He 80br (atomic number 35) nuclide decays either by β− decay or by electron capture. (masses of atoms: 80br =79.918528 amu; 80kr =79.916380 amu; 80se =79.916520 amu. neglect the mass of electrons involved because these are atomic, not nuclear, masses.) (a) write the balanced nuclear equations for each process below. use the isotope tool in the palette to enter both the mass numbers and the atomic numbers for each nuclide.

Answer : The balanced nuclear equations of each process.

The nuclide decay of by decay is as follows:

The nuclide decay of by the electron capture is as follows:

Explanation :

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."

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