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Chapter 12
The Atomic Nucleus
Enrico Fermi (1901 – 1954) Nobel 1938
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12.1 Discovery of the Neutron
• James Chadwick (1891 – 1974) Nobel 1935
discovered the neutron in 1932.
• Thus it was learned that the nucleus was
composed of two particles of approximately equal
mass, protons and neutrons.
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Can Electrons Exist inside the
Nucleus?
•
Before neutrons were discovered, it was thought
that the nucleus was composed of protons and
electrons.
• Can electrons exist inside the nucleus as
independent particles? No. Why not?
1. Nuclear size. Because of the small size of the
nucleus, the uncertainty principle puts a lower
limit on the kinetic energy that is greater than the
energy of any electron emitted from the nucleus.
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Can Electrons Exist inside the
Nucleus?
2. Nuclear spin.
3. Nuclear magnetic moment.
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Example 12.1
What is the minimum kinetic energy of a proton in
a medium-sized nucleus having a diameter of 8 x
10-15 m?
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12.2 Nuclear Properties
• The primary constituents of nuclei are the proton
and the neutron.
• Neutrons and protons are each made up of three
quarks.
• The nuclear mass is equal to the sum of the masses
of the constituent particles plus the binding
energy.
• The nuclear charge is + e times the number (Z) of
protons.
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Atomic Notation
AX
Z N
Z = atomic number (number of protons)
N = neutron number (number of neutrons)
A = mass number (Z + N)
X = chemical element symbol
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The Atomic Mass Unit
1 u = 1.66054 x 10-27 Kg
1 u = 931.494 Mev/c2
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Sizes and Shapes of Nuclei
• Nuclear radius
 Matter radius (physical size)
 Force radius (range of nuclear (strong) force.
 Charge radius (electrons are not affected by
nuclear force, but by electromagnetic field of
the nucleus.)
• Nuclear force radius  Mass radius  Charge
radius (experimentally determined)
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Nuclear Radius
The nuclear radius may be approximated from a
spherical charge distribution to be
R = r0 A1/3
where r0  1.2 x 10 –15 m.
10 –15 m is called a femtometer or in physics a
fermi (fm)
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Example 12.2
What is the nuclear radius of 40Ca? What energy
electrons and protons are required to probe the
size of 40Ca if one wants to “see” at least half the
radius?
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Example 12.3
What is the ratio of the radii of 238U and 4He?
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Intrinsic Spin
• Protons and neutrons are both fermions with spin
quantum numbers s = ± ½.
• They obey the same spin quantum rules as do
electrons.
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12.3 The Deuteron
• After the proton, the next simplest nucleus is the
deuteron: 21H.
• The deuteron mass is: 2.013553 u, and the mass
of the deuterium atom is 2.014102.
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Binding Energy
• The deuteron nucleus is bound by an energy Bd,
which represents mass energy.
• The mass of the deuteron nucleus is:
md = mp + mn – Eb /c2
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Binding Energy
Eb(ZAX) = [Nmn + ZM(1H) – M(ZAX)] c2
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12.4 Nuclear Forces
• Most straightforward techniques of studying
nuclear forces are scattering experiments.
• Two common ones are p – n and p – p scattering.
• Nuclear forces are attractive short range forces.
• Two nucleons within about 2 fm of each other feel
and attractive force.
• Outside 3 fm the strong force is essentially zero.
• Outside 3 fm the repulsive (proton – proton)
Coulomb force dominates.
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r  3 fm
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12.5 Nuclear Stability
• Equation 12.10 presented a method to determine
binding energies.
• If B is positive, the nucleus is stable against
dissociating into free neutrons and protons.
• A more general statement is a nucleus containing
A nucleons is stable if its mass is smaller than any
other possible combination of A nucleons.
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Nuclear Stability
The binding energy of a nucleus ZAX against
dissociation into any other possible combination
of nucleons, for example, R and S nuclei, is
Eb = [M(R) + M(S) – M(ZA X )] c2
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Example 12.4
Show that the nuclide 8Be has a positive binding
energy but is unstable with respect to decay into
two alpha particles.
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A  40, Z~N
A  40, N > Z
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Nuclear Coulomb Repulsion
Energy
3 Z (Z – 1) e2
DECoul =
5
4pe0 R
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Example 12.5
Show that Equation (12.18) can be written as
DECoul = 0.72 [Z(Z – 1)] A-1/3 MeV (12.19)
and use this equation to calculate the total
Coulomb energy of 238
U.
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Example 12.6
Calculate the binding energy per nucleon for 20
Ne,
10
56Fe, and 238U.
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Nuclear Models
• Physicists do not yet fully understand the nuclear
force or how nucleons interact inside the nucleus.
