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Question

Gold 19879Au undergoes β decay to an excited state of 19880Hg. If the excited state, decays by emission of a γ- photon with energy 0.412MeV, the maximum kinetic energy of the electron emitted in the decay is (This maximum occurs when the antineutrino has negligible energy. The recoil energy of the 19880Hg nucleus can be ignored. The masses of the neutral atoms in their ground states are 197.968225 u for 19879Au and 197.966752u for 19879Hg.)

A
0.412MeV
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B
1.371MeV
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C
0.961MeV
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D
1.473MeV
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Solution

The correct option is C 0.961MeV
Given reaction is:
19879Au 19880Hg+ 01e+γ
Energy obtained is, (M19879AuM19880Hg)c2=(197.968225u197.966752u)(3×108)2
And, 1u=1.6605×1027kg
Therefore, energy obtained in reduction of mass is 2.2×1013J, which is 1.373 MeV
Photon carries 0.412 MeV, which leaves electron with energy 1.3730.412MeV
Hence, electron has energy of 0.961 MeV.

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