A 1 MeV positron and a 1 MeV electron meet each moving in opposite directions. They annihilate each other by emitting two photons. If the rest mass energy of an electron is 0.51 MeV, the wavelength of each photon is:

sum of rest mass energy of electron and positron = 0.51 + 0.51 =1.02 MeV.

Initial energy of electron and positron = 1 + 1 = 2 MeV.

Total initial energy of electron and positron = 2 + 1.02 = 3.02 MeV

Therefore, the energy of the two photons = 3.02 MeV.

Hence, energy of each photon is E = 1.51 MeV = $1.51\times {10}^6 \mathrm{eV}$

Now, the wavelength of a photon of energy E ( in eV) is given by

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{E}\text{ (in }\mathrm{eV})}\text{Å}=\frac{12400 \mathrm{eV}}{1.51\times {10}^6 \mathrm{eV}}\text{Å} \\ =8.2\times {10}^{-3}Å \end{gathered}$$