Energy from White Dwarf Thermonuclear Runaway

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Homework Help Overview

The problem involves calculating the energy released during a thermonuclear runaway in a white dwarf, specifically when converting a specified mass of carbon to iron. The context is rooted in nuclear physics, particularly the reactions that occur in stellar environments.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of energy released based on rest mass energies and the number of particles involved in the reaction. Questions arise regarding the appropriate method for determining the number of carbon atoms needed to produce iron and the implications of nucleon conservation in nuclear reactions.

Discussion Status

The discussion is ongoing with participants exploring different interpretations of the nuclear reaction process. Some guidance has been offered regarding the preservation of nucleon count and the importance of using accurate atomic masses in calculations. There is no explicit consensus on the approach to take, as various perspectives on the reaction pathways are being considered.

Contextual Notes

Participants note that their prior experience with nuclear reactions is limited, primarily focusing on hydrogen fusion, which may affect their understanding of heavier element reactions. There is also mention of the need for clarity on the reaction specifics and the assumptions underlying the calculations.

kvl214
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Hello!

Homework Statement


The question is "Calculate the energy released if a 1 solar mass white dwarf undergoes a thermonuclear runaway and converts 0.5 solar mass of carbon (A=12) to iron (A=56)."

I've only ever done nuclear equations with hydrogen fusion, with four hydrogens to one helium. But I don't know the reaction for carbon to iron, since it's not evenly divisible.

Homework Equations


rest mass energy E = mc^2
energy released = E(reactants) - E(products)

The Attempt at a Solution


I calculated the number of carbon particles from 0.5 solar masses -- 7.4e56 C particles
And the rest mass energies of carbon and iron -- C=1.1e4 MeV, Fe=5.2e4 MeV

Since 56/12 = 4.66, do we just use a fraction amount for carbon? So 4.66 C → 1 Fe ?

I just don't know where to go from here.

Any help would be greatly appreciated! Thanks
 
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Nuclear reactions tend to operate slightly differently than chemical reactions.

Haven't you studied other nuclear reactions which occur in stars?
 
We only did hydrogen fusion, nothing with heavier elements...

What would you do for nuclear reactions?

Thanks!
 
I recalculated the rest mass energies with the atomic masses, since before i just used the integer number. So if i subtract 4.66*carbon energy from iron energy, does that work for the energy released in one reaction?

I also calculated that there are 1.6e56 reactions available from the mass of carbon...
 
I don't think you can take the 56/12 ratio to determine the quantity of iron produced. The nucleon count won't be preserved. Rather, reactions involve integral numbers of atoms. Of course, there are different pathways, but if the whole process only uses one pathway then there must be a whole number of carbon atoms to one iron atom.
 
haruspex said:
I don't think you can take the 56/12 ratio to determine the quantity of iron produced. The nucleon count won't be preserved. Rather, reactions involve integral numbers of atoms. Of course, there are different pathways, but if the whole process only uses one pathway then there must be a whole number of carbon atoms to one iron atom.

That's not right. You are thinking of chemistry. I think the only reasonable way to interpret the question is preserving the nucleon count. All you need to know about the reaction is that each nucleon will eventually wind up in an iron nucleus. So just find the mass per nucleon in carbon and use that to find the initial number of nucleons. Then find the mass difference between a nucleon in carbon and a nucleon in iron. And keep some extra decimal places around. The difference will be smallish.
 
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