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Goals of this chapter:
1. To understand some key properties of nuclei
2. To see how nuclear binding energy depends on the number of protons and neutrons
3. To investigate radioactive decay
4. To learn about hazards and medical uses of radiation
5. To analyze nuclear reactions
6. To investigate nuclear fission
7. To understand the nuclear reactions in our sun
1.
Chapter 43
Nuclear Physics
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by Wayne Anderson
Copyright © 2012 Pearson Education Inc.
2.
Goals for Chapter 43
• To understand some key properties of nuclei
• To see how nuclear binding energy depends on the
number of protons and neutrons
• To investigate radioactive decay
• To learn about hazards and medical uses of radiation
• To analyze nuclear reactions
• To investigate nuclear fission
• To understand the nuclear reactions in our sun
Copyright © 2012 Pearson Education Inc.
3.
Properties of nuclei
• The nucleon number A (also called the mass number) is the total number of
protons and neutrons in the nucleus. The nucleon mass is measured in atomic
mass units u, slightly less than the mass of the proton:
1 u = 1.660538782(83) 1027 kg.
• The radius of most nuclei is given by R = R0A1/3, where R0 is experimentally
determined as R0 = 1.2 10-15 m (1.2 fm)
• All nuclei have approximately the same density.
• Example 43.1: Common iron nuclei have mass number 56. Find the radius,
approximate mass, and density of an iron nucleus.
R R0 A1/3 (1.2 10 15 m)(56)1/3 4.6 fm
m (56 u)(1.66 10 27 kg) 9.3 10 26 kg
V 43 R 3 43 (4.6 10 15 m)3 4.110 43 m3
m 9.3 10 26 kg 17 3 Nucleus is 1013 times
2.3 10 kg/m
V 4.110 43 m3 the density of iron
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4.
Nuclides and isotopes
• Electron and nucleon masses (note, a 12 C nucleus is defined to have u = 12.0000)
• Proton: mp = 1.007276 u
• Neutron: mn = 1.008665 u
• Electron: me = 0.000548580 u
• The atomic number Z is the number of protons in the nucleus. The neutron
number N is the number of neutrons in the nucleus. Therefore A = Z + N.
• A nuclide is a single nuclear species having specific values for both Z and N,
16
e.g. 8 O
16
• The isotopes of an element have different numbers of neutrons, e.g. 8 O or 17
8 O
Why do you think changing N does not change the element, but changing Z does?
• Like electrons, nucleons have ½-integer spin angular momentum, obeying the
same relation as before: S s ( s 1) 4
3
• S z 12
The z-component is itself a quantum number, as before:
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5.
Magnetic Moments
• Since every nucleon has its own spin, you can imagine that putting them all into
a nucleus would make an overall spin of the vector sum of the individual ones,
and the result could be pretty chaotic.
• In fact, it turns out that the magnitude of the total angular momentum J of the
nucleus is also neatly quantized as:
J j ( j 1)
with quantized z-component
J z m j (m j 0, 1, 2,..., j )
• When A is even, j is an integer, but when A is odd, j is a half-integer.
• Associated with the nuclear angular moment is a magnetic moment. Recall the
Bohr magneton for electrons: B e In the case of a nucleus, the analogous
quantity is the nuclear magneton: 2me
e
n
2m p
• The z-component of the magnetic moment for the proton is: sz proton
2.7928 n
• Surprisingly, for a neutron it is non-zero: sz neutron
1.9130 n
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6.
Proton Spin Flips
• Recall that the potential energy associated with spin is U μ B z B .
• Example 43.2: Proton spin flips: Protons are placed in a 2.30-T magnetic field
that points in the positive z-direction. (a) What is the energy difference
between states with the x-component of proton spin angular momentum
parallel and anti-parallel to the field? (b) A proton can make a transition from
one of these states to the other by emitting or absorbing a photon with the
appropriate energy. Find the frequency and wavelength of such a photon.
(a) When the spin angular momentum is parallel to B,
U z B (2.7928)(3.152 10 8 eV/T)(2.30 T) 2.025 10 7 eV
When it is anti-parallel, it is just 2.025 x 10-7 eV. The energy difference is just
twice this.
