Question #a9f0d

Question

Question #a9f0d

1 Answer

Impact of this question

4413 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-02T00:02:20

$_{19}^{41} K$ $""_19^41"K"$

Explanation:

As you know, when a radioactive nuclide undergoes electron capture, an electron located in one of the inner shells is absorbed by the nucleus. In the process, a proton is converted to a neutron and an electron neutrino, $ν_{e}$ $nu_"e"$ , is emitted.

So right from the start, the fact that a proton is being converted to a neutron tells you to things

the atomic number of the nuclide decreases by $1$ $1$

the mass number of the nuclide remains unchanged

In your case, calcium, $Ca$ $"Ca"$ , has an atomic number equal to $20$ $20$ , which means that the resulting nuclide will have

$Z = 20 - 1 = 19$ $Z = 20 -1 = 19$

The element that has an atomic number equal to $19$ $19$ is potassium, $K$ $"K"$ . Since the mass number remains unchanged, you can say that the resulting isotope will be potassium-41.

So you can say that the nuclear equation that describes the electron capture of calcium-41 looks like this

$_{20}^{41} Ca +_{- 1}^{- 0} e \to_{19}^{41} K + ν_{e}$ $""_ 20^41"Ca" + ""_ (-1)^(color(white)(-)0)"e" -> ""_ 19^41"K" + nu_ "e"$

Notice that you have

$41 + 0 = 41 \to$ $41 + 0 = 41 " "" "->$ conservation of mass

$20 + (- 1) = 19 \to$ $20 + (-1) = 19 ->$ conservation of charge

Science

Math

Humanities

... and beyond

Question #a9f0d

1 Answer

Explanation:

Related questions

Impact of this question