Triple Alpha Reaction: DE's for He4, C12, O16, Ne20

[t_n_p]

Hi guys,
I posted this in the homework subforum but since its more intensive on the astrophysics side of things I thought I'd give it a shot here. Hope this isn't a problem mods.

Helium is burnt via the triple alpha reaction: 3He⁴ -> C¹² at a rate of λ_3α(He⁴)³, where He⁴ denotes the number abundance of He⁴. 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 He⁴,C¹², O¹⁶, Ne²⁰ abundances as a function of time.

Basically here's what I did.
d(He⁴)/dt = production - destruction = - λ_3α(He⁴)³
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

This is most likely wrong, but I'd thought I'd just give it a shot.
Any thoughts?
Thanks in advance

 

[gtoguy97]

.Your equations look correct to me. The only thing I would add is to make sure that you include the rate coefficients for the reactions (e.g. λ3α) as constants in your equations. You will also need to add a source of He4 if it is not being produced.