 Solution

ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2016/paper-1/4/e/solution

M c^{2}

, where

M = m_{p} + m_{n}

. The emitted photon carries momentum, requiring equal and opposite nuclear recoil. Some binding energy becomes deuteron kinetic energy.

Exact two-body kinematics in natural units gives

M = E_{γ} + m_{d}^{2} + E_{γ}^{2}, E_{γ} = \frac{M ^{2} - m _{d}^{2}}{2 M} .

Since

m_{d} = M - B

E_{γ} = B - \frac{B ^{2}}{2 M}, B - E_{γ} = \frac{B ^{2}}{2 M} .

With the stated rounded masses,

B - E_{γ} = \frac{( 2.224 ) ^{2}}{2 ( 938.272 + 939.565 )} MeV = 0.00131699 MeV = 1.317 keV .

2.222683 MeV

. At leading order the difference equals the nonrelativistic recoil estimate

p_{γ}^{2} / (2 m_{d}) ≃ B^{2} / (2 m_{d})

; exact kinematics fixes the small distinction between using

m_{d}

and

M

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