Question

Question #afc3d

Answer 1

Here's what I got.

Explanation:

For starters, grab a Periodic Table and look for potassium, `"K"`, and argon, `"Ar"`. Once you find them, make a note of their respective atomic numbers.

• `"For K: " Z_"K" = 19`
• `"For Ar: " Z_"Ar" = 18`

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

• `"potassium-40 " implies " " ""_19^40"K"`
• `"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 ->` conservation of mass

`19 = 18 + Z ->` 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

`""_19^40"K" -> ""_18^40"Ar" + ""_1^0beta + 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

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

This time, you have

`40 + A = 40 ->` conservation of mass

`19 + Z = 18 ->` conservation of charge

Solve the two equations to get

`A = 0" " 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

`""_ 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.