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NeTakaya
2 years ago
9

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.
Physics
1 answer:
erma4kov [3.2K]2 years ago
5 0

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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Explanation:

given,

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