How do i find quantities in fusion formulae

Main Question or Discussion Point

in the equation *C^12+C^12-->Mg^24+y(+13.93 MeV)* for what quantity is the energy assigned? and how would i plug in a certain amount of carbon?
 

Answers and Replies

mathman
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If I understand your question correctly, the energy given is for two atoms of C12 combining to form one atom of Mg24. If you want to plug in an amount of carbon, count the atoms in the quantity.
 
Astronuc
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The 13.93 MeV would be difference mass of the 2 C-12's and the Mg-24, and would be given off as a gamma ray for each nuclear reaction. Since it is based on 2 carbons, one would use the number of carbon atoms in half the carbon mass to determine a total energy yield.

The reaction is more likely when the carbon nuclei are traveling head on at each other.]

Keep in mind that other reactions are also possible, and perhaps more likely.

http://www.astrophysicsspectator.com/topics/stars/FusionCarbonOxygen.html

http://www.astro.cornell.edu/academics/courses/astro201/carbon_fusion.htm

http://physics.uoregon.edu/~jimbrau/astr122/Notes/ch20/carbon.html
 
Last edited:
thanks, but how then would i accurately find energy out come for a specific quantity?
 
Astronuc
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thanks, but how then would i accurately find energy out come for a specific quantity?
A specific quantity of carbon?

Well, one would have to know the branching ratios (or yield fractions) of the possible reactions, i.e. the fraction of 12C + 12C reactions that result in each of several products, and how much energy is produced for each reaction.

Then it's a matter of taking the mass of 12C, determining the number of 12C nuclei, divide that by half because 1 C-C fusion uses two nuclei, and multipy by the yield fraction and energy per reactions.

So the total energy would be N/2*(f1Q1 + f2Q2 + . . . + fiQi) where fi is the yield fraction of reaction i, and Qi is the energy release per reaction i.

The number of atoms (nuclei) is just the mass (kg)*/(atom mass in amu *1.66E-27 kg/amu).

It's not really possible to get an 'accurate' answer because carbon fusion occurs under rather complicated circumstances that we cannot reproduce experimentally on Earth. We can produce 50 keV carbon ions, but they will not be +12 ions, and the target will not be ionized either, so one would have to go to higher energies. However, we cannot reproduce densities found in heavy stars (2 x 108 kg/m3). So the best we can do is approximate with models of stars.
 

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