Question #afc3d

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Question #afc3d

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Answer 1 · 2017-09-20T23:39:17

Here's what I got.

Explanation:

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

$For K: Z_{K} = 19$ $"For K: " Z_"K" = 19$

$For Ar: Z_{Ar} = 18$ $"For Ar: " Z_"Ar" = 18$

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

$potassium-40 \Rightarrow_{19}^{40} K$ $"potassium-40 " implies " " ""_19^40"K"$

$argon-40 \Rightarrow_{18}^{40} Ar$ $"argon-40 " implies " " ""_18^40"Ar"$

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

$_{19}^{40} K \to_{18}^{40} Ar +_{Z}^{A} ?$ $""_19^40"K" -> ""_18^40"Ar" + ""_Z^A?$

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

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

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

Solve these equations to get

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

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

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

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

$_{19}^{40} K \to_{18}^{40} Ar +_{1}^{0} β + ν_{e}$ $""_19^40"K" -> ""_18^40"Ar" + ""_1^0beta + nu_e$

Keep in mind that an electron neutrino, $ν_{e}$ $nu_e$ is also emitted here.

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

$_{19}^{40} K +_{Z}^{A} ? \to_{18}^{40} Ar$ $""_19^40"K"+ ""_Z^A? -> ""_18^40"Ar"$

This time, you have

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

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

Solve the two equations to get

$A = 0 and Z = - 1$ $A = 0" " and " "Z = -1$

In this case, the unknown particle is an electron, or beta particle, $β^{-}$ $beta^(-)$ . 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

$_{19}^{40} K +_{- 1}^{- 0} β \to_{18}^{40} Ar + ν_{e}$ $""_ 19^40"K" + ""_ (-1)^(color(white)(-)0)beta -> ""_18^40"Ar" + nu_e$

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

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