Calculate the energy released by the electron-capture decay of Cobalt. Consider

Question

Calculate the energy released by the electron-capture decay of Cobalt. Consider only the energy of the nuclei (ignore the energy of the surrounding electrons). The following masses are given:
Cobalt: 56.936296 u
Iron: 56.935399 u

Express your answer in millions of electron volts (1 u= 931.5 MeV/c^2) to three significant figures.

Answer in MeV


Part B
A negligible amount of this energy goes to the resulting Iron atom as kinetic energy. About 90 percent of the time, after the electron-capture process, the Iron nucleus emits two successive gamma-ray photons of energies 0.122 MeV and 1.20×10-2 MeV in decaying to its ground state. The electron-capture process itself emits a massless neutrino, which also carries off kinetic energy. What is the energy of the neutrino emitted in this case?
Express your answer in millions of electron volts.
Answer in MeV.

Explanation / Answer

the energy released in electron capture, Q = (m2-m1)c^2 Q = (0.000897 amu)931.5 MeV = 0.8355555 MeV the energy released by the electron-capture decay of is 0.8355555 MeV. -------------------------------------------------------------------------------------- the energy released in electron capture, Q = 0.8355555 MeV this energy shared by two ?-rays and neutrino, hence Q = ?1+?2 + K.E of neutrino therefore K.E of neutrino = Q - ?1+?2 = (0.8355555 -0.134 - 0.0190 ) MeV Q = 0.6825555 MeV. the energy of the neutrino emitted in this case is 0.6825555 MeV.

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