Download Nuclear Physics

Physics 2018: Great Ideas in Science:
The Physics Module
Nuclear Physics Lecture Notes
Dr. Donald G. Luttermoser
East Tennessee State University
Edition 1.0
Abstract
These class notes are designed for use of the instructor and students of the course Physics 2018:
Great Ideas in Science. This edition was last modified for the Fall 2007 semester.
III.
Nuclear Physics
A. The Structure of Matter.
1. Matter exists in 3 different thermodynamic states:
a) A solid is a rigid body −→ takes a lot of energy to change
its shape. Solids can be classified into two types:
i)
ii)
Crystalline solids have atoms that are structured in an orderly fashion.
Amorphous solids have randomly arranged atoms.
b)
A liquid is fluid in nature −→ moderate energy required
to change its shape.
c)
A gas is also fluid in nature −→ little energy required to
change its shape.
2. If a gas gets hot enough, electrons circling the nucleus of the
atoms in the gas are “ripped” away from the nucleus =⇒ the gas
becomes ionized =⇒ ionized gas is called a plasma.
3. Matter consists of a distribution of particles (atoms and molecules).
a) Atoms consist of a nucleus surrounded by electrons (which
are negatively [–] charged). The nucleus consists of protons (positive [+] charge) and neutrons (no [0] charge).
b)
H, He, Li, Be, B, C, N, O, etc. are the elements of the
periodic table = atoms. The number of protons in the
nucleus defines each atom.
c)
For elements heavier than H, typically the number of neutrons is equal to the number of protons. Isotopes of
III–1
atoms contain different numbers of neutrons in the nucleus (e.g., 12 C, 13C, and 14 C are isotopes of carbon).
d)
Neutral Atoms: # protons = # electrons.
i)
If electrons are taken away from the atom such
that the number of protons exceeds the number of
electrons, the atom becomes a positive ion (e.g.,
H+ = H II = singly ionized hydrogen).
ii)
If the number of electrons exceeds the number of
protons in the nucleus, the atom becomes a negative ion (e.g., H− = negative hydrogen ion, a
hydrogen atom with two electrons instead of one).
iii) The ionization stage of an atom can be labeled
in a variety of different ways:
=⇒ Roman numerals: I = neutral atom (e.g.,
He I, neutral helium), II = singly ionized
(e.g., Fe II), III = doubly ionized (e.g., C III),
etc.
=⇒ ‘+’ exponents (positive ions): if no exponent appears, then we have a neutral atom
(e.g., He, neutral helium), ‘+’ = singly ionized (e.g., Fe+ ), ‘3+’ = triply ionized (e.g.,
C 3+ ), etc.
=⇒ ‘−’ exponents (negative ions): ‘−’ = one extra electron (e.g., H− ), ‘−−’ = two extra electrons (e.g., C−− ), etc.
e) Molecules are a collection of atoms that are bound together by molecular bonds.
i)
Salt: NaCl (1 sodium atom + 1 chlorine atom).
III–2
ii)
Water: H2 O (2 hydrogen atoms + 1 oxygen atom).
iii) Methane: CH4 (1 carbon atom + 4 hydrogen
atoms).
f ) Molecules can adhere to each other through chemical
bonds making a structured lattice =⇒ solids.
B. An Overview of Nuclear Reactions.
1. In order to get nuclear particles to interact with each other, one
needs to collide these particles with a very high kinetic energy
(i.e., high speed).
a) Since the nucleus is composed of both neutrons (zero charge)
and protons (positive charge), two nuclei close together
will experience a strong repulsive force due to Coulomb’s
electric force law.
b)
Remember that forces are derivatives of potential energy.
Hence if one wishes to bring two nuclei closer and closer
together, one must increase the kinetic energy of said nuclei to high enough values to overcome the electric potential field.
c)
From thermodynamics heat is nothing more than the average kinetic energy of the particles that make up a system
(i.e., matter).
d)
Hence, one way to get nuclei moving fast is to heat them
up.
2. Reactions in chemistry and physics can be one of two types:
a) Endothermic reactions absorb energy.
III–3
b)
Exothermic reactions release energy.
3. The element iron (Fe) has a nucleus that is the most stable of all
the atomic nuclei.
a) Elements lighter (i.e., less massive) than iron will produce
exothermic nuclear reactions when they fuse together to
make heavier elements =⇒ nuclear fusion.
b)
Elements heavier (i.e., more massive) than iron will produce exothermic nuclear reactions when they break apart
to make lighter elements =⇒ nuclear fission.
