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 $x\times {10}^{-3}\text{Å}$. The value of x is  [Take h = 6.62$\times$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

 $\begin{array}{c}\mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{E}}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{1.51\times {10}^6\times 1.6\times {10}^{-19}}==8.22\times {10}^{-13}\text{m}\\ =8.22\times {10}^{-3}\text{Å}\end{array}$