 Solution

ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2012/paper-1/4/iii/solution

The gamma width is

0.10 eV = 1 0^{- 7} MeV

, so the remaining proton width is

Γ_{p} = 0.20 - 0.15 - 1 0^{- 7} = 0.0499999 MeV

. At the peak of the same resonance the total-width denominator and gamma exit width cancel. alpha particles and carbon-12 have spin zero; the proton and the stated nitrogen level each have spin

1/2

. Hence

g_{α} / g_{p} = 4

, independently of the unspecified resonance spin. The proton reduced mass and momentum are

μ_{p} c^{2} = 879.223 MeV, p_{p}^{*} c = 2 (879.223) (2.113770) = 60.967 MeV .

\frac{σ _{α}}{σ _{p}} = \frac{g _{α}}{g _{p}} \frac{p _{p}^{* 2}}{p _{α}^{* 2}} \frac{Γ _{α}}{Γ _{p}} = 4 \frac{60.96 7 ^{2}}{205.79 0 ^{2}} \frac{0.15}{0.0499999} ≃ 1.053 .

The ratio compares each reaction at its own laboratory resonance energy. Factors of

ℏ^{2}

restoring the cross section's area units cancel. Only the supplied nitrogen spin magnitude is used: its printed positive parity is immaterial to this spin average, so no orbital-parity inference is made from that assignment.

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