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Nuclear Physics
Physics 12
Protons, Neutrons and Electrons
 The atom is composed of three subatomic
particles:
Particle
Charge
(in C)
Symbol
Mass
(in kg)
Electron
-1.602x10-19
e-
9.109 56x10-31
Proton
1.602x10-19
p+
1.672 614x10-27
Neutron
0
n0
1.674 920x10-27
Atomic Nucleus
 Atom described
using:
 X – atomic symbol
 A – atomic mass
number (nucleon
number)
 Z – atomic number
 Number of protons
and electrons = Z
 Number of
neutrons = A - Z
A
Z
X
Strong Nuclear Force
 The electrostatic forces inside a
nucleus would rip it apart if there was
not another force
 By the end of the 1930’s physicists
had determined that nucleons attract
each other
 This is the strongest force in the
known universe
Stability and the Nucleus
 Although the Strong Nuclear Force is
strong enough to hold a small nucleus
together, as the size of the nucleus
becomes larger, the electrostatic
forces begin to become more
important
 As a result, if we consider various
nuclei based on their Atomic Number
and Neutron Number we get the
following result:
Stability and the Nucleus
Each black dot
represents a
stable nucleus,
with the
number of
neutrons shown
on the vertical
axis and the
number of
protons on the
horizontal axis
Nuclides and Isotopes
 Nuclides are different combinations of
nucleons
 Isotopes occur when an element
(specific Atomic Number) has
different numbers of neutrons
(different Atomic Mass Numbers)
 For example, there are three common
isotopes of hydrogen:
Nuclides and Isotopes
Nuclear Binding Energy
 It takes 13.6 eV to separate an
electron from a hydrogen atom
 However, it takes more than 20 MeV
to separate a neutron from a helium4 atom
 The energy to separate all the
nucleons in a nucleus is called the
binding energy
Larger nuclei are
held together a
little less tightly
than those in the
middle of the
Periodic Table
Mass Defect
 If you were able to apply the 20 MeV required to
separate a neutron from helium-4, what would
happen to it?
 This is dealt with using Einstein’s Special Theory of
Relativity and the fact that mass and energy are
equivalent  E = mc2
 The mass of helium-4 (2p, 2n) is smaller than that
of helium-3 (2p, 1n) and a neutron
 The energy that was added to remove the neutron
was converted into mass
 The difference between the mass of a nuclide and
the sum of the masses of its constituents is called
mass defect
Atomic Mass Unit (u)
 When dealing with nucleons, it is often
more useful to deal with mass in unified
atomic mass units (u) instead of kilograms
Particle
Mass
(in kg)
Mass
(in u)
Electron
9.109 56x10-31
0.000 549
Proton
1.672 614x10-27
1.007 276
Neutron
1.674 920x10-27
1.008 665
Binding Energy Example
 Determine the binding energy in
electron volts and joules for an iron56 nucleus given that the nuclear
mass is 55.9206u
mnucleus  55.9206u
A  56
Z  26
N  AZ
N  30
Binding Energy Example
 Determine the binding energy in
electron volts and joules for an iron56 nucleus given that the nuclear
mass is 55.9206u
mtotal  Zmp  Nmn
mtotal  26(1.007276u )  30(1.008665u )
mtotal  56.449126u
m  56.449126u  55.9206u
m  0.5285u
Binding Energy Example
E  mc 2
E
E
E
E
 We would expect
the binding energy
 0.5285u(3.00 x 108 m / s )2
per nucleon to be
11
 7.888 x10 J
about 8MeV:
11
7.888 x10 J

8
19
4.924
x
10
eV
1.602 x10 J / eV
 8.79 x106 eV
56
 4.924 x108 eV
Radioactive Isotopes
 In discussing the nucleus, we looked
at a plot of stable nuclei
 It is also possible to have a nucleus
that is not stable (meaning that it will
fall apart)
 An unstable nucleus will decay
following a few very specific
processes
 We call this decay radioactivity and
classify it into one of three types
Radioactive Isotopes
Alpha Decay
 An alpha particle (α) is a helium
nucleus (two protons and two neutrons)
 A nucleus that emits an alpha particle will
lose the two protons and two neutrons
 Large nuclei will emit alpha particles
 They do not penetrate matter well and a
sheet of paper or 5cm of air will stop
most
 They can free electrons from atoms,
meaning they are a form of ionizing
radiation
Alpha Decay
Beta Decay
 When a nucleus emits a beta particle
(β), it appears to lose an electron or
positron from within the nucleus
 There are two types of beta decay (β- and
β+)
 Beta particles can penetrate matter to a
greater extent than alpha particles; they
can penetrate about 0.1mm of lead or
10m of air
 They are also a form of ionizing radiation
but less damaging than alpha particles
Beta Decay (β-)
 In this type of beta decay, a neutron
becomes a proton and a β- particle
(high energy electron) is emitted
 In addition an antineutrino ( ) is
emitted (antimatter) along with the
beta minus particle
 The nucleus’s atomic number
increases by one while the atomic
mass number remains the same
Beta Decay (β-)
Beta Decay (β+)
 In this type of beta decay, a proton
becomes a neutron and a β+particle
(high energy positron or antielectron)
is emitted
 In addition a neutrino (  ) is
emitted along with the beta plus
particle
 The nucleus’s atomic number
decreases by one while the atomic
mass number remains the same
Beta Decay (β+)
Gamma Decay (γ)
 When a nucleus goes through alpha or
beta decay, the daughter nucleus is often
left in an excited state
 In order to reduce the energy of the
nucleus, it will go through gamma decay
(high energy photon) to return to the
ground state
 Gamma radiation can pass through 10cm
of lead or 2km of air
 It is the most damaging of all due to the
energy of the gamma particle
Gamma Decay
Decay Series
 When a large nucleus decays by
alpha and beta radiation, the
daughter nucleus will be more stable
than the original nucleus
 However, the daughter nucleus may
still be unstable and will itself go
through alpha or beta radiation
 This leads to a decay series
Rate of Radioactive Decay
 It is impossible to predict when a specific
nucleus will decay
 You can describe the probability of decay
 The concept of half life is used with
radioactive decay: the time required for
half of the sample to decay
 Using the half life equation, it is possible to
determine how much of a sample would
remain after a given period of time
Half Life
 1
N  N0  
 2




N  sample remaining
N0  original sample
Δt  elapsed time
T  half life
t
T
Half Life