Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy of the surr

Question

Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy of the surrounding electrons). The following masses are given:
5727Co: 56.936296u
5726Fe: 56.935399u
Express your answer in millions of electron volts (1u=931.5MeV/c2) to three significant figures.
A negligible amount of this energy goes to the resulting 5726Fe atom as kinetic energy. About 90 percent of the time, after the electron-capture process, the 5726Fe nucleus emits two successive gamma-ray photons of energies 0.140MeV and 1.70 102MeV 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.

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Yến Oanh 1 year 2021-08-31T14:19:10+00:00 1 Answers 86 views 0

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    2021-08-31T14:21:08+00:00

    Answer:

    Explanation:

    ⁵⁷Co₂₇  + e⁻¹  =  ²⁷Fe₂₆

    mass defect = 56.936296 + .00055 – 56.935399

    = .001447 u

    equivalent energy

    = 931.5 x .001447 MeV

    = 1.3479 MeV .

    = 1.35 MeV

    energy of gamma ray photons = .14  + .017

    = .157 MeV .

    Rest of the energy goes to neutrino .

    energy going to neutrino .

    = 1.35 – .157

    = 1.193 MeV.

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