A) 1.02 MeV
B) 1.04 MeV
C) 2.08 MeV
D) 1.03 MeV
Correct Answer: B
Solution :
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).
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