Transcript Document

Lecture 11
Somak Raychaudhury
www.sr.bham.ac.uk/~somak/Y2SiU/
Guest Lecturer: Dr W J Chaplin
 Nuclear energy generation in stars
 Revision: Central temperature and pressure in a
star
 Can fusion occur at the centre of stars:
overcoming Coulomb repulsion
 Gamow’s solution
 The p-p chain
Revision
dP(r)
GM(r)

(r)
2
dr
r
• We derived the equation of hydrostatic equilibrium,
balancing gravity and thermal pressure
• Central
 pressure of a star like the Sun
• Central temperature of the Sun
Atomic nuclei
Usually we express the mass of nuclei in atomic
mass units u, defined to be 1/12 the mass of 12C
(the 12 is the mass number A.
12C
has 6 protons, 6 neutrons and 6 electrons)
1u=1.66054x10-27 kg = 931.494 MeV/c2
m p  1.00728u
mn  1.00866u
me  0.0005486u
• The mass of 1 proton + 1 electron is 1.0078285u
• Note 6 p + 6 n + 6 e- = 12.099.
Binding energy
The mass of an atom (protons+neutrons+electrons) is not equal to
the mass of the individual particles.
There is also a binding energy associated with the
nucleons themselves.
• The mass of the Carbon-12 atom:
6 p + 6 n + 6 e- = 12.099.
• The mass difference is 0.099u, equivalent to 92.22 MeV!
• This is the binding energy of the C-12 atom
Energy is released in fusion reaction if the sum of masses of initial nuclei is
larger that the mass of the final nucleus
hydrogen
mp +
mp
hydrogen
Positron (antielectron)
Deuterium
MD + me < 2 mp
M = 2 mp- MD - me
Energy released E = M c2
neutrino
Einstein’s relation: E = mc2
Deuterium has larger binding energy than protons (more tightly bound)
Energy and nuclear reactions
• Proton rest energy
When one proton and one neutron fuse to form a Deuteron
nucleus, the final mass is less than the sum of the mass of
the four particles. The deficit is the “binding energy”,
amounting to 2.2 MeV
Here 0.1% of the mass of the particles is being converted
to energy
|Ub|
There are no heavy elements
in the stars
Nuclear time-scale
• What if you could convert the entire mass of the Sun
into energy?
• At the current luminosity of the Sun, this would be
spent in
If 0.1% of the mass is converted to energy, the Sun could
still last for 1010 yr if powered by nuclear fusion energy.
The Sun is currently 4.5 x 109 yr old.
Nuclear energy: fusion
Nuclear energy is sufficient to sustain the Sun’s luminosity.
But can it actually occur naturally in the Sun?
Coulomb repulsion
The repulsive force between like-charged
particles results in a potential barrier that
gets stronger as the particles get closer:
• The strong nuclear force
becomes dominant on very
small scales, 10-15 m
• What temperature is
required to overcome the
Coulomb barrier?
Z1Z 2 e 2
UC 
40
r
1
Protons should be hot!
Statistical mechanics
• If the gas is in thermal equilibrium with
temperature T, the atoms have a range of
velocities described by the MaxwellBoltzmann distribution function.
• The number density of gas particles with
speed between v and v+dv is:
 m 
nv dv  n

 2kT 
3/ 2
e
mv 2

2 kT
4v 2 dv
kT
m
The most probable velocity:
v2  2
The average kinetic energy:
1
3
m v 2  kT
2
2
Overcoming the Coulomb barrier
•Fusion is possible if the average particle kinetic energy
(3/2 kT) is equal to or greater than the Coulomb
potential energy:
For two protons of z1=z2=1 separated by a typical
distance of r=10-15 m
This is much larger than the central temperature of the Sun
Maxwell-Boltzmann doesn’t help
The central temperature of the Sun is
The KE of a proton at this temperature is ≈ 2 keV
The electrostatic PE of two protons 10-15 m apart is 1 MeV
Energy
Could the protons at the tail end of the
Maxwell-Boltzmann distribution of
energies have sufficient kinetic energy
to overcome the Coulomb barrier?
The relative fraction of protons with thermal
energy of 1 MeV is only
Quantum mechanics to the rescue
The answer lies in quantum physics. The uncertainty principle states
that momentum and position are not precisely defined:

