# Question afc3d

Sep 20, 2017

Here's what I got.

#### Explanation:

For starters, grab a Periodic Table and look for potassium, $\text{K}$, and argon, $\text{Ar}$. Once you find them, make a note of their respective atomic numbers.

• $\text{For K: " Z_"K} = 19$
• $\text{For Ar: " Z_"Ar} = 18$

You can now write the isotope notation for your two isotopes.

• $\text{potassium-40 " implies " " ""_19^40"K}$
• $\text{argon-40 " implies " " ""_18^40"Ar}$

This means that the unbalanced nuclear equation that describes this radioactive decay process will look like this

""_19^40"K" -> ""_18^40"Ar" + ""_Z^A?

Your goal here will be to identify the unknown particle, which I labeled "?#".

As you know, in any nuclear reaction, mass and charge must be conserved. This implies that you can write two equations

$40 = 40 + A \to$ conservation of mass

$19 = 18 + Z \to$ conservation of charge

Solve these equations to get

$40 = 40 + A \implies A = 0$

$19 = 18 + Z \implies Z = 1$

The particle that has $Z = 1$ and $A = 0$ is the positron, which is the antiparticle of the electron. The positron is also known as a beta-plus particle, ${\beta}^{+}$.

This means that potassium-40 can decay to argon-40 by way of positron emission, or beta-plus decay.

The balanced nuclear equation that describes the positron emission of potassium-40 looks like this

${\text{_19^40"K" -> ""_18^40"Ar" + }}_{1}^{0} \beta + {\nu}_{e}$

Keep in mind that an electron neutrino, ${\nu}_{e}$ is also emitted here.

Now, you can also set up the unbalanced nuclear equation like this

$\text{_19^40"K"+ ""_Z^A? -> ""_18^40"Ar}$

This time, you have

$40 + A = 40 \to$ conservation of mass

$19 + Z = 18 \to$ conservation of charge

Solve the two equations to get

$A = 0 \text{ " and " } Z = - 1$

In this case, the unknown particle is an electron, or beta particle, ${\beta}^{-}$. You can thus say that potassium-40 can also decay to argon-40 by way of electron capture.

The balanced nuclear equation that describes the electron capture of potassium-40 looks like this

$\text{_ 19^40"K" + ""_ (-1)^(color(white)(-)0)beta -> ""_18^40"Ar} + {\nu}_{e}$

Once again, notice that the electron capture results in the emission of an electron neutrino.