Question

Consider the proton–proton cycle that occurs in most stars (including our own Sun): Step 1: 1H + 1H → 2H + e+ + νe Step 2: 2H + 1H → 3He + γ Step 3: 3He + 3He → 4He + 2 1H + γ Calculate the net energy released from the three steps. Do not ignore the mass of the positron in Step 1. (You may ignore the mass of the neutrino.) MeV

Answer 1

Step 1: ¹H + ¹H → ²H + e⁺ + $\nu$ + energy

Step 2: ²H + ¹H → ³He + γ + energy

Step 3: ³He + ³He → ⁴He + 2¹H + γ + energy

So Net reaction is ( step1 x 2 + step2 x 2 + step3)

4¹H →⁴He + 2 e⁺+ 2 ν + 3 γ + energy

Mass of a proton is 1.6726 × 10⁻²⁷kg , So the mass of 4 protons = 6.6904 × 10⁻²⁷kg

The mass of a ⁴He nucleus is 6.6447 × 10⁻²⁷kg

The mass of positron is 9.1094×10⁻³¹ kg , Somass of 2 positron = 18.2188 x 10^-31 kg = 0.0018 x 10^-27 kg

Missing mass($\Delta$ m) = Mass of 4 protons - (Mass of ⁴He + Mass of 2 positron) =

                                6.6904 × 10⁻²⁷ - (6.6447 × 10⁻²⁷ + 0.0018 x 10^-27 ) = 0.0439x10^-27 kg

The missing mass was converted into energy by Einstein’s famous equation
E = $\Delta$mc² = 0.0439x10^-27 x (3 x 10⁸ )² = 3.951 x 10^-12 J

1J = 6.242 x 10¹² Mev

Hence 3.951 x 10^-12 = 24.6602Mev

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