 Solution

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

For resonance channel threshold conversion, equate the compound-state energy measured from the two different channel thresholds. The proton-plus-nitrogen threshold exceeds the alpha-plus-carbon threshold by

Δ = (938.26 + 13973.30 - 3728.34 - 11177.76) MeV = 5.46 MeV .

Thus

T_{p}^{*} = T_{α}^{*} - Δ = 2.113770 MeV

and

T_{p, lab} = T_{p}^{*} \frac{938.26 + 13973.30}{13973.30} = 2.25570 MeV ≃ 2.26 MeV .

The positive centre-of-momentum energy verifies that the proton channel is open. These numbers use the stipulated nonrelativistic kinematics and rounded nuclear masses.

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