidly increases. Eventually when the
temperature becomes sufficiently large fusion takes place and energy is released. The
collapse of the gas (plasma) eventually ceases when the pressure due to the thermal
motion of the particles and the radiation pressure become sufficient to balance the
pressure due to the gravitational force. For the sun and other stars that are rich in
hydrogen, protons fuse to form deuterons with the release of a positron and a neutrino:
1
1
2

1 H 1H 1 D  e

Fusion takes place in the interior of the sun, where the core temperature is about 107 K.
The surface temperature is about 5,800 K, which is too low for fusion. Most collisions
between protons in the interior do not result in fusion, but the fusion process is aided by
the very high density in the core.
Additional fusion reactions in the sun eventually produce helium-4:
1
2
3
1 H  1 D  2 He  
1
3
4

1 H  2 He  2 He  e

3
3
4
1
2 He  2 He  2 He  2 1H
At the earth’s surface, the neutrino flux from the sun is very large, about 1011/cm2/s.
They interact very, very weakly with matter, so almost all the neutrinos that strike the
earth (and you) pass through undetected.
Fusion Reactors
Extensive research has been done and is continuing on the design of fusion reactors for
converting mass into energy. Fusion has many potential advantages over fission for
energy production. The mass change per nucleon in fusion is much greater than in fission
and there is little long-lived radioactive waste as a byproduct to worry about. In addition,
deuterium, a good fuel for fusion, is relatively abundant in nature and can be cheaply
extracted from water. The problem is how to produce the conditions such that significant
fusion can take place in a controllable manner in which energy can be usefully extracted.
A hydrogen bomb is an uncontrolled fusion reaction.
17
To emulate the density and temperature at the sun’s interior would require extreme
pressure. Additionally, no material could contain the high temperature plasma without
evaporating. A tokomak is a device that makes use of magnetic fields to confine a high
temperature plasma. Since the density of the plasma is much less than the density of the
sun’s core, the temperature of the plasma must be much larger than the sun’s core, ~ 108
K. In order for a fusion reactor employing a high-temperature plasma to be viable, the
energy released in the fusion process must exceed the energy required to heat the plasma
to the necessary temperature and maintain it at this temperature for a sufficient time. The
amount of time that the plasma must be held at the reaction temperature depends on the
density of the plasma.
The international community has agreed to build a new tokomak (ITER) in France that
would generate 10 times more energy than required to maintain the temperature of the
plasma at fusion conditions (http://www.iter.org/ ). The goal is to demonstrate the
viability of a fusion reactor.
Example:
Calculate the energy released in the following fusion reactions.
2
2
3
1
1 D  1 D  1T 1 H
2
3
4
1
1 D  1T  2 He  0 n
Solution:
For the first reaction,
 m  2md  mT  mH  2( 2.014102 u )  3.016049 u  1.007825 u
 0.00433 u
Q   mc 2  ( 0.00433u )( 931.5Mev / c 2u )c 2  4.03Mev
For the second reaction,
 m  md  mT  mHe 3  mn  2.014102 u  3.016049 u  4.002603 u  1.008665 u
 0.01888 u
Q   mc 2  ( 0.01888u )( 931.5Mev / c 2u )c 2  17.6Mev
The energy released in the second reaction is considerably more than in the first, but it
requires tritium, which is very rare. The moon, however, is thought to contain a
significant amount of tritium, and mining the moon for this fusion fuel has been proposed.
An alternative concept for a controlled fusion reactor is inertial laser confinement fusion.
In this approach a small pellet containing deuterium and tritium is blasted simultaneously
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from different angles by a large number of laser pulses. The hope is that the pulses can
generate sufficient pressure on the pellet so that the density and temperature reach fusion
conditions. Research on this approach is taking place at Lawrence Livermore National
Laboratory (https://lasers.llnl.gov ).
Elementary Particles
At one time it was thought that the most elementary particle was the atom. Then it was
assumed that the elementary particles consisted of the electron, proton, neutron, and