4. The amount of energy released duing these exothermic reactions
is proportional to the amount of the mass difference between
the parent (original) and daughter (offspring) particles through
Einstein’s famous equation
E = mc2 .
(III-1)
5. Conservation Laws.
a) If we assign a baryon number B of +1 to each baryon
(nucleon or hyperon) and –1 to an antinucleon or antihyperon, then in a closed system
X
b)
(III-2)
Similarly, if we assign a lepton number L of +1 to each
lepton (i.e., e− , µ, ν, etc.) and of –1 to antileptons (i.e.,
e+ , µ, ν, etc.), then in a closed system
X
c)
B = constant.
L = constant.
(III-3)
A similar conservation law does not exist for bosons —
the mesons and field particles (i.e., photons).
III–4
d)
Charge must be conserved in a nuclear reaction.
e) Mass-energy, via E = m c2 , must be conserved in a nuclear
reaction.
f ) Momentum must be conserved. Hence a matter-antimatter
reaction must create two photons to conserve momentum
(e.g., e− + e+ → 2γ).
C. Thermonuclear Reactions.
1. In 1938, it became clear that the long-term energy source for
stars must be thermonuclear fusion reactions. In these reactions,
lighter elements burn to form heavier elements =⇒ nucleosynthesis.
2. Two nuclei will fuse to form one nuclei if they come within
10−13 cm of each other — but they must be moving fast enough to
overcome the Coulomb repulsion that exists between like charged
particles.
a) Particles must be at a high temperature to be moving fast.
b)
This high temp completely ionizes all of the nuclei.
c)
Temps must build even more to get the kinetic energy to
overcome the Coulomb barrier.
3. In main sequence stars, H is fused into He. Since H is composed
of 1 baryon and He, 4 baryons (2p + 2n), 4 H nuclei must be
used to construct one He nuclei.
4 × mH = 4 × 1.0078 amu = 4.0312 amu
= 4.0026 amu
−mHe
∆m
= 0.0286 amu
III–5
a) This mass deficit, ∆m, is converted into energy:
E = ∆m c2 = (0.0286) × (1.66 × 10−24 gm)
×(9.00 × 1020 cm2/s2 )
= 4.3 × 10−5 erg
b)
From this calculation, we see that the energy release efficiency, η, of this H→He reaction is 0.0286/4.0312 = 0.0071
=⇒ only 0.71% of the original mass of H is converted to
energy!
c)
With this in mind, we can rewrite Einstein’s famous equation as
E = η m c2 ,
(III-4)
where η is the efficiency of the reaction and m is the initial
mass in the reaction.
d)
We can calculate the total energy the Sun will release
during its main sequence lifetime. Since the reactions are
only being carried out in the core of the Sun and this core
contains about 10% of the Sun’s mass, the total energy
release will be
Etot = 0.1M η c2
= 1.28 × 1051 ergs.
e) The present luminosity of the Sun is 3.90 × 1033 erg/s. If
the Sun’s luminosity remains somewhat constant while on
the main sequence, we can determine its main sequence
lifetime:
tMS () = Etot/L = 3.28 × 1017 sec = 1.04 × 1010 yrs,
the Sun’s main sequence lifetime is about 10 billion years.
Since the Sun is currently 5 billion years old, it is at midlife.
III–6
4. As previously mentioned, high temperatures are needed to overcome the Coulomb repulsion of the charged nuclei. But how high
a temperature?
a) From classical physics, we can set the average kinetic energy of the particles involved equal to the thermal energy
of the particles and solve for the temperature (see page
334 of Carroll and Ostlie). This gives
2Z1 Z2 e2
Tclassical =
,
3kB r
(III-5)
where Z1e is the charge on particle 1, Z2 e is the charge on
particle 2, and r is the distance where a nuclear reaction
will occur (about the size of a nucleus, 10−13 cm).
b)
For 2 protons coming together (Z1 = Z2 = 1), this gives
a temperature of 1010 K, whereas the Sun’s central temperature is only 1.58 × 107 K.
c)
We could also investigate this in terms of energy. For
the Sun’s central temperature, each proton will have a
thermal energy of 1 keV, whereas the Coulomb potential
barrier is 1000 keV (1 MeV)! Not all the particles have
this energy, some are moving at much quicker velocities,
hence have higher thermal energies and temperatures following the Maxwell-Boltzmann distribution of velocities.
Unfortunately, the number of particles in the tail of this
Maxwellian is insufficient to account for the Sun’s luminosity.
