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 or a positron is 0.51 MeV, the wavelength of each photon is x×10−3Å. The value of x is  [Take h = 6.62×10-34]

Energy released in annihilation = 0.51 + 0.51 =1.02 MeV.Initial energy = 1 +1 = 2 MeV. Therefore, the energy of the two photons is 1.02 + 2 = 3.02 MeV. Hence energy of each photon is E = 1.51 MeV. The wavelength of a photon of energy E (in eV) is given by λ=hcE=6.62×10−34×3×1081.51×106×1.6×10-19==8.22×10−13m                      =8.22×10−3Å