 Solution

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

Using the supplied masses,

μ_{α} c^{2} = \frac{3728.34 ( 11177.76 )}{14906.10} = 2795.801 MeV, T_{α}^{*} = 10.10 \frac{11177.76}{14906.10} = 7.573770 MeV .

Therefore

p_{α}^{*} = 205.790 MeV / c ≃ 206 MeV / c

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