Energy Released in Fusion Reactions (HL IB Physics)

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

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

(overall reaction)

nucleus	rest mass / u
hydrogen-1	1.007825
helium-4	4.002603

 

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

mass defect:  Δm = 4(1.007825u) − 4.002603u = 0.028697u

• Where atomic mass unit, u = 1.66 × 10⁻²⁷ kg
• Using mass-energy equivalence, the energy released by one reaction is

ΔE = Δmc² = 0.028697 × (1.66 × 10⁻²⁷) × (3 × 10⁸)²

energy released:  ΔE = 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

energy released:  ΔE = 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 =

number of reactions = 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

= 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