Question #1292d

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Question #1292d

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Answer 1 · 2018-02-08T00:12:32

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 $1$ $1$ 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 $1$ $1$ . At the same time, the nuclide emits an electron, also called a beta particle, $β$ $beta$ , and an electron antineutrino, ${¯ ν}_{e}$ $bar(nu)_"e"$ .

![https://cnx.org/contents/5529301b-e00a-4bb0-8e97-9efec34edf4c@3]( useruploads.socratic.org )

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

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

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

$40 = A + 0 + 0 \to$ $40 = A + 0 + 0 " " ->$ conservation of mass

$19 = Z + (- 1) + 0 \to$ $19 = Z + (-1) + 0 " " ->$ conservation of charge

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

This means that you have

$_{Z}^{A} X =_{20}^{40} Ca$ $""_Z^A"X" = ""_20^40"Ca"$

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

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

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Question #1292d

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