Physics

Physics

Problem

In stellar interiors, four protons ultimately combine to form one alpha particle through a sequence of reactions. Neglecting the small masses of neutrinos, the net reaction can be written as 4 11H  ⟶  24He+2e++2γ4\,_{1}^{1}\text{H}\;\longrightarrow\;_{2}^{4}\text{He}+2e^{+}+2\gamma Given nuclear and particle masses (in u): • m(11H)=1.007825 um(_{1}^{1}\text{H}) = 1.007825\,\text{u} • m(24He)=4.002603 um(_{2}^{4}\text{He}) = 4.002603\,\text{u} • m(e+)=0.0005486 um(e^{+}) = 0.0005486\,\text{u} Constants: 1 u=931.5 MeV/c21\,\text{u}=931.5\,\text{MeV}/c^{2}; photons may be treated as massless. a. Compute the total initial mass and total final mass for the net reaction. b. Determine the reaction energy released (Q-value) in MeV. c. Express the result from part (b) as energy per kilogram of hydrogen converted. Use Avogadro’s number NA=6.022×1023 mol−1N_{A}=6.022\times10^{23}\,\text{mol}^{-1}. d. Briefly explain why even a relatively small mass defect can power a star for billions of years, making explicit reference to your numerical result from part (c).

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