The proton-proton cycle consists of the following fusion reactions: 1_H + 1_H ri
ID: 1509716 • Letter: T
Question
The proton-proton cycle consists of the following fusion reactions: 1_H + 1_H rightarrow 2_H + e^+ + V_e 1_H + 2_H rightarrow 3_He + gamma 3_He + 3_He rightarrow 4_He + 1_H + 1_H Calculate the energy released by the reaction burning 1_H to produce 2_H. (Note: the mass energy of the positron is released, because it quickly annihilates with an electron.)_____MeV Calculate the energy released by the reaction burning 1_H and 2_H to produce 3_He. _____MeV Calculate the energy released by the reaction burning 3_He to produce 4_He and more 1_H.____MeV What is the total energy released by the cycle?_____MeVExplanation / Answer
mass are as given
1H = 1.00782 amu
2H = 2.01410 amu
3H = 3.01605 amu
3He = 3.01603 amu
4He = 4.00260 amu
energy released = (change in mass*velocity of light^2)
(a) change in mass= (2*1.00782)-(2.01410)) amu
energy released = change in mass in amu*1.66*10^(-27)*c^2/(1.6*10^(-19))=1.44 MeV
(b) change in mass=(1.00782+2.01410)-( 3.01605)=0.00587 amu
energy released=change in mass in amu*1.66*10^(-27)*c^2/(1.6*10^(-19))=5.48 MeV
(c) change in mass=(2*3.01605)-(4.00260 +(2*1.00782))=0.01386 amu
energy released=change in mass in amu*1.66*10^(-27)*c^2/(1.6*10^(-19))=12.94 MeV
(d) total energy released=sum of all above energy=19.86 MeV
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