Astrophysics - triple alpha reaction

[t_n_p]

Homework Statement

Helium is burnt via the triple alpha reaction: 3He⁴ -> C¹² at a rate of \lambda_3α(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

 

[hikaru1221]

Is there He4 production here? Btw, C12 is a stable isotope
It's like a motion problem: you know v(x), you can find x(t)

 

[t_n_p]

ok, since there is no He4 production I'm left with..

d(He4)/dt = 0 - destruction.

and destruction is occurring at a rate of \lambda_3α(He⁴)³

Therefore...

d(He4)/dt = - \lambda_3α(He⁴)³

Thus, this would be my answer?

 

[hikaru1221]

I don't know what \lambda_3α(He⁴)³ means but basically it's just like that.

 

[t_n_p]

I don't know what \lambda_3α(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 λ_12αHe⁴C¹². Then, once O¹⁶ has been produced, we also get O¹⁶(α,γ)Ne²⁰ which occurs at the rate λ_16αHe⁴O¹⁶.

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_3α(He⁴)³ - λ_12αHe⁴C¹²

d(O¹⁶)/dt = production - destruction = λ_12αHe⁴C¹² - λ_16αHe⁴O¹⁶

d(Ne²⁰)/dt = production - destruction = λ_16αHe⁴O¹⁶

The only thing that bothers me is that the abundances of some of the other elements is dependent on the amounts of other elements...

 

[hikaru1221]

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.

 

[t_n_p]

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⁺ + ν_e occurs at a rate: r_pp=n_p²σ_pp/2
p + d -> He³ + γ occurs at a rate: r_pd=n_pn_dσ_pd
He³ + He³ -> He⁴ + 2p occurs at a rate: r₃₃ = n₃₃²σ₃₃/2

where r_ij = n_in_jσ_ij/(1+ δ_ij), therefore r_pp is the reaction rate between 2 protons, r_pd 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 = r_pp - r_pd = λ_ppH²/2 - λ_pdHD

I understand how they get formation rate - destruction rate = r_pp - r_pd, 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!