• Current research stresses the constituent quarks
that make up the nucleus.
• Because of this lack of knowledge, physicists have
used models to explain nuclear behavior.
• Nuclear models fall into two categories.
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Nuclear Models
1. Independent Particle Models:
Nucleons move nearly independently in a
common nuclear potential. The shell model has
been the most successful of these.
2. Strong Interaction Models:
Nucleons are strongly coupled together. The
liquid drop model has been quite successful in
explaining nuclear masses as well as nuclear
fission. Accounts for nuclear binding energy.
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Shell Model
• The shell model is based on the fact that each
nucleon moves in a well-defined orbital state
within the nucleus in an averaged field produced
by the other nucleons.
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Neutron energy levels slightly lower than the proton
levels because of the additional Coulomb repulsion.
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Liquid Drop Model
• The semi-empirical binding energy formula (Carl
F. von Weizsäcker – 1935) is an example of the
liquid drop model.
• This model treats the nucleons as though they
were molecules in a drop of liquid.
• The nucleons strongly interact with each other and
undergo frequent collisions as they jiggle around
within the nucleus.
• The jiggling is analogous to thermal vibrations of
the molecules in a liquid drop.
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Semi – Empirical Binding Energy
Formula
Eb = avA –
aAA2/3
2
(N – Z)2
3
Z
(Z
–
1)
e
–
- aS
+d
A
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4pe0 R
aV = 14 MeV Volume
aA = 13 MeV
Surface
aS = 19 MeV Symmetry
d = + D even-even nuclei
d = 0 odd-A (e-o, o-e)
d = - D odd-odd nuclei
where D = 33 MeV • A-3/4
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Semi – Empirical Binding Energy
Formula
• avA: The volume term indicates that the binding
energy is approximately the sum of all the
interactions between nucleons.
• aAA2/3: This is a correction term due to the surface
effect. That is, that nucleons on the surface do not
feel the effect of being completely surrounded by
other nucleons.
• The third term is the Coulomb energy term.
• The d term is due to the various ways nucleons
can pair. It shows that the nucleus is more stable
for even-even nucleus.
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Even-Even
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12.6 Radioactive Decay
• Unstable nuclei decay spontaneously to some
other combination of A nucleons that has a lower
mass.
• We will cover the three primary forms of
radioactive decay:
 a (alpha)
 b (beta)
 g (gamma)
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Definitions
•
•
•
•
Activity (R) = decays per unit time.
1 Bq = 1 decay/s
N (t) = number of unstable nuclei at time t.
The decay constant (l) is the probability per unit
time that any given nucleus will decay.
• N0 = number of nuclei in the sample at t = 0.
• R0 = activity at t = 0.
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Decay Formulas
N(t) = N0 e-lt
R(t) = R0 e-lt
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Half - Life
• The half-life of a given element t1/2 is the time it
takes for half the nuclei in a sample to decay.
0.693
t1/2 =
l
• The mean lifetime t is defined as 1/l.
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Example 12.7
A sample of 210Po that a decays with t1/2 = 138
days is observed by a student to have 2000
disintegrations/s (2000 Bq).
(a) What is the activity in mCi for this source?
(b) What is the mass of 210Po?
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Example 12.8
A sample of 18F is used internally as a medical
diagnostic tool to look for the effects of the
positron decay (t1/2 = 110 min). How long does it
take for 99% of the 18F to decay?
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Example 12.9
What is the alpha activity of a 10-kg sample of
235U that is used in a nuclear reactor?
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12.7 Alpha, Beta, and Gamma
Decay
• All these methods of decay had been observed by
the early 20th century.
• When a nucleus decays, all the conservation laws
must be conserved: mass-energy, linear
momentum, angular momentum, and electric
charge.
• An additional decay law must be added – the law
of conservation of nucleons.
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Law of Conservation of Nucleons
• The total number of nucleons (A, the mass
number) must be conserved in a low-energy
nuclear reaction (say, less than 100 MeV) or
decay.
• Neutrons may be converted into protons, and vice
versa, but the total number of nucleons must
remain constant.
• At higher energies enough rest energy may be
available to create nucleons, but the other
conservation laws still apply.
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Radioactive Decay
• Radioactive decay may occur for a nucleus when
some other combination of the A nucleons has a
lower mass.
• Applying the law of conservation of energy:
Daughter Lighter
Parent
M (ZAX ) = MD + My + Q/c2
Q is the energy released and is equal to the total kinetic
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energy of the reaction products.
Energy Released (Disintegration
Energy)
Q = [M (ZAX ) - MD - My] c2
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Example 12.10
Show that 230
92U does not decay by emitting a
neutron or proton.