III–7
d)
We can now turn to quantum mechanics to solve the problem. As discussed in the subsection on particle spin, in
reality elementary particles are not little billiard balls colliding with each other as a result of following trajectories.
Instead, they follow probability distributions described by
their wave functions. In quantum mechanics, there is a
small probability that wave functions can penetrate energy barriers that are higher than the energy of the wave
function. This effect is known as quantum tunneling.
e) Using quantum mechanics, we can describe a temperature needed to produce a sufficient number of tunneling
events to sustain a nuclear reaction (see page 335 of the
textbook) as
Tquantum =
4µm Z12 Z22 e4
,
3kB h2
(III-6)
where µm is the reduced mass of the colliding “particles”
and h is Planck’s constant.
f ) In this equation, two protons can come together (i.e., fuse)
at a temperature of 107 K, which is consistent with the
central temperature deduced for the Sun.
g) A more detailed calculation from statistical mechanics
shows that the bulk of the energy is being liberated by
reactions involving particles in the high energy tail of a
Maxwellian distribution.
i)
Particles with energies at the Gamow Peak will
be the ones that supply most of the energy through
thermonuclear reactions.
ii)
The Gamow Peak corresponds to a local maximum in the two probability functions: the e−E/kB T
III–8
Maxwell-Boltzmann distribution term and the e−bE
quantum tunneling penetration term, where,
−1/2
2
23/2π 2 µ1/2
m Z1 Z2 e
.
b≡
h
iii) The Gamow Peak for a given temperature will
occur at the energy of
bkB T
E◦ =
2
!2/3
,
(III-7)
for the Sun, the Gamow Peak is at 6 keV.
5. By making use of statistical mechanics in conjunction with quantum mechanics, stellar interior modelers set up power laws that
describe the energy production rate per unit mass of the form
ε = ε◦ Xi Xxρα T β ,
(III-8)
where the X’s are the mass fractions of the fusing particles, and
ε◦ , α. and β are constants that depend upon the reactions involved (more to come on this).
D. Various Reaction Chains
1. Two different fusion processes convert H into He, the first is
7
<
important for stars with Tc <
∼ 1.8 × 10 K (M ∼ 1.3M , ∼F5 V
star) and is called the proton-proton chain.
a) The first of this reaction chain is called the PP I chain:
Energy
Reaction
Reaction
Released
Time
(MeV)
1
1
2
+
H + H −→ H + e + νe
1.442
1.4 × 109 yr
1
H + 2H −→ 3 He + γ
5.493
6 sec
3
3
4
1
1
He + He −→ He + H + H
12.859
106 yr
III–9
1
i)
ii)
H = hydrogen atom (1 proton).
2
H = heavy hydrogen (1 proton + 1 neutron) =
deuterium.
iii)
3
iv)
4
v)
vi)
He = light helium (2 protons + 1 neutron).
He = helium (2 protons + 2 neutron) = alpha
particle.
γ = Gamma ray photon.
e+ = positron (positive charge) = anti-electron
(antimatter). This positron interacts with the free
electrons in the core virtually immediately which
produces 2 additional gamma ray photons.
vii) νe = electron neutrino (neutral particle). The
neutrino’s absorption cross section is negligible and
leaves the stellar core (and star) immediately without further interaction. The energy loss from the
neutrino is 0.263 MeV which has not been included
in the Energy Released column.
viii) In this PP I change, please note that the following reaction can take place 1.4% of the time that
the first reaction takes place
1
H + e− + 1 H −→ 2 H + νe
the so-called “pep” (proton-electron-proton) reaction which releases 1.4 MeV and loses an additional
1.4 MeV in energy loss from the neutrino.
III–10
ix)
Note that the Energy Released column is a combination of the energy of any gamma rays created,
the kinetic energy = thermal energy picked up by
the resulting nuclei as a result of this reaction, and
the energy gained by the positron annihilation.
x)
The times listed for the Reaction Time column is
that for the Sun’s central temperature.
xi)
Note that the first two reaction steps must occur
twice before the last one can take place.
xii) The last step of this reaction chain is occurs
69% of the time in comparison to the other two
PP chains in the production of 4 He in the Sun.
xiii) This PP I chain dominates the other PP chains
in stars with central temperatures of T <
∼ 1.6 ×
107 K.
xiv) Note the long average time it takes for the first
and third reaction steps to take place for a single
particle. However when normalized by the total
number of particles in the Sun’s (or star’s) core,
about 9.0 × 1037 of these reactions take place per
second!
b)
A second chain, called PP II, also can occur (31% of the
time in the Sun) in the production of 4 He once the first
two steps of the PP I chain occur.
