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: 56.936296u5726Fe: 56.935399uExpress your answer in millions of electron volts (1u=931.5MeV/c2) to three significant figures.

Question

Larissa Hansen

Physics

22 December, 21:09

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: 56.936296u5726Fe: 56.935399uExpress your answer in millions of electron volts (1u=931.5MeV/c2) to three significant figures.

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Reese Briggs · Answer 1

The mass difference is

Δm = m2 - m1

56.936296 u - 56.935399 u

= 0.000897‬ u

The energy released by the electron-capture decay

E = Δmc²

(0.000897‬ u) c² (931. 5 MeV / c²: 1 u)

= 0.8355555 MeV