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.
Given: u = 6.5 m/s, initial velocity a = 1.5 m/s², acceleration s = 100.0 m, displacement
Let v = the velocity attained after the 100 m displacement. Use the formula v² = u² + 2as v² = (6.5 m/s)² + 2*(1.5 m/s²)*(100 m) = 342.25 (m/s)² v = 18.5 m/s
White dwarfs are likely to be much more common. The number of stars decreases with increasing mass, and only the most massive stars are likely to complete their lives as black holes. There are many more stars of the masses appropriate for evolution to a white dwarf.
