A proton beam is fully stopped in a thick target ( )

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Nuclear Physics · Nuclear reactions

The Nucleus Resolved, Ch Reactions: four-vectors, amplitudes and partial waves

Problem

A proton beam is fully stopped in a thick 63Cu^{63}\mathrm{Cu} target (M=63.0 g mol−1M = 63.0\,\mathrm{g\,mol^{-1}}), entering at E1=8.00 MeVE_1 = 8.00\,\mathrm{MeV} and slowing to E2=2.00 MeVE_2=2.00\,\mathrm{MeV} before the reaction channel of interest switches off. Take the nuclear radius R=1.2×631/3 fmR = 1.2\times63^{1/3}\,\mathrm{fm}, the proton rest energy mpc2=938.3 MeVm_pc^2 = 938.3\,\mathrm{MeV}, and ℏc=197.327 MeV fm\hbar c = 197.327\,\mathrm{MeV\,fm}; treat the protons non-relativistically (E≪mpc2E \ll m_pc^2) and use the de Broglie wavenumber k=2mpc2E/ℏck=\sqrt{2m_pc^2E}/\hbar c. ((a)) Calculate kRkR at E1E_1 and at E2E_2, and state how many partial waves (ℓ=0\ell=0 up to the nearest integer) contribute to the reaction at each energy. ((b)) At E1E_1 the elasticity parameters for the contributing waves are η0=0.20\eta_0=0.20, η1=0.55\eta_1=0.55, η2=0.85\eta_2=0.85, η3=0.98\eta_3=0.98; at E2E_2 only η0=0.20\eta_0=0.20 and η1=0.55\eta_1=0.55 contribute (higher waves do not reach the nucleus). Calculate the reaction cross section σreac=(π/k2)∑ℓ(2ℓ+1)(1−∣ηℓ∣2)\sigma_{\rm reac}=(\pi/k^2)\sum_\ell(2\ell+1)(1-|\eta_\ell|^2) at each energy. ((c)) Using the mean of the two cross sections as a trapezoidal estimate of the excitation function, and a mean stopping power (−dE/dx)/ρ=60.0 MeV cm2 g−1(-\mathrm{d}E/\mathrm{d}x)/\rho = 60.0\,\mathrm{MeV\,cm^2\,g^{-1}} over this energy range, estimate the thick-target yield Y≈NAM σˉreac E1−E2(−dE/dx)/ρY \approx \frac{N_A}{M}\,\bar\sigma_{\rm reac}\,\frac{E_1-E_2}{(-\mathrm{d}E/\mathrm{d}x)/\rho} in reactions per incident proton, with NA=6.022×1023 mol−1N_A = 6.022\times10^{23}\,\mathrm{mol^{-1}}.

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