University Physics IV · Nuclear Physics · 13.9

Fusion & the Gamow Peak

The Sun's core is five hundred times too cold to climb the proton–proton barrier, and it burns anyway. This topic locates the narrow window of energies where a rare fast particle meets a barrier thin enough to tunnel, and turns that window into a reaction rate.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Two nuclei feel the strong force only at a couple of femtometres, and reaching that separation costs a Coulomb energy of order an MeV, while the mean thermal energy in the Sun's core is 1.29 keV. Classically the Sun cannot shine. Two exponentials rescue it, and neither is worth anything alone.

Barrier penetration contributes exp(−b/√E), fantastically small at kT but climbing steeply with energy; the Maxwell–Boltzmann distribution contributes exp(−E/kT), which collapses just as steeply. Their product is a hump — the Gamow peak — centred at E₀ = (E_G(kT)²/4)¹⁄³, far out on the thermal tail and still far below the barrier top, and essentially the whole reaction rate comes from the few keV around it. Locating that hump is the entire calculation.

It says which energies a cross-section measurement must reach, why the rate follows a steep local power law T^((τ−2)/3) rather than an Arrhenius exponential, and why multiplying Z₁Z₂ by seven turns hydrogen burning from a T⁴ process into a T²⁰ one and hands the Sun's job to a different cycle in a heavier star. The cost is two assumptions: a non-degenerate Maxwellian plasma, which a white dwarf does not have, and a smooth non-resonant cross-section, which D–T does not have.

Simple definition: The Gamow peak is the band of collision energies — high on the Maxwell tail, yet far below the Coulomb barrier — that contributes almost all of a thermonuclear reaction's rate.
Example: For p+p in the Sun's core, kT = 1.29 keV and the barrier is about 720 keV, yet the peak sits at E₀ = 5.9 keV, is Δ = 6.4 keV wide, and is suppressed by e^(−13.7) ≈ 1.1×10⁻⁶.

Non-resonant cross-sectionσ(E) = S(E) · e^(−b/√E) / E

Peels off the two known energy dependences, leaving S(E) flat enough to extrapolate over decades.

E is centre-of-mass energy in keV; S(E) in keV⋅barn; b in keV^½

Gamow energy and Gamow constantE_G = 2μc²(π α Z₁Z₂)², b = √E_G

One number carries the whole pair, since b scales as Z₁Z₂√μ.

μ = reduced mass. p+p: E_G = 493 keV, b = 22.2 keV^½; D+T: b = 34.4 keV^½

Gamow peak energyE₀ = (b⋅kT/2)²⁄³ = (E_G(kT)²/4)¹⁄³

The one energy that dominates the rate integral — neither kT nor the barrier height.

In keV. Solar core kT = 1.29 keV gives E₀ = 5.9 keV for p+p

Peak width, height, and slopeΔ = 4√(E₀kT/3), τ = 3E₀/kT, ⟨σv⟩ ∝ T^((τ−2)/3)

Δ is the band a laboratory must cover; e^(−τ) is the entire barrier suppression.

Δ/E₀ = 4/√τ. Solar p+p: Δ = 6.4 keV, τ = 13.7, exponent n = 3.9

Rate per unit volumer₁₂ = n₁ n₂ ⟨σv⟩ / (1 + δ₁₂)

The delta halves p+p and D+D so each identical pair is counted once, not twice.

m⁻³ s⁻¹, with n in m⁻³ and ⟨σv⟩ in m³ s⁻¹; δ₁₂ = 1 for identical nuclei

Lawson triple product (D–T)n T τ_E ≳ 3×10²¹ m⁻³ keV s

The reactor version of the same physics: you must hold the plasma, not merely heat it.

τ_E is the energy confinement time; T in keV, near the 15 keV operating optimum

Classically, the Sun is dark

Two protons feel the strong force only when their centres are a couple of femtometres apart, and getting there costs Coulomb energy: with e²/4πε₀ = 1.44 MeV⋅fm, pushing them to 2.0 fm costs 0.72 MeV. The Sun's core sits at 1.5×10⁷ K, so kT = 1.29 keV, and the barrier stands 557 thermal energies above the mean. A classical activation estimate — the fraction of pairs with enough energy to go over — gives e⁽⁻⁵⁵⁷⁾ ≈ 10⁽⁻²⁴²⁾. Multiply by every proton–proton collision the Sun has had in 4.6 billion years, of order 10⁸², and you still predict 10⁽⁻¹⁶⁰⁾ reactions. The Sun shines, so this is not a picture that needs a correction factor; it is the wrong picture. Whatever supplies the rate must send particles through the barrier, not over it.

