How Much Energy is Released in the First Step of the CNO Cycle?

[Confundo]

Homework Statement

The first step of the CNO cycle is
$$p \rightarrow C_{6}^{12} \rightarrow N_{7}^{14} + \gamma$$

Estimate the energy of the CNO cycle, state any assumptions.

Homework Equations

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The Attempt at a Solution

I was looking at mass difference,


(1.00728u + 12u) - 13.00574u = 0.00154u to convert into energy for the gamma
1u = 931.494MeV/c^2
0.00154u = 1.4345MeV/c^2

therefore E(gamma) = 1.45 MeV

assumptions: energy and momentum are conserved.

When I looked at the wiki page for the CNO they used Hydrogen and had an energy for the gamma of 1.95MeV, I' got the 1.95MeV value using the mass of the hydrogen. I'm just a little concerned that I didn't take into account any kinetic energy the proton may of had before collision.

 

[Jhero]

Thank you for your question on the energy of the CNO cycle. You are correct in your calculation that the energy of the gamma ray in the first step of the CNO cycle is approximately 1.45 MeV.

As for your concern about not taking into account the kinetic energy of the proton, it is true that the proton may have some kinetic energy before it undergoes fusion in the CNO cycle. However, this kinetic energy is negligible compared to the energy released in the fusion reaction itself. Therefore, for the purposes of this calculation, it is acceptable to assume that the proton has no kinetic energy before the fusion reaction.

Additionally, it is important to note that the energy released in the CNO cycle is not solely determined by the mass difference between the reactants and products. The energy also depends on the binding energy of the nuclei involved. This is why the value of 1.95 MeV is used in the wiki page, as it takes into account the binding energy of the hydrogen nucleus.

I hope this helps clarify any concerns you may have had. Thank you for your interest in the CNO cycle.