III–11
Energy
Reaction
Reaction
Released
Time
(MeV)
3
4
7
He + He −→ Be + γ
1.586
1.0 × 106 yr
7
Be + e− −→ 7 Li + νe
0.861
0.4 yr
7
1
4
4
Li + H −→ He + He
17.347
104 yr
7
i)
7
ii)
iii)
c)
Be = beryllium-7 atom (4 protons + 3 neutrons).
Li = lithium-7 atom (3 protons + 4 neutrons).
The neutrino energy loss in this PP chain is 0.80 MeV.
iv)
This PP II chain dominates the other PP chains
<
in stars with central temperatures of 1.6×107 <
∼T ∼
2.5 × 107 K.
v)
The Energy Released and the Reaction Time have
the same meaning as they did for the PP I chain.
A third and final PP chain, called PP III, occurs only
0.3% of the time in the Sun in the production of 4 He once
the first two steps of the PP I chain occurs and the first
step of the PP II chain occurs.
Reaction
7
Be + 1 H −→
8
B −→
8
Be −→
8
B+γ
Be + e+ + νe
4
He + 4 He
8
III–12
Energy Reaction
Released
Time
(MeV)
0.135
70 yr
17.98
1 sec
0.095
1 sec
i)
ii)
8
B = boron-8 atom (5 protons + 3 neutrons).
8
Be = beryllium atom (4 protons + 4 neutrons).
iii) The neutrino energy loss in this PP chain is 7.2 MeV.
The Davis solar neutrino experiment, which detected only 1/3-rd of the predicted solar neutrinos
was most sensitive to these 8 B beta decay neutrinos.
Recently, neutrinos have been found to oscillate between the 3 known neutrino states which accounts
for the low detection rate of the Davis experiment.
iv)
This PP III chain dominates the other PP chains
7
in stars with central temperatures of T >
∼ 2.5×10 K
in the production of helium.
v)
The Energy Released and the Reaction Time have
the same meaning as they did for the PP I chain.
7
>
2. For more massive stars (Tc >
∼ 1.8 × 10 K, M ∼ 1.3M ,
∼F5 V star), the CNO cycle is the dominant reaction chain.
a) This reaction chain uses carbon as a catalyst:
Reaction
12
C + 1H
13
N
13
C + 1H
14
N + 1H
15
O
15
1
N+ H
i)
12
−→
−→
−→
−→
−→
−→
13
N+γ
13
C + e+ + νe
14
N+γ
15
O+γ
15
N + e + + νe
12
C + 4 He
Energy
Released
(MeV)
1.95
2.22
7.54
7.35
2.71
4.96
Reaction
Time
1.3 × 107
7 min
2.7 × 106
3.2 × 108
82 sec
1.1 × 105
C = carbon-12 (6 protons + 6 neutrons).
III–13
yr
yr
yr
yr
13
ii)
C = carbon-13 (6 protons + 7 neutrons).
iii) 13N = nitrogen-13 (7 protons + 6 neutrons) [radioactive].
14
iv)
15
v)
vi)
N = nitrogen-14 (7 protons + 7 neutrons).
N = nitrogen-15 (7 protons + 8 neutrons).
15
O = oxygen-15 (8 protons + 7 neutrons) [radioactive].
vii)
16
O = oxygen-16 (8 protons + 8 neutrons).
viii) The neutrino energy loss in the 13N beta decay is 0.710 MeV and the neutrino energy loss is
1.000 MeV for the second 15 O beta decay.
ix)
The Energy Released and the Reaction Time have
the same meaning as they did for the PP I chain.
b)
Note that this reaction sequence does not make any new
elements other than He!
c)
For the last step in the CNO cycle, an additional set of
reactions can take place:
III–14
15
N + 1H
16
O + 1H
17
F
17
O + 1H
i)
ii)
17
Energy
Reaction
Reaction
Released
Time
(MeV)
16
−→
O+γ
12.126
1.0 × 107 yr
−→ 17F + γ
0.601
3.0 × 1010 yr
−→ 17O + e+ + ν
2.762
3 min
−→ 14N + 4He
1.193
2.0 × 1011 yr
F = fluorine-17 (9 protons + 8 neutrons).
18
O = oxygen-16 (8 protons + 10 neutrons).
iii) The neutrino energy loss in the 17 F beta decay is
0.94 MeV.
iv)
The resulting 14 N isotope can then be used back
in the
14
N + 1 H −→
15
O+γ
reaction in the primary CNO cycle.