(b) The corresponding frequency and wavelength are:
E 4.05 10 7 eV
f 15
9.79 107 Hz = 97.9 MHz
h 4.136 10 eV-s
c
3.06 m
f
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7.
NMR and MRI
• Nuclear magnetic resonance and MRI use strong magnetic fields to align the
nuclear spins, then flips the spins with radio waves. When the radio waves
cease, the spins flip spontaneously and emit radio photons that are measured.
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8.
Nuclear binding energy
• The mass of the 12C atom, made up of 6 protons and 6 neutrons, defines the mass
unit u, i.e. it has a mass of exactly 12 u. The individual masses of the protons and
neutrons is 6(1.007276 u) + 6(1.008665 u) = 12.095646 u. The difference,
0.0956 u, when converted to energy E = mc2, is the binding energy EB of the
nucleus. It is convenient to use the mass-energy equivalent of c2, which is 931.5
MeV/u, so that 0.0956 u => 89.1 MeV is the binding energy of 12C. It is the
energy that must be added to separate the nucleons. The quantity EB/c2 is called
the mass
• Note that defect.
the nucleus is a collection of positive charges, which all mutually repel
each other due to the electrostatic force. Therefore, there must another, attractive
force acting in the nucleus to overcome this repulsion, and this is called the
nuclear strong force. It binds together both protons and neutrons, and is
responsible for the drop in energy (mass) of bound nuclei.
• The exact nature of this force is still not fully understood, but we can characterize
the binding energy as:
EB = (ZMH + Nmn – )c2ZA. M
A
Note that MH is the mass of a hydrogen atom, not just its proton, and Z M also
includes the electrons of the atom.
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9.
Nuclear binding energy
• Example 43.3: Find the mass defect, total binding energy, and binding energy per
62
nucleon of 28 Ni , which has the highest binding energy per nucleon of all nuclides.
62
The neutral atomic mass of28 Ni is 61.928349 u.
We have Z = 28, MH = 1.007825 u, N = A – Z = 34, mn = 1.008665 u, and
A
Z M = 61.928349 u. The mass defect is then M = 0.585361 u. The
corresponding binding energy is just EB = (0.585361 u)(931.5 MeV) = 545 MeV.
The binding energy per nucleon is 545 MeV/62 = 8.795 MeV/nucleon.
• At left is the binding energy/nucleon
for all nuclides. H has the smallest, but
notice that He is unusually stable
compared to its neighbors. The peaks
show that pairing of nucleons improves
binding energy.
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10.
The nuclear force
• Important characteristics of the nuclear force:
It does not depend on charge. Protons and neutrons are bound. It has a
short range, of the order of nuclear dimensions.
Because of its short range, a nucleon only interacts with those in its
immediate vicinity. The lack of long-range interaction is called
saturation. The electromagnetic force and gravity do not show such
saturation.
It favors binding of pairs of protons or neutrons with opposite spins and
with pairs of pairs (a pair of protons and a pair of neutrons, each pair
having opposite spins).
Surface tension term Repulsion term N-Z parity term Pairing term
• We will discuss two models of the nucleus, the liquid-drop model and the
shell model. The liquid drop model is phenomenological (no detailed
theory for the numbers, only tweaking to make it fit observations). In this
2
model: E C A C A2/3 C Z ( Z 1) C ( A 2Z ) C A 4/2
B 1 2 3 4 5
A1/3 A
where C1 = 15.75, C2 = 17.80, C3 = 0.7100, C4 = 23.69, C5 = 39, all in MeV.
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11.
Nuclear models
• For the shell model, we start with the concept of
the potential energy function for nucleons. From
this perspective, the potential well for protons is
shallower than for neutrons (see figure at right).
• We can use these potentials to calculate wave
functions, which leads to the idea of filled or
partially filled shells and subshells, just as for
electrons around atoms. Before, we found stable
electron configurations for values Z = 2, 10, 18, 36,
54 and 86.
