Answer
1.592 MeV
$\lambda=7.81\times10^{-13}m$
Work Step by Step
The photon’s energy equals the sum of the total energy of the other 2 particles.
$$E=2(KE+mc^2)=2(0.285MeV +0.511MeV)=1.592MeV$$
Now find the wavelength using equation 27–6.
$$\lambda=\frac{hc}{E}$$
$$\lambda=\frac{(6.63\times10^{-34}J \cdot s) (3.00\times10^8m/s)}{(1.592\times10^{6}eV)(1.60\times10^{-19}J/eV) }$$
$$\lambda=7.81\times10^{-13}m$$