Somak Raychaudhurywww.sr.bham.ac.uk/~somak/Y2SiU/

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

Lecture 11Guest Lecturer: Dr W J Chaplin

Revision

• 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

dP(r)

drGM(r)

r2 (r)

Atomic nucleiAtomic 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

um

um

um

e

n

p

0005486.0

00866.1

00728.1

• The mass of 1 proton + 1 electron is 1.0078285u• Note 6 p + 6 n + 6 e- = 12.099.

Binding energyBinding energy

There is also a binding energy associated with the nucleons themselves.

• The mass difference is 0.099u, equivalent to 92.22 MeV!• This is the binding energy of the C-12 atom

• The mass of the Carbon-12 atom: 6 p + 6 n + 6 e- = 12.099.

The mass of an atom (protons+neutrons+electrons) is not equal to the mass of the individual particles.

Energy is released in fusion reaction if the sum of masses of initial nuclei is larger that the mass of the final nucleus

mp + mp

MD + me < 2 mpDeuterium

Positron (antielectron)

neutrino

Deuterium has larger binding energy than protons (more tightly bound)

M = 2 mp- MD - meEnergy released E = M c2

Einstein’s relation: E = mc2

hydrogen

hydrogen

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

There are no heavy elements in the stars

|Ub|

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 couldstill last for 1010 yr if powered by nuclear fusion energy. The Sun is currently 4.5 x 109 yr old.

Nuclear energy: fusionNuclear energy: fusion

Nuclear energy is sufficient to sustain the Sun’s luminosity. But can it actually occur naturally in the Sun?

Coulomb repulsionCoulomb repulsion

r

eZZUC

221

04

1

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?

Protons should be hot!

Statistical mechanicsStatistical mechanics

• If the gas is in thermal equilibrium with temperature T, the atoms have a range of velocities described by the Maxwell-Boltzmann distribution function.

• The number density of gas particles with speed between v and v+dv is:

dvvekT

mndvn kT

mv

v22

2/3

42

2

The most probable velocity:m

kTv 22

The average kinetic energy: kTvm2

3

2

1 2

Overcoming the Coulomb barrierOvercoming 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 helpMaxwell-Boltzmann doesn’t help

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

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

Quantum mechanics to the rescueQuantum mechanics to the rescue

The answer lies in quantum physics. The uncertainty principle states that momentum and position are not precisely defined:

2

xpx

•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 rescueQuantum mechanics to the rescue

What temperature is required for two protons to come within one de Broglie wavelength of each other?

deBroglie h

3kTmH

Quantum tunnelingQuantum tunneling

•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.

Approximately: tunneling is possible if the protons come within 1 de Broglie wavelength of each other:

h

ph

v

h

3kTmH

For two protons, at T~107 K m

T7

3

1057.31013.1

Nuclear reactionsNuclear 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)Proton-proton chain (PPI)

HHeHeHe

HeHH

eHHH e

11

42

32

32

32

11

21

21

11

11

2

The net reaction is: 2224 42

11

eeHeH

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:

eenp Energy

This occurs via the weak force. The rate of this reaction determines the rate of Helium production

Proton-proton chain (PPII and PPIII)Proton-proton chain (PPII and PPIII)

HeHLi

LieBe

BeHeHe

e

42

11

73

73

74

74

42

32

2

Alternatively, helium-3 can react with helium-4 directly:

HeBe

eBeB

BHBe

e

42

84

84

85

85

11

74

2

Yet another route is via the collision between a proton and the beryllium-7 nucleus

This reaction only occurs 0.3% of the time in the Sun.

In the Sun, this reaction occurs 31% of the time; PPI occurs 69% of the time.

The triple-alpha processThe triple-alpha process

The burning of helium occurs via the triple alpha process:

CHeBe

BeHeHe126

42

84

84

42

42

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

kgWeTY T /1009.51

8027.4438

325

113

W/kg1085.3 0.418

325

83 TY

Note the very strong temperature dependence. A 10% increase in T increases the energy generation by a factor 50.

NucleosynthesisNucleosynthesis

At the temperatures conducive to helium burning, other reactions can take place by the capturing of -particles (He atoms).

NeHeO

OHeC2010

42

168

168

42

126

NucleosynthesisNucleosynthesis

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

10,000 years

Neutrino have zero or very small mass and almost do not interact with matter

Neutrino image of the Sun

The Davis experiment

400,000 liters of perchlorethyleneburied 1 mile deep in a gold mine

About 1 Chlorine atom per day is converted into Argon as a result ofinteraction with solar neutrino

Much more difficult than finding a needle in a haystack!!

There are 1032 Cl atoms in a tank!

Sudbury neutrino observatory: 1000 tons of heavy water D2O

32,000 ton of ultra-pure water13,000 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 massParticle physics models should be modified