Calculating Energy Released in Triplealpha Process

[Ayame17]

I was working at the energy released through the triplealpha process, and came to a slight problem. I worked out for Helium first, using:

(2 x 1.007825u) + (2 x 1.008665u) - 4.002602u = 0.030378u

which when x3 and converted gives ~84.9MeV.

Doing the same for carbon, but using 12.011u for the atomic mass (as is given in most places), I end up with ~81.9MeV, when I should be getting ~92.2MeV. But when I do the same equation using 12u, I get the right answer! What's going on?

 

[malawi_glenn]

12u is DEFINED to be the mass of one carbon-12 atom, it can't be anything else! 12.011u is the average atomic mass..(averaging over all C-isotopes and their relative abundances).

You should also realize that the triple alpha process takes place inside hot stars, so you must be careful to check if you have the correct number of electrons on LHS as RHS (not saying that this is the error you made, but one should always check that)

 

[Ayame17]

But surely 4.002602u is the average atomic mass of helium, so you'd have to use 4u in the first equation and then get a total of ~92.2MeV, which is wrong...

Or is it okay to use 4.002602u for the helium and 12u for the carbon?

 

[malawi_glenn]

One alpha atom (He-4) has mass of 4.002602u.

You have to DEFINE the concept of u (atomic mass unit), so one has DEFINED it to be the mass of ONE C-12 atom.

 

[Ayame17]

New problem now - when doing the same from 12C + 4He -> 16O, I get ~10.25MeV. I think it's meant to be 7.162MeV. Where am I going wrong this time?

 

[malawi_glenn]

New problem now - when doing the same from 12C + 4He -> 16O, I get ~10.25MeV. I think it's meant to be 7.162MeV. Where am I going wrong this time?

How am I supposed to know that when you don't show your calculations and what values you used?

Even though this is not posted in HW-thread, it really smells like homework. And even if you are only checkning these things for fun, show some detailed calculations, please!

 

[Ayame17]

Good point, sorry! Not homework (this time!) just some really crap lecture notes, not explaining things.

C is actually formed from 3xHe (the original reaction), so O must be formed from 4xHe.
Taking Q(He) = 0.030378u, Q(4xHe) = 0.121512 ~113.19MeV

Oxygen: (8 x 1.007825u) + (8 x 1.008665u) - 15.9994u = 0.13252u

which is ~123.44MeV. Last time it was simply the difference between the two, but this gives me ~10.25MeV. I looked at it as Q(4xHe) = Q(3xHe) + Q(C), but that made the difference even larger. I really can't see where she's got her numbers from.

 

[malawi_glenn]

in the reaction C-12 + He-4 -> O-16, we have the same amount of electrons on LHS as RHS, so we can use atomic masses.

O-16 atom has mass: 15.9949u
He-4 atom has mass: 4.00260u
C-12 atom has mass 12.0000u

mass difference = 0.0077u

1u = 931.49MeV

Energy released in this reaction = 7.17MeV

You can't put C-12 = 3*He-4 ! If the reaction is C-12 + He-4 -> O-16; then you must use these nuclei!