• For nuclei, the rules are different because U(r) is
different, and also the spin-orbit coupling is much
stronger. It is found that N or Z = 2, 8, 20, 28, 50,
82 or 128 are particularly stable. These are called
magic numbers. Doubly magic nuclides are:
4 16 40 48 208
2 He, 8 O, 20 Ca, 20 Ca, 82 Pb
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12.
Nuclear stability and radioactivity
• Radioactivity is the decay of unstable
nuclides by the emission of particles and
electromagnetic radiation.
• Figure 43.4 (right) is a Segrè chart
showing N versus Z for stable nuclides.
Notice that as A increases, the stable
nuclides leave the N = Z line and prefer
more neutrons that protons. This is
probably due to the increasing influence
of repulsion among the protons.
• There are no stable nuclides with A = 5
8
or 8. The 4 Be nuclide decays
immediately into two He nuclei.
• The highest A is 209. There are so-
called islands of stability above this.
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13.
Alpha decay
• An alpha particle is a 4He nucleus, which is very stable. Large nuclei can
decay by splitting into a smaller nucleus and an alpha particle, e.g.
226 222
88 Ra 86 Rn 42 He
• Alpha decay is possible whenever the parent nuclide is more massive than the
sum of the two daughter products, e.g. Example 43.5: Show that decay is
possible in the reaction above, where the Ra isotope is 226.025403 u, the Rn
isotope is 222.017571 u, and He is 4.002603 u.
The mass difference is (226.025403 – 222.017571 – 4.002602) u = +0.005229 u
The energy equivalent is (0.005229)(931.5) MeV = 4.871 MeV. The kinetic
energy of the particle can be determined from K = ½ mv2 = p2/2m. The
momentum is conserved, so the particle gets 222/(222+4) of the total energy.
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14.
Beta and gamma decay
• There are three types of beta decay: beta-minus, beta-plus, and electron capture.
beta-minus decay n p e
•
• beta-plus decay p n e
• electron capture p n e
• Note the second reaction must take place within a nucleus, or else it is energetically
impossible, i.e. it takes external energy (as from an excited nucleus) to make it
happen.
• A beta-minus – particle is an electron, beta-plus is a positron.
• When radium decays to emit an particle, the can have kinetic energy of either
4.784 MeV, or 4.602 MeV. The latter is due to the product radon nucleus being left
in an excited state. The radon nucleus will then relax to the nuclear ground state,
and emit a photon of energy 0.186 MeV. The photon is called a gamma ray.
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15.
Natural radioactivity
• Certain nuclei decay naturally,
which is called natural
radioactivity. Figure 43.7 (right)
shows a Segrè chart for the 238U
decay series.
• Each decay is labeled by a time,
called the half-life—note that
some are very long (billions of
years) and some are extremely
short (minutes or seconds). This
is related to the potential barrier
through which the decay product
has to tunnel.
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16.
Activities and half-lives
• The half-life is the time for the number of
radioactive nuclei to decrease to one-half of their
original number.
• Because the number of decays is proportional to
the number of atoms available to decay, e.g.
dN Ndt
(the minus sign indicates a loss), the number of
remaining nuclei decreases exponentially (see
figure at right). The solution to the above
equation is found by rearranging and integrating:
dN N
dt ln t N N 0e t
N N0
• To find the half-life, just determine when
ln 2 An established unit of
N = N0/2, which is when e t
12 , t t1/2
or .
radiation activity (dN/dt) is
called the Curie:
• If you start with N0 nuclei, after 1 half-life you 1 Ci = 3.70x1010 decay/s.
will have N0/2, and after another half-life you In SI units, 1 decay/s is called
will have N0/4, etc.
the becquerel (Bq).
The© 2012
quantity 1/Inc.is called the mean lifetime.
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17.
Activities and half-lives
• Example 43.8: Activity of 57Co. The isotope 57Co decays by electron capture to
57Fe with a half-life of 272 d. The 57Fe nucleus is produced in an excited state,
and it almost instantaneously emits gammy rays that we can detect. (a) Find the
mean lifetime and decay constant for 57Co. (b) If the activity of a 57Co radiation
sources is now 2.00 Ci, how many 57Co nuclei does the source contain?
