# Question #1292d

Feb 8, 2018

Here's what I got.

#### Explanation:

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, ${\overline{\nu}}_{\text{e}}$.

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

$\text{_ 19^40"K" -> ""_ Z^A"X" + ""_ (-1)^(color(white)(-)0)"e" + ""_ 0^0bar(nu)_"e}$

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

$40 = A + 0 + 0 \text{ } \to$ conservation of mass

$19 = Z + \left(- 1\right) + 0 \text{ } \to$ 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

$\text{_Z^A"X" = ""_20^40"Ca}$

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

$\text{_ 19^40"K" -> ""_ 20^40"Ca" + ""_ (-1)^(color(white)(-)0)"e" + ""_ 0^0 bar(nu)_"e}$