E5-1. Stellar Fusion & Energy Release HL

1. The Physics of Stellar Fusion

Definition and Core Mechanism

  • Nuclear Fusion is the joining of two small, light atomic nuclei to produce a larger, more massive nucleus. Because the product has a higher binding energy per nucleon, a massive amount of energy is released.
  • In the core of a stable star like our Sun, hydrogen atoms undergo continuous nuclear fusion to form helium. This process provides the outward radiation pressure that counteracts the inward crush of gravity, keeping the star in a stable state known as hydrostatic equilibrium.
  • The classic D-T fusion reaction (which also occurs in experimental fusion reactors on Earth) involves isotopes of hydrogen:
    $${}^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \longrightarrow {}^{4}_{2}\text{He} + {}^{1}_{0}\text{n} + \text{Energy}$$
Deuterium-Tritium Fusion Reaction + Deuterium (²H) + Tritium (³H) FUSION + + Helium-4 (⁴He) Neutron (¹n) Energy (17.6 MeV)

The Coulomb Barrier

  • Because atomic nuclei are positively charged, they experience an immense electrostatic force of repulsion as they approach one another.
  • To fuse, nuclei must get incredibly close (within $\sim 10^{-15}\text{ m}$) so that the attractive strong nuclear force can overcome the electrostatic repulsion.
  • Therefore, stellar fusion requires extreme temperatures (providing massive kinetic energy to the particles) and extreme pressures/densities (increasing collision frequency). This state of matter is called a plasma.

2. Advanced HL Examples

Example 1: The Thermodynamics of the Coulomb Barrier (University Bridge)

Problem: Estimate the minimum temperature required to force two bare protons to undergo nuclear fusion.
(a) First, calculate the electrostatic potential energy (in Joules) between two protons separated by exactly $1.0 \times 10^{-15}\text{ m}$ (the effective range of the strong nuclear force).
(b) Assuming the classical kinetic theory of gases applies ($E_k = \frac{3}{2}k_B T$), calculate the temperature $T$ corresponding to this energy. Why does fusion in the Sun occur at "only" $\approx 1.5 \times 10^7\text{ K}$, much lower than this classical calculation implies?

Solution:

Part (a): Electrostatic Potential Energy
Using Coulomb's Law for potential energy ($E_p = \frac{k q_1 q_2}{r}$): $$E_p = \frac{(8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \times (1.60 \times 10^{-19}\text{ C})^2}{1.0 \times 10^{-15}\text{ m}}$$ $$E_p = \frac{(8.99 \times 10^9) \times (2.56 \times 10^{-38})}{1.0 \times 10^{-15}} = \mathbf{2.30 \times 10^{-13}\text{ J}}$$ (This is roughly equivalent to $1.44\text{ MeV}$).

Part (b): Temperature Calculation & Quantum Tunneling
Equate the required potential energy to the average thermal kinetic energy: $$2.30 \times 10^{-13}\text{ J} = \frac{3}{2} k_B T = \frac{3}{2} (1.38 \times 10^{-23}\text{ J/K}) T$$ $$T = \frac{2 \times (2.30 \times 10^{-13})}{3 \times (1.38 \times 10^{-23})} \approx \mathbf{1.11 \times 10^{10}\text{ K}}$$ The Quantum Reality: This classical calculation suggests we need $11$ billion Kelvin to ignite fusion! However, the core of the Sun is only 15 million Kelvin. Fusion works at this "low" temperature due to two factors: 1) The Maxwell-Boltzmann distribution dictates that a tiny, statistically significant fraction of particles are moving much faster than the average speed. 2) Quantum Tunneling: Particles exhibit wave-particle duality. There is a non-zero quantum probability that protons can "tunnel" through the Coulomb barrier even if they don't possess the required classical kinetic energy to breach it.

Example 2: The Sun's Mass Loss Rate (Classic IB Exam Concept)

Problem: The total power output (Luminosity) of the Sun is $L_\odot = 3.83 \times 10^{26}\text{ W}$. This energy is generated entirely by the mass defect of hydrogen fusion in its core.
(a) Calculate the rate at which the Sun is losing mass in $\text{kg s}^{-1}$.
(b) If the Sun has a total mass of $M_\odot = 1.99 \times 10^{30}\text{ kg}$, but only the inner $10\%$ of its mass is hot enough to undergo fusion, estimate the maximum remaining lifespan of the Sun (in years) assuming its luminosity remains constant. (Note: The hydrogen-to-helium proton-proton chain converts approximately $0.71\%$ of the fused hydrogen mass into pure energy).

Solution:

Part (a): Mass Loss Rate ($\Delta m / \Delta t$)
Using Einstein's mass-energy equivalence $E = \Delta m c^2$, we can differentiate with respect to time to get Power: $P = \frac{\Delta m}{\Delta t} c^2$. $$\frac{\Delta m}{\Delta t} = \frac{P}{c^2} = \frac{3.83 \times 10^{26}\text{ J/s}}{(3.00 \times 10^8\text{ m/s})^2}$$ $$\frac{\Delta m}{\Delta t} = \frac{3.83 \times 10^{26}}{9.00 \times 10^{16}} = \mathbf{4.26 \times 10^9\text{ kg s}^{-1}}$$ The Sun destroys over 4 million metric tons of its own mass every single second!

Part (b): Estimating Solar Lifespan
First, find the total mass of Hydrogen available for fusion (10% of total mass): $$M_{\text{core}} = 0.10 \times 1.99 \times 10^{30}\text{ kg} = 1.99 \times 10^{29}\text{ kg}$$ Only $0.71\%$ of this hydrogen mass is actually converted into pure energy (mass defect) during the fusion process. So, the total available mass that can be "lost" as energy is: $$\Delta M_{\text{available}} = 0.0071 \times 1.99 \times 10^{29}\text{ kg} = 1.41 \times 10^{27}\text{ kg}$$ Now, divide the available mass to be lost by the rate of mass loss calculated in Part (a): $$t = \frac{\Delta M_{\text{available}}}{\text{Rate of mass loss}} = \frac{1.41 \times 10^{27}\text{ kg}}{4.26 \times 10^9\text{ kg/s}} = 3.31 \times 10^{17}\text{ s}$$ Convert seconds to years ($1\text{ yr} \approx 3.15 \times 10^7\text{ s}$): $$t = \frac{3.31 \times 10^{17}}{3.15 \times 10^7} \approx \mathbf{1.05 \times 10^{10}\text{ years}}$$ Conclusion: The Sun has a total main-sequence lifespan of roughly 10.5 billion years. Since it is currently ~4.6 billion years old, it has about 5 to 6 billion years left.