In stellar interiors

PHYSICS · Multi-part · CBSE · Physics · Nuclei · Nuclei

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\,_{1}^{1}\text{H}\;\longrightarrow\;_{2}^{4}\text{He}+2e^{+}+2\gamma

Given nuclear and particle masses (in u): •

m(_{1}^{1}\text{H}) = 1.007825\,\text{u}

•

m(_{2}^{4}\text{He}) = 4.002603\,\text{u}

•

m(e^{+}) = 0.0005486\,\text{u}

Constants:

1\,\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

N_{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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