3. The thermonuclear reaction rate, ε (in erg/gm/s), is very sensitive to temperature. For the two hydrogen to helium reaction
chains (i.e., the proton-proton chain and CNO cycle), we can
write Eq. (IV-48) as two separate equations:
εpp = ε◦ ρ X
2
T
106
εcc = ε◦ ρ XXCN
!β
T
106
(III-9)
!β
,
(III-10)
where X is the mass fraction of hydrogen (as defined on page IV-8
of the course notes), XCN is the weighted average of the combined
mass fraction of carbon and nitrogen (since these species are the
III–15
two lead-off species of the two CNO cycles), note that typically
XCN =
1
Z ,
3
(III-11)
with the Z being the metalicity mass fraction. Finally, ε◦ and α
are temperature dependent constants given in the following table
(from Bosmas-Crespin, Fowler, Humblet 1954, Bull. Soc. Royale
Sciences Liege, No. 9-10, 327).
6
T /10
4–6
6 – 10
9 – 13
11 – 17
16 – 24
εpp
log ε◦ α
T /106
–6.84 6 12 – 16
–6.04 5 16 – 24
–5.56 4.5 21 – 31
–5.02 4 24 – 36
–4.40 3.5 36 – 50
εcc
log ε◦
–22.2
–19.8
–17.1
–15.6
–12.5
α
20
18
16
15
13
In this table, for overlapping temperatures, a weighted average
is used to get the final rate.
4. We will see shortly, that when a main sequence uses up all of its
H fuel in its core, the now He-rich core will contract and heat
up. When temperatures exceed 108 K, helium fusion can begin.
Helium fuses via the triple-α process. The ash of this reaction
is carbon.
III–16
a) The reaction is as follows:
Energy
Reaction
Released
(MeV)
4
4
8
He + He ←→ Be + γ
-0.0921
8
4
12
Be + He −→
C+γ
7.37
i)
Since this reaction chain is not occurring in the
Sun, we have not reported on any reaction times
here since they are very temperature dependent.
ii)
As can be seen, three α-particles (i.e., He nuclei)
fuse to become one carbon nuclei.
iii) 8Be is unstable and quickly decays, so there is
not much berylium around for the 2nd chain to take
place. For every 1 berylium nuclei, there are 1010
α-particles, however this ratio is more than enough
to release enough energy to power a red giant star.
Note that this reaction actually drains energy away
from the gas =⇒ it is an endothermic reaction.
All reactions we have mentioned up to now have
been exothermic =⇒ they release energy.
b)
The 3α process has an even greater sensitivity to temperature than the CNO cycle:
−8
ε3α ≈ 10 ρ Y
3
T
106
!30
erg/gm/s,
(III-12)
where Y is the fractional abundance of helium per unit
mass.
c)
This is the way the Universe makes carbon. As such,
the C atoms that make up our DNA were created in an
III–17
ancient red giant star that no longer exists. To quote Carl
Sagan, we are star stuff!
d)
Should the core of a red giant obtain temperatures that
exceed a few hundred million Kelvins, another reaction
can take place via an alpha (α) capture:
12
C + 4He −→
16
O+γ
which releases 7.161 MeV of energy.
e) Most of the
16
O in the Universe is made in this fashion.
5. Finally, if somewhat higher temperatures are ever encountered
inside a star, which happens during stellar evolution of massive
stars, even heavier elements can be created from the fusion of
additional α-particles and α-particle by-products:
Energy
Minimum
Reaction
Released
Temperature
(MeV) Required (106 K)
14
O + 4 He −→ 20Ne + γ
4.730
700
20
4
24
Ne + He −→
Mg + γ
9.317
1500
24
4
28
Mg + He −→
Si + γ
9.981
1800
28
4
32
Si + He −→
S+γ
6.948
2500
32
4
36
S + He −→
Ar + γ
6.645
3500
12
12
24
C + C −→
Mg + γ
13.930
800
16
16
32
O + O −→
S+γ
16.539
2000
a) Various silicon burning reactions can occur at temperatures exceeding 3 × 109 K. Silicon burning produces the
iron (Fe) group elements.
b)
Once Fe is formed, reactions that produce heavier elements are all endothemic and have a tough time forming
via the standard thermonuclear burning. Such elements,
III–18
and the elements not built upon α-particles, are created
via the r- (for rapid neutron capture) and s- (for slow
neutron capture) processes. These processes will be discussed in the supernovae section of the course notes (i.e.,
§VII).
c)
The reaction times of this heavy element nucleosynthesis
will be discussed in the stellar evolution sections of the
notes.
III–19