(c) What will be the activity after 1 year?
(a) Since the half-life t1/2 = ln 2/ = (272 d)(86400 s/d) = 2.35 107 s, the decay
constant is = ln 2 / 2.35 107 s = 2.95 10-8 s-1. Tmean = 1/ = 3.39 107 s =
392 d.
(b) The activity
dN
N 2.00 Ci (2 10 6 )(3.70 1010 decay/s) 7.4 10 4 decay/s
dt
so N = 7.4 104 decay/s / 2.95 10-8 s-1 = 2.51 1012 nuclei.
(c) After 1 year, the number of nuclei remaining will be
t (2.9510 8 s 1 )(3.156107 s)
N (t ) N 0 e N 0e 0.394 N 0
The activity will decline by a like factor, or 0.394 (2.00 Ci) = 0.788Ci
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18.
Activities and half-lives
• Example 43.9: Radiocarbon Dating. The isotope 14C decays via beta-minus decay to 14N
with a half-life of 5730 y. Before 1900 the activity per unit mass of atmospheric carbon
due to the presence of 14C averaged about 0.255 Bq per gram of carbon. (a) What fraction
of carbon atoms were 14C? (b) In analyzing an archaeological specimen containing 500
mg of carbon, you observe 174 decays in one hour. What is the age of the specimen?
(a) Since the half-life t1/2 = ln 2/ = (5730 y)(3.156 107 s/y) = 1.808 1011 s, the decay
constant is = ln 2 / 1.808 1011 s = 3.83 10-12 s-1.
We get the number of atoms from the activity
dN dN / dt 0.255 s 1
N N (t ) 12 1
6.65 1010 atoms
dt 3.86 10 s
This is the number of 14C atoms, but how many C atoms are there in 1 gram (1/12.011
mol)? Avogadro’s number is 6.022 1023 atoms/mol, so the total in 1 gram is 6.022
1023 atoms/mol * 1/12.011 mol = 5.01 1022 atoms. The fraction of 14C atoms was 1.33
10-12.
dN
(b) The activity when it died would be N (3.83 10 12 )(6.65 1010 ) / 2 0.127 Bq
dt
Now it is 174 decays/3600 s = 0.048 Bq, or a factor of 0.379 of what it was. Thus,
N(t)/N0 = 0.379. The age is then:
ln( N / N 0 ) ln(0.379) 11
t 12 1
2.53 10 s 8020 y
3.83 10 s
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19.
Biological effects of radiation
• Follow the text discussion of the biological effects of
radiation.
• Table 43.3 gives the RBE for several types of radiation.
• Figure 43.9 shows sources of U.S. radiation exposure.
• Follow Example 43.10 about
a medical x ray.
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20.
Nuclear reactions
• A nuclear reaction is a rearrangement of
nuclear components due to bombardment by
a particle rather than a spontaneous natural
process.
• The difference in masses before and after
the reaction corresponds to the reaction
energy Q.
• Follow Example 43.11, which looks at
exoergic and endoergic reactions.
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21.
Nuclear fission
• Nuclear fission is a decay
process in which an
unstable nucleus splits into
two fragments (the fission
fragments) of comparable
mass.
• Figure 43.11 (right) shows
the mass distribution of the
fission fragments from the
fission of 236U*.
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22.
Liquid-drop model
• The liquid-drop model
helps explain fission.
See Figure 43.12 below.
• Figure 43.13 (right)
shows the potential
energy function of two
fission fragments.
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23.
Chain reactions
• The neutrons released by fission can cause a chain reaction (see
Figure 43.14 below).
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24.
Nuclear reactors
• A nuclear reactor is a system in which a controlled
nuclear chain reaction is used to liberate energy.
• Figure 43.15 below illustrates a nuclear power plant.
• Follow Example 43.12.
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25.
Nuclear fusion
• In a nuclear fusion reaction, two or more light nuclei
fuse to form a larger nucleus.
• Figure 43.16 below illustrates the proton-proton chain,
which is the main energy source in our sun.
• Follow Example 43.13.
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