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Alpha Decay
AX
Z
A-4D
Z-2
+a
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Disintegration Energy for Alpha
Decay
4He)]c2
Q = [M(AZX) – M(A-4
D)
–
M(
2
Z-2
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Beta (b ) Decay
-
• Special example:
n
p + b +n
-
• General case:
AX
Z
A
Z+1D
+ b +n
-
n is an antineutrino
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Disintegration Energy b Decay
-
Q=
[M(ZAX
)-
A
M(Z+1D
)]c2
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b (Positron) Decay
+
• Special example:
p
n + b +n
+
• General case:
AX
Z
A
Z-1D
+ b +n
+
n is an neutrino
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Disintegration Energy b Decay
+
Q=
[M(AZX
)-
A
M(Z-1D
) – 2me]c2
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Electron Capture
• Special case for a proton:
p + e-
n=n
• General case:
A
ZX
+
e-
A
Z-1D
+n
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Disintegration energy for Electron
Capture
Q=
[M(AZX
) – M(
A
2
D)]c
Z-1
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Example 12.11
Show that the relations expresses for the
disintegration energy Q in Equations (12.38),
(12.41), and (12.44) are correct.
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Example 12.12
Show that 55Fe may undergo electron capture, but
not b + decay.
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Example 12.13
Find whether alpha decay or any of the beta
decays are allowed for 226
Ac.
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Gamma Decay
AX*
AX
+g
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Example 12.14
Consider the g decay from the 0.072-MeV excited
state to the ground state of 226Th at rest in Figure
12.15. Find the exact expression for the gammaray energy by including both the conservation of
momentum and energy. Determine the error
obtained by using the approximate value in
Equation (12.46).
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12.8 Radioactive Nuclides
• “Standard Model” tells us the universe was
created in the Big Bang about 13 billion (13 x 109)
years ago.
• Neutrons and protons fused together in the first
few minutes to form deuterons and other light
nuclei.
• Unstable nuclei that were formed exhibit natural
radioactivity.
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Radioactive Nuclides
• There are many natural radioactive nuclides left
on Earth with lifetimes long enough to be
observed.
• Since the Earth (and solar system) is about 4.5
billion years old, only those nuclides with halflives longer than a few billion years can now exist.
• Most, but not all, have A > 150.
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Decay Modes
• In addition to nuclear fission, heavy radioactive
nuclides can change their mass number (A) only
by alpha decay.
• They can change their charge number (Z) by either
alpha or beta decay.
• As a result, there are only four paths that heavy
naturally occurring radioactive nuclides may take
to decay to stable end products.
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The Four Paths of Decay
• The four paths have mass numbers expressed by
either 4n, 4n +1, 4n +2, 4n +3 (n = integer).
• This is because only alpha decay can change the
mass number.
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Dating Using Lead Isotopes
•
204Pb
is not radioactive and no other nuclides
decay to it. Thus is abundance is constant.
• 206Pb and 207Pb are stable, but are the end products
of decay of 238U and 235U respectively.
• 235U has a relatively short half-life (0.70 x 109
years), most 235U has already decayed to 207Pb, the
the ratio 207Pb/ 204Pb has been relatively constant
for the past 2 billion years.
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Dating Using Lead Isotopes
• Since the half-life of 238U (4.47 x 109 years) is
about the same of the age of the Earth, only about
half the original abundance has decayed to.
• Therefore, the ratio 206Pb/ 204Pb is still increasing.
• A plot of the abundance of 206Pb/ 204Pb versus
207Pb/ 204Pb can be a sensitive indicator of the age
of lead ores.
• Using this technique has been used to show that
moon rocks are as old as the solar system.
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Example 12.15
Assume that all the 206Pb found in a given sample
of uranium ore resulted from the decay of 238U and
that the ratio 206Pb/ 238U is 0.60. How old is the
ore?
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Radioactive Carbon Dating
• Radioactive 14C is produced in our atmosphere by
bombardment of 14N by neutrons produced by
cosmic rays.
n + 14N
14C
+p
• A natural equilibrium of 14C to 12C exists in
molecules of CO2 in the atmosphere that all living
organisms take in.
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Radioactive Carbon Dating
• When living organisms die, their intake of 14C
ceases, and the ration of 14C/ 12C decreases as 14C
decays.
• The half-life of 14C is 5730 years. Thus the age of
objects can be determined up to about 45,000
years.
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Example 12.16
A bone suspected to have originated during the
period of the Roman emperors was found in Great
Britain. Accelerator techniques gave its 14C/ 12C
ratio as 1.1 x 10-12. Is the bone old enough to have
Roman origins?
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