Tunnelling supplies the factor e^(−b/√E)

Run WKB across the Coulomb barrier: the transmission is exp(−(2/ħ)∫√(2μ(V−E)) dr) between the classical turning points. For E well below the barrier top that integral collapses to 2πη, with η = Z₁Z₂αc/v the Sommerfeld parameter. Substituting E = ½μv² converts it to exp(−b/√E), where b = √E_G and E_G = 2μc²(παZ₁Z₂)². Everything about the pair — charges and reduced mass — is compressed into b, which scales as Z₁Z₂√μ: 22.2 keV^½ for p+p, 25.6 for p+d, 34.4 for D+T, 212 for p+¹⁴N. The function is savage at low energy. At E = 1.29 keV it is 3.3×10⁻⁹; at 5.9 keV, 1.1×10⁻⁴; at 30 keV, 1.7×10⁻². Merely doubling the energy from 5.9 to 11.8 keV multiplies it by about 15.

Folding the two exponentials gives one hump

Multiply the population factor e^(−E/kT), which collapses as E rises, by the penetration factor e^(−b/√E), which explodes. The combined exponent is f(E) = E/kT + b/√E; setting df/dE = 1/kT − b/(2E³⁄²) to zero gives the Gamow peak at E₀ = (b⋅kT/2)²⁄³ = (E_G(kT)²/4)¹⁄³. For p+p in the Sun that is (22.2 × 1.29/2)²⁄³ = 5.9 keV — 4.6 kT, and still 122 times below the barrier. Expanding f about E₀ gives a Gaussian of full width Δ = 4√(E₀kT/3) = 6.4 keV, with peak value e^(−τ) where τ = f(E₀) = 3E₀/kT = 13.7. Note that Δ/E₀ = 4/√τ, so the solar pp 'peak' is about as wide as its own centre energy: a broad hump, not a spike, and the Gaussian step is only good to a factor of order unity. Only when τ is large — CNO's 62 — is it genuinely narrow.

The S-factor is what a laboratory can report

Write the cross-section as σ(E) = S(E)e^(−b/√E)/E. The 1/E is the geometric πƛ² factor and the exponential is the barrier; whatever remains, S(E), is nuclear structure, and for a non-resonant reaction it varies slowly enough to extrapolate across decades. The numbers show why that matters. S(0) for p+p is 4.0×10⁻²⁵ MeV⋅b, absurdly small because a proton must weak-decay into a neutron during the collision; for ³He+³He it is 5.4 MeV⋅b, twenty-five orders of magnitude larger. At the solar Gamow peak the p+p cross-section is around 10⁻²⁶ barn — it has never been measured and never will be, and the rate is computed from weak-interaction theory instead. Accelerators reach 0.1–1 MeV, underground laboratories such as LUNA a few tens of keV, and S bridges the gap. The trap is resonance: D+T has a broad ⁵He level at 64 keV, so its S is anything but flat and the smooth Gamow formula misleads.

Two hydrogen-burning routes, two temperature exponents

Doing the Gaussian integral gives ⟨σv⟩ ∝ (kT)^(−2/3)e^(−τ), and since τ ∝ (kT)^(−1/3) the local logarithmic slope is n = d ln⟨σv⟩/d ln T = (τ−2)/3. For solar p+p, τ = 13.7 gives n = 3.9. The CNO cycle's slowest link, ¹⁴N(p, γ)¹⁵O, has Z₁Z₂ = 7 and b = 212 keV^½, so at the same temperature E₀ = 26.6 keV, τ = 61.7 and n = 19.9. Both routes do identical book-keeping — 4¹H → ⁴He + 2e⁺ + 2νₑ, Q = 26.73 MeV with about 0.6 MeV leaving as neutrinos — but they answer temperature completely differently: a core 10% hotter multiplies CNO by 6.7 and pp by only 1.45. That is why the two cross near 1.7×10⁷ K, why stars above roughly 1.3 M☉ burn by CNO, and why the Sun at 1.5×10⁷ K draws only about 1% of its luminosity from CNO, a fraction Borexino has now measured. Remember n is a local slope, not a law: it falls as T rises.

