In the fusion reaction, we have talked about the CNO cycle. Please calculate the
ID: 1828699 • Letter: I
Question
In the fusion reaction, we have talked about the CNO cycle. Please calculate the energy released or absorbed in each step of the reaction as described below and calculate the net energy of the total reaction:
In the fusion reaction, we have talked about the CNO cycle. Please calculate the energy released or absorbed in each step of the reaction as described below and calculate the net energy of the total reaction: C12 + H rightarrow N13 N13 rightarrow C13 + e+ C13 + H rightarrow N14 N14 + H rightarrow O15 O15 rightarrow N15 + e+ N15 + H rightarrow C12 + He4 Total Useful rest mass energies H = 938.789 MeV He = 3,728.42 MeV C12 = 11,178 MeV C13 = 12,112.6 MeV N13 = 12,114.8 Mev N14 = 13,043.9 MeV O 15 = 13,973.3 MeV N15 = 13,972.6 MeV Electron: e - = 0.51 MeV Positron: e+ = -0.51 MeVExplanation / Answer
C12 + H---> N13
energy released =11,178 +938.789 -12114.8 =1.989MeV
N13 ---->C13 +e+
energy released = 12114.8 -12112.6 -(-0.51) = 3.31MeV
C13 +H --->N14
energy released = 12112.6 +938.789 -13043.9 = 7.489MeV
N14 +H --->O15
energy released =13043.9 +938.789 -13973.3 = 9.389MeV
O15 --->N15 +e-
energy released =13973.3 -13972.6-0.51 = 0.19MeV
N15 + H --->C12 + He
energy released =13972.6 +938.789 - 11178-3728.42 = 4.969MeV
Net energy released = 27.336 MeV
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