 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2019/10/b/solution

Use the squared total energy-momentum four-vector, expressed in energy-squared units. With the proton initially at rest,

W^{2} = (m_{p} c^{2} + E_{γ})^{2} - E_{γ}^{2} = m_{p}^{2} c^{4} + 2 m_{p} c^{2} E_{γ} .

W = (m_{p} + m_{π}) c^{2}

. Solving gives

E_{γ}^{th} = m_{π} c^{2} (1 + \frac{m _{π}}{2 m _{p}}) = 144.715 MeV .

The numerical value uses the supplied

938

and

135 MeV

rest energies. The threshold exceeds the pion's rest energy because the outgoing system must carry the incident photon's nonzero laboratory momentum.

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