What is the daughter nucleus produced when 64Cu undergoes beta decay?

1 Answer
Apr 2, 2016


#""_64^30Zn# or #""_64^28Ni# depending on the type of β-decay.


β-decay is when either a proton decays into a neutron, releasing a positron, or a neutron decays into a proton, releasing an electron.

By looking at the periodic table, we find that copper has an atomic number of 29, so we would write copper-64 as #""_64^29Cu#.

Assuming that we are dealing with neutron decay, the mass number does not change, because protons and neutrons weigh the same, and the atomic number increases by one, as a new proton is being formed from a neutron. Because the atomic number changes, this is actually a new element: number 30, since one more proton is added to number 29.

On the periodic table, this is zinc, written as #""_64^30Zn#.

As an equation, this looks like

#""_64^29Cu -> ""_64^30Zn + ""_0^-1β#

Check that the numbers on the top and bottom add up so mass and charge are conserved.

#29 = 30 + (-1)#
#64 = 64 + 0#

If we are looking at a proton decaying into a neutron, emitting a positron, the equation would be

#""_64^29Cu -> ""_64^28Ni + ""_0^1β#