Exercises

1. $kT = 8.617 \times 10^{-5}\ \text{eV/K} \times 1.5 \times 10^{7}\ \text{K}$ = 1.293 keV.
2. $\mu$ = 0.94016 u, so $\mu c^2$ = 875.75 MeV, and $E_G$ = 45.1 MeV (45 106 keV).
3. $E_0$ = 26.6 keV and $\Delta$ = 13.5 keV: the window runs from about 20 to 33 keV. $E_0$ is about 21 times $kT$: the reactions come from the far tail of the thermal distribution, where protons are few but the barrier is a little thinner.
4. The atlas gives 27.4 keV and 14.1 keV at 15.7 MK; scaled to 15 MK they become 26.6 keV and 13.5 keV, as above. The rate explorer shows 26.5 ± 6.8 keV: it takes the mass numbers (14 and 1) for the reduced mass, and a $\mu$ 0.7% smaller moves $E_0$ by 0.2% ($E_0 \propto \mu^{1/3}$).
5. $\sigma(E) = \dfrac{S}{E}\, e^{-\sqrt{E_G/E}}$, so $\dfrac{\sigma(E_0)}{\sigma(70\ \text{keV})} = \dfrac{70}{E_0}\, e^{\sqrt{E_G/70} - \sqrt{E_G/E_0}}$ ≈ 3.7 × 10⁻⁷: about 2.7 million times smaller. That is why, at the Sun's energies, the cross section is extrapolated (exercise 3) rather than measured.

1. $\sigma = \dfrac{1.7\ \text{keV b}}{70\ \text{keV}}\, e^{-\sqrt{E_G/70\ \text{keV}}}$ = 2.3 × 10⁻¹³ b, that is 0.23 pb (1 pb = 10⁻³⁶ cm²).
2. 200 µA is 1.25 × 10¹⁵ protons per second. Reactions per second: $\sigma\, n_t\, N_p$ = 2.3 × 10⁻³⁷ cm² × 10¹⁸ cm⁻² × 1.25 × 10¹⁵ s⁻¹ = 2.9 × 10⁻⁴ s⁻¹, or 25 a day. Detected: 0.25 a day at 1% efficiency, about 12 a day at 50%. This is what the atlas means by reactions that happen “only a few times per day”.
3. At the surface: 1 cm⁻² min⁻¹ × 100 cm² × 1440 min = 144 000 muons a day. Underground, a million times fewer: 0.14 a day, about one a week.
4. With 12 signal events a day, the muon background equals it as soon as one muon in 11 630 (9 × 10⁻⁵) lands in the window. A muon that crosses the whole crystal leaves tens of MeV in it, above the window; but those that clip a corner, and the electrons, photons and neutrons the muons make in the shielding and the walls, spread energy over the whole spectrum. Keeping fewer than one in ten thousand in a window a few hundred keV wide is not realistic at the surface: that is why the muon background “limits the sensitivity for γ-ray energies up to several MeV” (methods). Underground the problem disappears: even if every muon landed in the window, it would be 0.14 events a day against 12. The background that remains comes from the radioactivity of the rock and of the set-up, nearly all of it below 2.6 MeV (the most energetic γ-ray line of natural radioactivity, from ²⁰⁸Tl), and from beam-induced reactions on target contaminants. Felsenkeller in Dresden is the middle way: 45 m of rock reduces the muons by a factor of 40, and with an active veto the background is about twice that of a deep laboratory (timeline).
5. At the Gamow peak, $E_0$ = 26.6 keV, $\sigma$ ≈ 8 × 10⁻²⁰ b: with the same beam, target and 50% efficiency, 2 × 10⁻³ reactions a year, one every 604 years or so. No laboratory measures there: the Sun's energies are reached by extrapolating the data with R-matrix theory, which is what exercise 3 is about.

1. At the peak, $E = E_R$: $\sigma \propto 4\Gamma_\gamma/\Gamma_p$, so it falls by 34/40 = 0.85, −15%. The width of the peak is $\Gamma \approx \Gamma_p$: +18%. Far from the resonance, $(E-E_R)^2 \gg \Gamma^2/4$ and $\sigma \propto \Gamma_p \Gamma_\gamma/(E-E_R)^2$: the tail rises by 18%. A wider, lower peak with higher wings.
2. and 3. With AZURE2 on this evaluation (computed on 13 October 2026; only $\Gamma_p$ changed, everything else as in the model file):

The peak and the width do what the one-level formula says. The model's χ²/N is already far above 1 on this page: the model file is a preliminary version that does not yet reproduce the published best fit (the note on the evaluation page explains). What counts is the change, not the absolute value. And the change is huge: the data fix $\Gamma_p$ far more tightly than the 6 keV you moved it by. 4. The LUNA data sit on the low-energy flank of the resonance, where the $S$-factor rises by some 10–14% (at 300 keV from 6.72 to 7.64 keV b), and their uncertainties are around 1%: on average every point moves by about 17 standard deviations (the square root of Δχ²/N), and the χ² grows with the square of the shift. At stellar energies, though, the $S$-factor grows by only 9%, not 18%: there it is not only the resonance's tail but also direct capture, which does not depend on $\Gamma_p$, and the two amplitudes add (interfere) before they are squared. Only the resonant part grows. 5. With the energy moved by 5 keV the χ² of all the data reaches 304 958, more than for the change of width: on the steep flanks of a resonance 34 keV wide, moving it by 5 keV changes the $S$-factor by many times the uncertainty of the data measured there. That is why the energy of a well-measured resonance is the most tightly constrained parameter of an R-matrix fit.