Question #1292d

The idea here is that when a radioactive nuclide undergoes beta-minus decay, its atomic number will increase by #1# and its mass number will remain unchanged.

In beta-minus decay, a neutron located inside the nucleus is converted to a proton, which is what causes the atomic number to increase by #1#. At the same time, the nuclide emits an electron, also called a beta particle, #beta#, and an electron antineutrino, #bar(nu)_"e"#.

So if you take #""_Z^A"X"# to be the nuclide that results from the beta-minus decay of potassium-40, you can say that

#""_ 19^40"K" -> ""_ Z^A"X" + ""_ (-1)^(color(white)(-)0)"e" + ""_ 0^0bar(nu)_"e"#

Now, in every nuclear reaction, mass and charge must be conserved. This means that you have

#40 = A + 0 + 0 " " -># conservation of mass

#19 = Z + (-1) + 0 " " -># conservation of charge

Solve these two equations to get #A = 40# and #Z = 20#. A quick look at the Periodic Table will show that the resulting nuclide is calcium-40.

This means that you have

#""_Z^A"X" = ""_20^40"Ca"#

The balanced nuclear equation that describes the beta-minus decay of potassium-40 looks like this

#""_ 19^40"K" -> ""_ 20^40"Ca" + ""_ (-1)^(color(white)(-)0)"e" + ""_ 0^0 bar(nu)_"e"#