Download Lecture 30/3 Nuclear processes Ulf Torkelsson 1 Nuclear reactions

Lecture 30/3
Nuclear processes
Ulf Torkelsson
1
Nuclear reactions
Consider the reaction
I (Ai , Zi ) + I (Aj , Zj ) → K (Ak , Zk ) + L (Al , Zl )
(1)
The amount of energy released in this reaction is
Qijk = (Mi + Mj − Mk − Ml ) c2 .
(2)
By summing over all possible nuclear reactions that have reaction rates Rijk we can get the amount
of energy released per unit mass as
q=
ρ X 1 Xi Xj
Rijk Qijk .
m2H
1 + δij Ii Aj
(3)
ijk
Strictly speaking this is an upper limit to how much energy that can heat the star, because some
of the energy might be carried away by neutrinos, that pass straight through the star.
The energy that is perhaps released in a nuclear reaction is due to that the constituent nucleons
are bound more tightly by the strong force in the new nucleus than in the old nuclei. For light
nuclei the binding energy per nucleon is in general increasing with the size of the nucleus up to
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Fe, in which they are the most strongly bound, and in heavier nuclei the binding energy per
nucleon decreases with the size of the nucleus. Therefore there are two ways of deriving energy
from atomic nuclei, by fusing light nuclei, and by fissioning heavy nuclei.
The nuclei that are involved in a fusion reaction are repelling each other through the Coulomb
force, since they are both positively charged. This Coulomb repulsion is so strong that the kinetic
energy of the nuclei is insufficient in bringing them any closer together than a thousand times the
radius of a nucleus, which is the distance scale at which the strong force works, however because of
quantum tunneling there is a small probability that the nuclei will anyway react with each other.
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Hydrogen burning
The most common element in the universe is hydrogen, so it is natural to assume that this is the
main fuel in the nuclear reactions. There are two ways in which hydrogen can be fused into helium.
2.1
p-p chain
Two protons cannot form a bound system so the first step in hydrogen burning is the weak reaction
p + p →2 D + e+ + ν,
(4)
in which one of the protons is converted to a neutron. The deuteron can then react with a new
proton
2
D + p →3 He + γ.
(5)
After this there are two possible reactions
3
He +3 He →4 He + 2p,
(6)
3
(7)
which completes the p-p I chain, and
He +4 He →7 Be + γ.
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In the p-p II chain we then get
7
Be + e−
7
Li + p
→
7
Li + ν
→ 24 He,
(8)
(9)
and in the p-p III chain
7
8
Be + p →
8
8
B
Be
8
→
B+γ
(10)
+
Be + e + ν
(11)
4
(12)
→ 2 He.
The energy that is released in this reaction is
Qp−p = (4mp − m4 He ) c2 = 26.73 MeV.
(13)
The amount of energy that is lost through the neutrino emission varies between 0.26 MeV during
the production of deuterium to 7.2 MeV in the boron decay, which occurs only in a small fraction
of the reactions. For this reason the amount of energy that on average becomes available to the
star is 26 MeV, and
qp−p ∝ ρT 4 .
(14)
2.2
The CNO-cycle
Hydrogen can also be converted to helium via the CNO-cycle that uses carbon
12
C +1 H
→
13
N
→
13
C+e +ν
(16)
C +1 H
→
14
N+γ
(17)
→
15
O+γ
(18)
O →
15
C + He,
(20)
O+γ
(21)
13
13
14
1
N+ H
15
1
N+γ
+
+
N+e +ν
4
(15)
(19)
15
N+ H
→
12
14
N +1 H
→
15
O →
15
N+e +ν
(22)
→
16
O+γ
(23)
→
17
F+γ
(24)
F →
17
O + e+ + ν
(25)
14
N +4 He
(26)
or nitrogren
15
15
16
1
N+ H
1
O+ H
17
17
O +1 H
→
+
to catalyse the reaction. It is important here that as the rate of the proton reactions will increase
with temperature, the rates of the β-decays are independent of time, thus at sufficiently high
temperatures the reaction rate will be limited by the β-decays. The energy that is available to the
star is slightly lower, 25 MeV, than in the p-p chain. That is because more energy is lost via the
neutrinos in this case. The energy generation rate can be written as
3
qCNO ∝ ρT 16 .
