Question

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 1

Answer : The balanced nuclear equations of each process.

The nuclide decay of $_{35}^{80}\textrm{Br}$ by $\beta ^{-}$ decay is as follows:

$_{35}^{80}\textrm{Br}\rightarrow _{-1}^{0}\beta + _{36}^{80}\textrm{Kr}$

The nuclide decay of $_{35}^{80}\textrm{Br}$ by the electron capture is as follows:

$_{35}^{80}\textrm{Br} + _{-1}^{0}\textrm{e}\rightarrow _{34}^{80}\textrm{Se}$

Explanation :

• Beta decay is the process that unstable atoms can use to become more stable atoms. There are two types of Beta decay, Beta-plus and Beta-minus.
• In Beta-minus decay, a neutron in an atom's convert into proton, an electron and an antineutrino.

• Electron capture is the process that unstable atoms can use to become more stable atoms.

• In Electron capture, an electron in an atom's inner shellis drawn into the nucleus where it combines with a proton and forming a neutron & neutrino.