Two protons repel each other electrically. To fuse, they must come within about a femtometer ($10^{-15}$ m), where the strong nuclear force grabs them. You saw in estimating a star's center that the Sun's core is at about 15.7 million K. Is that hot enough? Classically, not even close. The rescue is quantum tunneling.
💡 Why it works
The Coulomb barrier at separation $r$ is $U = kZ_1Z_2e^2/r$, with $ke^2 = 1.44$ MeV·fm. For two protons at about 1 fm, $U \approx 1.4$ MeV. In the core, $kT = (8.617 \times 10^{-5}\text{ eV/K})(1.57 \times 10^{7}\text{ K}) = 1.35$ keV, about a thousand times less. Even counting the fast tail of the Maxwell distribution, the fraction of protons with 1 MeV goes as $e^{-E/kT} = e^{-740}$: zero for all practical purposes. Quantum mechanics changes the question. A proton's wave function leaks into the barrier, and the probability of getting through falls off as $$P \approx e^{-\sqrt{E_G/E}}, \qquad E_G = 2m_rc^2(\pi\alpha Z_1Z_2)^2,$$ where $m_r$ is the reduced mass and $\alpha \approx 1/137$. For two protons $E_G \approx 493$ keV.
🔑 Key idea
The Sun's core is about a thousand times too cool for protons to climb over the Coulomb barrier. Protons fuse only by tunneling through it, and mostly at energies of a few times $kT$ (the Gamow peak, about 6 keV in the Sun), where enough protons are fast and the tunneling chance is no longer hopeless.
The potential energy of two protons against their separation: electric repulsion outside about 1.5 fm, nuclear attraction inside. The dashed line is the typical thermal energy in the Sun's core.
Worked examples
✏️ Worked example: How far short is the core?
Compare the barrier for two protons at 1.0 fm with $kT$ at 15.7 million K.
Answer: $U/kT \approx 1060$. Classically, protons would need about $1.7 \times 10^{10}$ K to reach the barrier with typical energies.
✏️ Worked example: The tunneling factor
Using $E_G = 493$ keV for two protons, find the tunneling factor at $E = kT = 1.35$ keV and at $E = 6$ keV.
At 1.35 keV: $\sqrt{493/1.35} = 19.1$, so $P \approx e^{-19.1} = 5 \times 10^{-9}$.
At 6 keV: $\sqrt{493/6} = 9.1$, so $P \approx e^{-9.1} = 1 \times 10^{-4}$.
Answer: Tiny, but not zero, and much larger for the faster protons. With about $10^{56}$ protons in the core, tiny is enough.
✏️ Worked example: A quantum estimate of the temperature needed
Suppose fusion becomes possible when a proton's de Broglie wavelength $h/p$ reaches the closest-approach distance $ke^2/E$. With $E = p^2/(2m_r) = \tfrac32kT$ and $m_r = m_p/2$, what temperature is needed?
Answer: About $1 \times 10^{7}$ K, right where the Sun's core is. Quantum mechanics lowers the requirement by a factor of about a thousand.
⚠️ Common mistake
Thinking the Sun's core is hot enough for protons to climb over the Coulomb barrier. Typical protons there have about 1 keV, against a barrier near 1 MeV. When Arthur Eddington proposed in 1920 that stars run on hydrogen fusion, critics objected that stars are not hot enough; tunneling, worked out by George Gamow in 1928 and applied to stars soon after, answered them. Check: compare $kT$ in keV with the barrier in MeV before calling anything "hot enough".
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.