xp x 
2
•The uncertainty in the position
means that if two protons can
get close enough to each other,
there is some probability that
they will be found within the
Coulomb barrier.
 This is known as tunneling.
 The effectiveness of this
process depends on the
momentum of the particle
Quantum mechanics to the rescue
What temperature is required for two protons to come
within one de Broglie wavelength of each other?
deBroglie 

h
3kTm H
Quantum tunneling
Approximately: tunneling is possible if the
protons come within 1 de Broglie wavelength
of each other:
For two protons, at
T~107 K
h
h
 

p v
h
3kTm H
1.13 103

3.57 107 m
T
•So the protons don’t need to get anywhere near 10-15m
before they can begin to tunnel past the barrier
• Without this quantum effect, fusion would not be possible in
the Sun and such high luminosities could never be achieved.
Nuclear reactions
So – what are the specific reactions are we talking about??
 The probability that four H atoms will collide at once to form a single He atom
is exceedingly small. Even this simple fusion reaction must occur via a
number of steps.
Proton-proton cycle
Proton-proton chain (PPI)
1
1
H 11H 12H  e    e
2
1
H 11H  23He  
3
2
He  23He  24He  211H
The net reaction is:
411H 24He  2e   2e  2
But each of the above reactions occurs at its own rate. The first
step is the slowest because it requires a proton to change into a
neutron:
Energy  p   n  e   e
This occurs via the weak force. The rate of this reaction determines
the rate of Helium production
Proton-proton chain (PPII and PPIII)
Alternatively, helium-3 can react with
helium-4 directly:
3
4
7
He

He

2
2
4 Be  
7
4
Be  e  37 Li   e
7
3
Li 11H 2 24He
In the Sun, this reaction occurs 31% of the time; PPI
occurs 69% of the time.
Yet another route is via the collision
between a proton and the beryllium-7
nucleus 7
1
8
4
Be 1 H 5 B  
B  48Be  e    e
8
5
Be 2 24He
8
4
This reaction only occurs 0.3% of the
time in the Sun.
The triple-alpha process
The burning of helium occurs via the triple alpha process:
4
2
He  24He  48Be
8
4
Be  24He 126C  
The intermediate product 8-beryllium is very unstable, and will decay if not
immediately struck by another Helium. Thus, this is almost a 3-body
interaction
11 2 3 3 44.027T81
3
5
8

 5.09 10  Y T e
 3  3.85 10  Y T
8
2
5
3
41.0
8
W / kg
W/kg
Note the very strong temperature dependence. A 10% increase in T
increases the energy generation by a factor 50.
Nucleosynthesis
At the temperatures conducive to helium burning, other reactions
can take place by the capturing of -particles (He atoms).
C  24He 168O  
12
6
20
O  24He 10
Ne  
16
8
Nucleosynthesis
The binding energy per nucleon describes the stability of a nucleus. It is easier to break up a
nucleus with a low binding energy.
The solar neutrino problem
Neutrino have zero or very small mass and almost do not interact with matter
10,000 years
Neutrino image of the Sun
The Davis experiment
400,000 liters of perchlorethylene
buried 1 mile deep in a gold mine
About 1 Chlorine atom per day is
converted into Argon as a result of
interaction with solar neutrino
There are 1032 Cl atoms in a tank!
Much more difficult than finding
a needle in a haystack!!
Sudbury neutrino observatory: 1000 tons of heavy water
D2O
32,000 ton of ultra-pure water
13,000 detectors
Observed neutrino flux is 2 times lower than the theoretical prediction!
The problem has been finally solved just recently:
Neutrinos “oscillate”!
They are converted into other flavors: mu and tau neutrinos
Neutrinos should have mass
Particle physics models should be modified