(27)
He +4 He →8 Be,
(28)
Helium burning
The reaction
4
2
in itself does not work for burning helium, because the life time of 8 Be is only 2.6 × 10−16 s.
However this is still longer than the mean collision time of α-particles at 108 K. There is therefore
a non-negligible probability that
8
Be +4 He →12 C
(29)
during this short time. As a matter of fact this probability is enhanced by there being a resonant
energy level at 7.65 MeV in 12 C, which agrees closely with the energy that is released in this triple-α
reaction, 7.275 MeV. Because this is essentially a three-particle interaction the energy generation
rate is
ρ3α ∝ ρ2 T 40 .
(30)
As the carbon abundance increases one can also have
12
4
C +4 He →16 O.
(31)
Carbon and oxygen burning
Carbon burning starts at 5 × 108 K
12
C +12 C
24
Mg + γ
(32)
12
C+
C
→
23
Mg + n
(33)
12
C +12 C
→
23
Na + p
(34)
12
C +12 C
→
20
Ne + α
(35)
→
16
O + 2α
(36)
12
C+
12
→
12
C
and oxygen burning starts at 109 K
16
O +16 O →
32
S+γ
(37)
O →
31
S+n
(38)
O +16 O →
31
P+p
(39)
O →
28
Si + α
(40)
O →
24
Mg + 2α.
(41)
16
O+
16
16
16
16
O+
16
O+
16
On average 13 MeV is released in the carbon burning and 16 MeV in oxygen burning. Note that
the light nuclei that are released in these reactions will quickly react with other heavy nuclei in
the environment.
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Nuclear statistical equilibrium
As the temperature increases further, photodissociation of the nuclei starts to become important.
For instance at 1.5 × 109 K we have
20
Ne + γ →16 O + α,
(42)
and at 3 × 109 K even silicon can photodisintegrate. Thus we enter a regime in which a statistical
equilibrium is established between the fusion reactions that build up heavier nuclei, and photodisintegration processes. However there is still a small drift towards the iron group nuclei, which do
not photodisintegrate until the temperature is 7 × 109 K.
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Neutron capture
Elements heavier than iron cannot be formed through fusion reactions since such reactions would
be endothermic. Instead heavier elements are formed through neutron capture
I (A, Z) + n
→ I1 (A + 1, Z)
3
(43)
I1 (A + 1, Z) + n
→ I2 (A + 2, Z)
(44)
I2 (A + 2, Z) + n
→ I3 (A + 3, Z) .
(45)
Eventually the nucleus will have gained so many neutrons that it becomes unstable
IN (A + N, Z) → J (A + N, Z + 1) + e− + ν̄,
(46)
and the new element may then continue to go through new β-decays or catch new neutrons.
Depending on the rate of the neutron capture compared to that of the β-decay, the nucleus may
either decay as soon as it becomes unstable the first time, the slow (s)-process, or it might capture
more neutrons before it decays, the rapid (r)-process. The r-process will in general produce more
neutron rich nuclei.
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Pair production
When the photons are sufficiently energetic, two photons can produce an electron-positron pair.
The simple estimate of the temperature based on that kT = 2hν = 2me c2 , yields a temperature
of 1.2 × 1010 K. However, some of the photons are sufficiently energetic already at temperatures
above 109 K.
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Iron photodisintegration
At high temperatures there are also photons that can drive the reaction
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Fe → 13 4 He + 4n,
(47)
and at 7 × 109 K there are even photons that can disintegrate helium into protons and neutrons.
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