# What nuclide undergoes alpha decay to produce sodium-24?

Sep 30, 2015

$\text{_13^28"Al}$

#### Explanation:

You know that alpha decay takes place when an $\alpha \text{-particle}$ is being ejected from the nucleus of a radioactive isotope. An $\alpha \text{-particle}$ 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 $\alpha \text{-particle}$ 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 $\left(2 +\right)$, 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

${\text{_Z^A"X" -> ""_11^24"Na" + }}_{2}^{4} \alpha$

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

$A = 24 + 4 \text{ }$ and $\text{ } Z = 11 + 2$

The unknown isotope will thus have

$\left\{\begin{matrix}A = 28 \\ Z = 13\end{matrix}\right.$

A quick look in the periodic table will show you that the element that has 13 protons in its nucleus is aluminium, $\text{Al}$. 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

${\text{_13^28"Al" -> ""_11^24"Na" + }}_{2}^{4} \alpha$