Question

What nuclide undergoes alpha decay to produce sodium-24?

Answer 1

`""_13^28"Al"`

Explanation:

You know that alpha decay takes place when an `alpha"-particle"` is being ejected from the nucleus of a radioactive isotope.

An `alpha"-particle"` is simply the nucleus of a helium-4 atom. A helium-4 atom has a total of two protons and two neutrons in its nucleus, and two electrons surrounding that nucleus.

In that case, if an `alpha"-particle"` is the nucleus of a helium-4 atom, then it must have a mass number equal to `4`, since it has two protons and two neutrons, and a net charge of `(2+)`, since it no longer has the two electrons that the helium-4 atom has.

So, you know that an unknown radioactive isotope decays via alpha decay to yield a sodium-24 nucleus.

A sodium-24 nucleus contains 11 protons and 13 neutrons. This means that you can write

`""_Z^A"X" -> ""_11^24"Na" + ""_2^4alpha`

Here `A` and `Z` represent the unknown element's mass number and atomic number, respectively.

So, if you take into account the fact that mass number and atomic number must be conserved in a nuclear reaction, you an say that

`A = 24 + 4" "` and `" "Z = 11 + 2`

The unknown isotope will thus have

`{(A = 28), (Z= 13) :}`

A quick look in the periodic table will show you that the element that has 13 protons in its nucleus is aluminium, `"Al"`. This means that you're dealing with aluminium-28, an isotope of aluminium that has 15 protons in its nucleus.

The complete nuclear equation will be

`""_13^28"Al" -> ""_11^24"Na" + ""_2^4alpha`