From a stellar rate to a reactor criterion

The rate per unit volume is r₁₂ = n₁n₂⟨σv⟩/(1 + δ₁₂), the Kronecker δ halving the identical-particle cases p+p and D+D so no pair is counted twice. With nₚ ≈ 3×10³¹ m⁻³ in the solar core this yields a central power density near 280 W m⁻³ — less than a compost heap. The Sun is bright because it is enormous, not because its core is fierce. A reactor has no such option: it runs D–T at kT ≈ 15 keV, where ⟨σv⟩ is some 10²⁷ times the solar p+p value, and the problem becomes holding the energy in, which is the Lawson condition nTτ_E ≳ 3×10²¹ m⁻³ keV s. Two assumptions sit underneath all of this. The plasma must be non-degenerate and Maxwellian — in a white dwarf it is neither, and reactions become pycnonuclear, driven by density rather than temperature — and electrons must not screen the barrier, which in the laboratory they do, lifting measured rates by a few per cent and biasing exactly the low-energy data you most want.

Change one variable at a time

Make the relationship visible.

Interactive model

Plasma temperature T15 10⁶ K

Gamow constant b22.2 √keV

Push T from 15 to 40 and the peak slides right and lifts — that is the Tⁿ law made visible. Then set b = 34.4, the D–T value, against 22.2 for p+p: the peak moves right but sinks, because more charge means a taller barrier and a rarer, higher-energy window.

PEAK ENERGY E₀5.90 keV

PEAK WIDTH Δ6.38 keV

EXPONENT τ = 3E₀/kT13.7

RATE SLOPE n3.9

Live interpretationPEAK ENERGY E₀: 5.90 keV. PEAK WIDTH Δ: 6.38 keV. EXPONENT τ = 3E₀/kT: 13.7. RATE SLOPE n: 3.9

Catch the common trap

Explain before calculating.

Misconception check

The classic error is to keep the classical picture and bolt tunnelling on in the wrong place — writing the rate as e^(−E_C/kT) with E_C the barrier height, or, more subtly, evaluating the cross-section at the mean thermal energy and multiplying by the number of particles. Both fail for the same reason: the rate is an integral, and its integrand is a product of two functions pulling in opposite directions, so neither endpoint dominates it. In the Sun the integrand at 5.9 keV beats the one at kT = 1.29 keV by a factor of about 900, even though vastly more protons sit near kT. The other half of the error is to imagine the Gamow peak is a resonance, or a property of the nuclei. It is neither. It is the maximum of a product of two perfectly smooth monotonic functions, and it shifts whenever the temperature shifts, as (kT)²⁄³.

The Sun's core has kT = 1.29 keV, and for p+p (b = 22.2 keV^½) the Gamow peak sits at E₀ = 5.9 keV. A more massive star's core runs twice as hot, kT = 2.59 keV, with the same reaction. Where is E₀ there?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe Sun's core is at T = 1.5×10⁷ K. Using k = 8.617×10⁻⁸ keV K⁻¹ and e²/4πε₀ = 1.44 MeV⋅fm, find kT, estimate the Coulomb barrier for two protons brought to r = 2.0 fm, and show that a classical Boltzmann estimate leaves the Sun dark.

Thermal energy: kT = (8.617×10⁻⁸ keV K⁻¹)(1.5×10⁷ K) = 1.29 keV. This is the mean of a distribution, not a threshold — the Maxwell tail runs far above it.
Barrier: E_C = Z₁Z₂e²/4πε₀r = (1)(1)(1.44 MeV⋅fm)/(2.0 fm) = 0.72 MeV = 720 keV.
Ratio: E_C/kT = 720/1.29 = 557, so the barrier stands 557 thermal energies above the mean.
Classical estimate: the fraction of pairs above the barrier goes as e⁽⁻⁵⁵⁷⁾ = 10^{(−557/2.303)} = 10⁽⁻²⁴²⁾.
The Sun has run of order 10⁸² proton–proton collisions in 4.6 Gyr, so this predicts 10⁽⁻¹⁶⁰⁾ reactions — exactly zero. The barrier has to be tunnelled, not climbed.

