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Solutions.

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Presentation on theme: "Solutions."— Presentation transcript:

1 Solutions

2 Cross section vs. experimental yield
[for special case of “thin” target] Example: H + 1H 2H + e+ +  (first step of pp chain; solar Gamow peak: 6 keV) cross section: theo=8·10-48 cm2=8·10-24 barn at Elab=1 MeV [Ecm=0.5 MeV] proton beam: total charge (Q), current (I), elementary charge N0/t=(Q/t)/q=I/q=1 Ampere/1.6·10-19 C=6·1018 incident protons/s proton target: dense hydrogen gas [Nt=1020 protons/cm2] ® Nr=(8·10-48 cm2)(1020 protons/cm2)(6·1018 protons/s)=5·10-9/s [1 reaction in 6 years!] We need very high beam currents of very low energies!

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4 From: Newton, J. R. , Longland, R. , and Iliadis, C
From: Newton, J.R., Longland, R., and Iliadis, C. Matching of experimental and statistical-model thermonuclear reaction rates at high temperatures. Phys. Rev. C 78, , 8 pages (2008). Se also: Newton, J.R., Iliadis, C., Champagne, A.E., Coc, A., Parpottas, Y., and Ugalde, C. Gamow peak in

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