Calculating Energy Released in Triplealpha Process

Click For Summary

Discussion Overview

The discussion revolves around calculating the energy released during the triple alpha process and subsequent reactions involving carbon and helium. Participants explore the use of atomic masses, the definitions of mass units, and the implications of using average atomic masses versus defined atomic masses in their calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the energy released in the triple alpha process using helium and carbon atomic masses but encounters discrepancies when using different values for carbon's atomic mass.
  • Another participant clarifies that 12u is defined as the mass of one carbon-12 atom, while 12.011u is the average atomic mass, suggesting that using average masses can lead to incorrect results.
  • There is a discussion about whether it is acceptable to mix average atomic masses and defined atomic masses in calculations, with some participants questioning the consistency of this approach.
  • A new problem is introduced regarding the reaction of carbon and helium to form oxygen, with a participant expressing confusion over the calculated energy release and the values used in their calculations.
  • Another participant provides specific atomic masses for the involved nuclei and calculates the energy released in the carbon and helium reaction, arriving at a different value than previously mentioned.
  • Concerns are raised about the lack of detailed calculations in the posts, with one participant suggesting that the discussion resembles a homework problem despite being labeled otherwise.

Areas of Agreement / Disagreement

Participants express differing views on the use of atomic masses in calculations, with no consensus reached on the appropriateness of using average versus defined masses. Additionally, there is disagreement on the correct energy release values for the reactions discussed, indicating unresolved issues in the calculations presented.

Contextual Notes

Participants highlight the importance of ensuring the correct number of electrons on both sides of the reaction and the definitions of atomic mass units, which may affect the calculations. There are also references to potentially incomplete lecture notes that may contribute to the confusion.

Ayame17
Messages
43
Reaction score
0
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?
 
Physics news on Phys.org
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)
 
Last edited:
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?
 
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.
 
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?
 
Ayame17 said:
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!
 
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.
 
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!
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
13K