Astrophysics - triple alpha reaction

In Summary, the triple alpha reaction creates He4 at a rate of \lambda3α(He4)3. Deuterium is produced and destroyed in proportion to the amount of He4 present.
  • #1
t_n_p
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Homework Statement



Helium is burnt via the triple alpha reaction: 3He4 -> C12 at a rate of [tex]\lambda[/tex](He4)3, where He4 denotes the number abundance of He4.

Write the differential equation for He4 as a function of time.

The Attempt at a Solution



d(He4)/dt = production - destruction.

I'm unfamiliar with how to find production and destruction rates...
Any help appreciated
 
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  • #2
Is there He4 production here? Btw, C12 is a stable isotope :smile:
It's like a motion problem: you know v(x), you can find x(t) :smile:
 
  • #3
ok, since there is no He4 production I'm left with..

d(He4)/dt = 0 - destruction.

and destruction is occurring at a rate of [tex]\lambda[/tex](He4)3

Therefore...

d(He4)/dt = - [tex]\lambda[/tex](He4)3

Thus, this would be my answer?
 
  • #4
I don't know what [tex]\lambda[/tex](He4)3 means but basically it's just like that.
 
  • #5
hikaru1221 said:
I don't know what [tex]\lambda[/tex](He4)3 means but basically it's just like that.

I don't what what it means either, but we are told that's the rate so what can you do!

part b) and c) gets a little harder...

Following part a) and once there is some C12 present then we can get C12(α,γ)O16 which occurs at the rate λ12αHe4C12. Then, once O16 has been produced, we also get O16(α,γ)Ne20 which occurs at the rate λ16αHe4O16.

Write down DE's for C12, O16, Ne20 abundances as a function of time.

I think this is right, but I want to double check:
d(C12)/dt = production - destruction = λ(He4)3 - λ12αHe4C12

d(O16)/dt = production - destruction = λ12αHe4C12 - λ16αHe4O16

d(Ne20)/dt = production - destruction = λ16αHe4O16

The only thing that bothers me is that the abundances of some of the other elements is dependent on the amounts of other elements...
 
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  • #6
If λ... is some constant, the DE can be solved for He4, then plug that into the first DE, you have DE for C12. Solving it and plugging it into the 2nd one...
Anyway I don't get what the question expects.
 
  • #7
I understand what you mean. If lambda is a constant, I can use separation of variables to solve for He4, which can then be subbed into the d(C12)/dt equation to solve C12 using same method so on and so forth until the end.

I have a somewhat similar example, but I have trouble understanding it.
It goes as such...

p + p -> d + β+ + νe occurs at a rate: rpp=np2σpp/2
p + d -> He3 + γ occurs at a rate: rpd=npndσpd
He3 + He3 -> He4 + 2p occurs at a rate: r33 = n332σ33/2

where rij = ninjσij/(1+ δij), therefore rpp is the reaction rate between 2 protons, rpd is the reaction rate between a proton and deuteron etc

E.g.
The DE for deuterium is then..
d(D)/dt = formation rate - destruction rate = rpp - rpd = λppH2/2 - λpdHD

I understand how they get formation rate - destruction rate = rpp - rpd, but how they do the next step to get the lambdas I have no idea.

from what I gather, lambda = de broglies wavelength.

Still very, very confused!
 
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What is the "triple alpha reaction" in astrophysics?

The triple alpha reaction is a nuclear fusion process that occurs in the cores of stars, specifically in red giants and other low-mass stars. It involves the fusion of three helium-4 nuclei (alpha particles) to form a carbon-12 nucleus.

Why is the triple alpha reaction important in astrophysics?

The triple alpha reaction is important because it is one of the primary processes that generates energy in stars. It also plays a crucial role in the production of heavier elements, as carbon is a building block for all life and essential for the formation of planets.

What are the conditions required for the triple alpha reaction to occur?

The triple alpha reaction requires extremely high temperatures (around 100 million degrees Celsius) and pressures (billions of times higher than Earth's atmospheric pressure) to overcome the repulsive forces between positively charged nuclei and fuse them together.

How does the triple alpha reaction impact the evolution of stars?

The triple alpha reaction is responsible for the formation of a stable carbon core in low-mass stars, which allows them to continue burning and releasing energy. In higher-mass stars, the triple alpha reaction can lead to a chain of fusion reactions that produce heavier elements, ultimately leading to supernova explosions.

Can the triple alpha reaction be replicated on Earth?

Currently, the triple alpha reaction cannot be replicated on Earth due to the extreme conditions required for it to occur. However, scientists are working on ways to artificially recreate this process in controlled environments for potential use in energy production.

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