photon. It is now known that a large variety of short-lived sub-atomic particles can be
generated by high energy collisions of other particles and by the decay of other shortlived particles. Physicists are continually trying to determine which of these particles are
the basic building blocks of nature. That is, which are truly ‘elementary’.
An important property of particles is the force that they can experience. There are four
basic forces in nature, given in the table below.
Force
Strong
Electromagnetic
Weak
Gravitational
Relative
Strength
1
10-2
10-6
10-43
Range
Short (~ 1 fm)
Long (~ 1/r2)
Sort (~10-3 fm)
Long (~ 1/r2)
Mediating
Field Particle
Gluon
Photon
W ±, Z
Graviton
The forces vary widely in strength. The strongest is the short range “strong force” that
acts between nucleons in the nucleus. The “electromagnetic force” is the electric and
magnetic force that is based on a particle’s charge and is long-range. The “weak” force is
a short range force that is involved in beta decay. The “gravitational” force is the
weakest of all and is long range.
It is believed that these forces are mediated by the exchange of ‘virtual’ particles listed in
the right column of the table. This would be somewhat analogous to the bonding of the
two protons in a hydrogen molecule that is caused by the ‘exchange’ of the two orbital
electrons. When two particles interact by one of these fundamental forces, a virtual field
particle is being continually emitted by one particle and absorbed by the other. It is
called virtual since it is never directly detected. These virtual particles only exist during
the short time during which they are being exchanged. The existence and non-existence
of these particles would seem to violate conservation of energy; however, the uncertainty
principle, Et  /2, allows this to happen to a degree that depends on the time during
which the energy is determined. Because of the short range of the interactions and the
speed of the virtual particles, their lifetime can be short and the energies of the virtual
particles can be significant.
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Classifications of Particles
All particles other than the photon can be classified as either a lepton or a hadron.
Leptons
The leptons include the electron (e-) , muon (-) , and tau ( -) particles and their
associated neutrinos - e, ,  . These six particles also have their antiparticles. All but
the muon and tau particle are stable. The muon and tau masses are about 207 and 3490
times the electron mass. The leptons have a spin of ½.
Hadrons
The hadrons include mesons and baryons.
Mesons
There are a large number of mesons with a wide range of masses. Some have charge and
some don’t. Examples of mesons are the pion (0,  -, +) and the kaon (K0, K-, K+).
Mesons have integer spin and all are unstable.
Baryons
There are also a large number of baryons. The proton and the neutron are baryons.
Others include the lambda (0) and sigma (0, -, +) particles. Baryons have halfinteger spin (1/2, 3/2, …), and all are unstable except the proton. A free neutron has a
lifetime of about 15 min. When bound in the nucleus it is stable.
Particles can also be classified by their spin. Half-integer spin particles (leptons and
baryons) are referred to as fermions. Integer spin particles (0, 1, 2, ..) are referred to as
bosons. The mesons and the photon (spin = 1) are bosons.
Quarks
The leptons are considered to be elementary particles. The have no internal structure.
The hadrons, however, are not considered to be elementary particles and are thought to
consist of bound states of quarks. A quark has a charge of 1/3e or 2/3e. There are six
types of quarks and their antiparticles: up (u), down (d), charmed (c), strange (s), top (t),
and bottom (b). Quarks are fermions with a spin of ½. Isolated quarks have never been
observed. The mass and charge of the quarks are listed below. The antiquarks have
opposite signs.
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Quark
u
d
c
s
t
b
Rest mass
energy
360
360
1500
540
173
5
Charge
(e)
+2/3
-1/3
+2/3
-1/3
+2/3
-1/3
Mesons consist of 2 quarks and baryons consist of 3 quarks. The composition of the
proton is uud . Its charge is 2/3e + 2/3e – 1/3e = e. The neutron is udd and its charge is
2/3e - 1/3e – 1/3e = 0. The + is du and its charge is 1/3e+2/3e = e.
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