The fusion reaction that generates energy in the sun is based on the fusion of 4

Question

The fusion reaction that generates energy in the sun is based on the fusion of 4 protons (ionized Hydrogen) into an alpha particle (doubly ionized Helium):
41H+1+ 2e- --> 4He+2 + 2v +6

The phontons are massless while the neutrons v have an extremely small mass which we can neglect. The mass of 4He is 4.002603 atomic mass units while that of 1H is 1.007825 AMU. One atomic mass unit corresponds to 931.5 M ev/c2. Determine the energy level released in this reaction. This energy is carried off by the neutrons and photons and motion of Helium.

Explanation / Answer

I am little confused with the reaction, but the procedure is given below 41H+1+ 2e- --> 4He+2 + 2v +6? Mass defect = 4 (1.007825 ) + 2 (0.0005486) - 4.002603 -2(1.008665) - 6(1.007276) = 4.031304 + 0.0010972 - 4.002603 - 2.01733 - 6.043656 = -8.0311878 u Therefore E = Mc^2 E = - 8.0311878 x c^2 x 931.5 MeV/c^2 = - 7481 MeV

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