• # question_answer A $\gamma -$ray photon produces an electron-positron pair, each moving with a K.E of 0.01 MeV The energy of the$\gamma -$ray photon is A)  1.02 MeV                   B)  1.04 MeV       C)    2.08 MeV        D)  1.03 MeV

Rest mass energy of an electron $={{m}_{0}}{{C}^{2}}=(9.1\times {{10}^{-11}})\times {{(3\times {{10}^{8}})}^{2}}J$ $=\frac{(9.1\times {{10}^{-31}})\times {{(3\times {{10}^{8}})}^{2}}}{1.6\times {{10}^{-13}}}\text{MeV}$ The total energy of an electron $=E={{m}_{0}}{{C}^{2}}+kE$ $\therefore$Total energy of two electrons $=2{{m}_{0}}{{C}^{2}}+2kE$ $\therefore$Energy of incident $\gamma -$ray photon             $=(2\times 0.51+2\times 0.01)\,MeV=1.04\,\text{MeV}$ Hence, the correction option is (b).