Richard - Nucleosynthesis Flashcards
Explain the following observations from the stellar abundance curve:
1. Exp decrease from H to A~100
2. Rarity of D, Li, Be, B compared with neighbours H, He, C, N, O
3. High abundance of alpha particle nuclei relative to neighbours
4. Strongly marked peak at Fe-56
- Increasing rarity of synthesis for increasing A, reflecting that stellar evolution to advance stages to build high A is not common.
- Inefficient production, also consumed in stellar interiors even at relatively low temps.
- He burning and alpha process more productive than H burning and a process in this region.
- e process; stellar evolution to advanced stage where maximum energy is released.
Why does hydrogen burning occur at much lower temps than theoretical calculations suggest?
Theory suggest it should occur at 10^10K to overcome Coulomb repulsion
The actual temperature is 10^7K
Quantum tunnelling allows ‘tunnelling’ through the potential barrier hence not requiring the full energy. Nuclei also have discrete energy levels and if given the right amount of energy the reaction can be sped up (resonance).
What is the equation for cross-section and what is it telling you? Provide units etc
Cross-section(sigma) = (# reactions X(a,b)Y per second)/(flux of incident particles a)
Effective surface area for particle X interacting with particle a producing particles Y and b.
Measured in barns = 10^-24 cm2
Write the full reaction for the shorthand notation X(a,b)Y. Do the same for 12C(p,gamma)13N
X + a —> Y + b
12C + p —> 13N + gamma
What is the astrophysical S-factor S(E), how is it used? How can nuclear properties such as resonance affect calculations?
Characterises the reaction probability between two colliding particles at a specific energy E.
S(E) must be determined from experiments — these experiments have to be done at high energies then extrapolated down to stellar conditions (much lower energies) — existence of resonances can hamper this extrapolation
Resonance represents specific energy levels in the nucleus where the cross-section for a nuclear reaction increases sharply due to the influence of excited states within the nucleus.
What is the first significant stage of nuclear burning? What type of interaction is involved?
4•1H —> 4He + 2e+ + 2•neutrinos
Weak interaction (p —> n)
Q-value = 26.734MeV
There are three main ways of producing He-4 via protons. Write down the final reaction for each stage and it’s Q value.
PP1: 3He + 3He —> 4He + 2•1H
(Above is slow and a build up of 3He occurs)
Q1 = 26.20MeV
PP2: 7Li + 1H —> 4He + 4He
Q2 = 25.66 MeV
PP3: 8Be —> 4He + 4He
Q3 = 19.76MeV
What is the CNO cycle and how is it split up?
CNO uses heavy elements to synthesise 4He — split into the CN branch and ON branch
CN BRANCH
12C —> 13N —> 13C —> 14N —> 15O —> 15N + 1H —> 12C + 4He
(Less efficient than pp1 or pp2 **full reactions not shown)
ON BRANCH
(Smaller chance of occurring)
15N —> 16O —> 17F —> 17O + 1H —> 14N + 4He
Which cycle occurs when? PP chain or CNO cycle?
Both happen at the same time
The RATE of the reactions is different
Cross over point is ~ 2x10^7K
(Sun is dominant PP chain)
Other chains occur at even higher temps and pressures | important for nuclei synthesis | not for energy production
What are temps required for helium fusion? What can it produce? Why is it significant?
~ 10^8K temps required
(Significant Coulomb barrier and fast reactions required)
4He + 4He interchanges 8Be
• lifetime of 8Be is around 10^-16s
•small conc builds up (allowing one more 4He capture to occur)
4He + 8Be —> 12C* —> 12C + gamma
12C* is important as it’s an example of resonance
What is the Q value for helium fusion
Q value = 7.275 MeV
What occurs after Helium burning? Temps, Q value
Majority of molecules are 12C and 16O
Carbon fusion occurs next
Requires approx 5•10^8K
(Star needs at least 8 solar masses)
Most common decay modes result in Q values of:
24Mg* —> 20Ne + alpha | Q = 4.616MeV
24Mg* —> 23Na + p | Q = 2.238MeV
What’s the final few stages after carbon burning?
Neon burning ~ 1.5x10^9K
Mostly 16O and 24Mg produced
Oxygen burning ~ 2x10^9K
Mostly 28Si and 32S produced
Silicon burning ~ 3x10^9K
At high enough temps (4x10^9K) forward and reverse reactions can be in a state of equilibrium.
End result is mostly 56Fe
Any further reactions would produce a negative Q value. Hence the fun stops here
What is the difference between the s-process and r-process?
s-process (slow neutron capture)
• Time of decay < Time of neutron capture
• unstable nuclei beta decay then neutron capture
• Existing nuclei capture neutrons —> decay into protons —> moves along the valley of stability —> steadily creating heavier stable isotopes
• typically occurs in lower neutron flux regions 10^7-10^10 neutrons/cm^3 (time 10^4-10^5 yrs)
• starting point is iron
r-process (rapid process)
• Time to decay > Time to neutron capture
• absorbs neutron before beta decay occurs
• sudden influx of neutrons allows neutron absorption before beta decay —> producing neutron Rich isotopes which are often highly unstable —> these then undergo rapid beta decay forming heavy elements beyond iron
• neutron flux 10^21-10^24 n/cm3 (short time scale)
What is the termination point for the s-process?
Half life’s beyond Pb are very short — hence will decay before neutron is able to be absorbed.
Hence slow neutron captures can’t produce anything beyond Pb.
* Heavier nuclei are unstable and return to Pb (lead)
