Energy Released in Fusion Reactions (DP IB Physics: SL): Revision Note

In the Sun, fusion occurs via a process known as the proton-proton chain.

It is predicted that 80% of the total power output of the Sun is produced through the following cycle:

$\begin{array}{r}{}_1{}^1\mathrm{H}+{}_1{}^1\mathrm{H}\to {}_1{}^2\mathrm{H}+{}_1{}^0\mathrm{e}^{+}+{\nu }_{\mathrm{e}}\\ {}_1{}^1\mathrm{H}+{}_1{}^2\mathrm{H}\to {}_2{}^3\mathrm{He}+\mathrm{\gamma }\\ {}_2{}^3\mathrm{He}+{}_2{}^3\mathrm{He}\to {}_2{}^4\mathrm{He}+{}_1{}^1\mathrm{H}+{}_1{}^1\mathrm{H}\end{array}\}4{}_1{}^1\mathrm{H}\to {}_2{}^4\mathrm{He}+2{}_1{}^0\mathrm{e}^{+}+2{\nu }_{\mathrm{e}}$ (overall reaction)

The neutrinos produced in the first step carry away 2% of the energy released by the process.

Determine the mass of hydrogen-1 that must be fused each second to produce this output.

Luminosity of the Sun = 3.85 × 10²⁶ W.

Answer:

Step 1: Determine the energy released per overall fusion reaction

• In the overall reaction, 4 hydrogen-1 nuclei fuse into a helium-4 nucleus, so the mass defect is:

$∆m$ = 4(1.007825u) − 4.002603u

$∆m$ = 0.028697u

• Where atomic mass unit, u = 1.66 × 10⁻²⁷ kg

• Using mass-energy equivalence, the energy released by one reaction is:

$∆E=∆m{c}^2$

$∆E$ = 0.028697 × (1.66 × 10⁻²⁷) × (3 × 10⁸)²

$∆E$ = 4.287 × 10⁻¹² J

Step 2: Determine the energy released minus the energy that is carried away by neutrinos

• Per reaction, neutrinos carry away 2% of 4.287 × 10⁻¹² J, so 98% of the energy contributes to the luminosity of the Sun

$∆E$ = 0.98 × (4.287 × 10⁻¹²) = 4.201 × 10⁻¹² J

Step 3: Determine the number of fusion reactions that happen each second

• This process accounts for 80% of the luminosity of the Sun,

• So, the total power output of the reaction = 0.8 × (3.85 × 10²⁶) W

• The number of fusion reactions each second is:

number of reactions = $\frac{\mathrm{power}\mathrm{output}}{\mathrm{energy}\mathrm{released}\mathrm{per}\mathrm{reaction}}$

number of reactions = $\frac{0.8\times \left(3.85\times {10}^{26}\right)}{4.201\times {10}^{-12}}$ = 7.332 × 10³⁷ s⁻¹

Step 4: Determine the mass of hydrogen that fuses each second

• Every reaction fuses 4 hydrogen-1 nuclei, so the mass per reaction is 4 × 1.007825u

• The mass of hydrogen-1 that fuses each second in this process is

mass of hydrogen-1 = 4u × number of reactions

mass of hydrogen-1 = 4 × 1.007825 × (1.66 × 10⁻²⁷) × (7.332 × 10³⁷)

mass of hydrogen-1 = 4.91 × 10¹¹ kg s⁻¹