AnswerkT = 1.29 keV, E_C ≈ 0.72 MeV = 557 kT, and e⁽⁻⁵⁵⁷⁾ ≈ 10⁽⁻²⁴²⁾. Classical activation under-predicts the Sun by roughly 160 orders of magnitude, so stellar fusion is a tunnelling problem.

MediumFor p+p the Gamow constant is b = 22.2 keV^½. At the solar core temperature (kT = 1.29 keV) locate the Gamow peak, find its width and the exponent τ, and compare the rate integrand at the peak with its value at the mean thermal energy.

The integrand is e^(−E/kT) × e^(−b/√E) = e^(−f), with f(E) = E/kT + bE^(−1/2).
Set df/dE = 1/kT − b/(2E³⁄²) = 0, giving E₀³⁄² = b⋅kT/2, so E₀ = (b⋅kT/2)²⁄³.
Numerically b⋅kT/2 = 22.2 × 1.29/2 = 14.3, and 14.3²⁄³ = 5.90 keV — that is 4.6 kT, and 122 times below the 720 keV barrier.
Exponent: τ = f(E₀) = 3E₀/kT = 3(5.90)/1.29 = 13.7, so the peak height is e^(−13.7) = 1.1×10⁻⁶. Width: Δ = 4√(E₀kT/3) = 4√(5.90 × 1.29/3) = 4√2.54 = 6.4 keV.
At E = kT the integrand is e^{(−1 − 22.2/√1.29)} = e^{(−1 − 19.5)} = e^(−20.5) = 1.2×10⁻⁹.
Ratio: 1.1×10⁻⁶ ÷ 1.2×10⁻⁹ ≈ 9×10². The mean thermal energy contributes nearly a thousand times less than the peak, despite holding far more particles.

AnswerE₀ = 5.9 keV (4.6 kT), Δ = 6.4 keV, τ = 13.7. The integrand at E₀ exceeds the one at kT by about 9×10², which is why the rate is set by a band around 5.9 keV rather than by kT itself.

HardThe CNO cycle's slowest link is ¹⁴N(p, γ)¹⁵O, with reduced mass μc² = 875 MeV and Z₁Z₂ = 7. Taking πα = 0.02293 and the solar kT = 1.29 keV, find its Gamow energy, peak energy, τ and local temperature exponent, and use them to say why the Sun runs on the pp chain while a 1.5 M☉ star does not.

Gamow energy: E_G = 2μc²(παZ₁Z₂)² = 2(875 MeV)(0.02293 × 7)² = 1750 × 0.02575 = 45.1 MeV, so b = √(45 100 keV) = 212 keV^½ — 9.6 times the p+p value.
Peak: E₀ = (b⋅kT/2)²⁄³ = (212 × 1.29/2)²⁄³ = 137²⁄³ = 26.6 keV, four and a half times the pp peak at the same temperature.
Exponent: τ = 3E₀/kT = 3(26.6)/1.29 = 61.7, so the barrier costs e^(−61.7) ≈ 1.5×10⁻²⁷, against 1.1×10⁻⁶ for p+p — CNO is handicapped by 21 orders of magnitude at this temperature.
Temperature slope: n = (τ − 2)/3 = (61.7 − 2)/3 = 19.9, against 3.9 for p+p.
A core 10% hotter therefore multiplies CNO by 1.10^19.9 = 6.7 and pp by only 1.10^3.9 = 1.45. Compounding that, the two rates cross near 1.7×10⁷ K.
A 1.5 M☉ star has a core hotter than that, so it burns hydrogen by CNO; and because n ≈ 20 makes its energy generation hair-trigger in T, the released power is concentrated enough to drive a convective core, which the Sun does not have.

AnswerE_G = 45 MeV, b = 212 keV^½, E₀ = 26.6 keV, τ = 61.7, n ≈ 20 — against 493 keV, 22.2 keV^½, 5.9 keV, 13.7 and n ≈ 3.9 for p+p. CNO loses badly at 1.5×10⁷ K but gains 6.7× per 10% in T, so it takes over above about 1.7×10⁷ K.