# What particle is needed to complete this nuclear reaction? "_86^222Rn->"_84^218 Po + underline ?

Nov 30, 2016

An alpha particle.

#### Explanation:

The thing to remember about nuclear equations is that mass and charge must always be conserved.

In other words, in any nuclear equation

• the overall mass number remains unchanged
• the overall atomic number remains unchanged

Your unbalanced nuclear equation looks like this

$\text{_ (color(white)(1)color(blue)(86))^color(darkgreen)(222)"Rn" -> ""_ (color(white)(1)color(blue)(84))^color(darkgreen)(218)"Po" + ""_ (color(blue)(Z))^color(darkgreen)(A)"?}$

The goal here is to find the atomic number, $\textcolor{b l u e}{Z}$, and the mass number, $\textcolor{\mathrm{da} r k g r e e n}{A}$, of the unknown particle.

Since the overall mass number must be conserved, you can say that

$\textcolor{\mathrm{da} r k g r e e n}{222} = \textcolor{\mathrm{da} r k g r e e n}{218} + \textcolor{\mathrm{da} r k g r e e n}{A} \to$ conservation of mass

This will get you

$\textcolor{\mathrm{da} r k g r e e n}{A} = 222 - 218 = 4$

The overall atomic number is also conserved, so you can say that

$\textcolor{b l u e}{86} = \textcolor{b l u e}{84} + \textcolor{b l u e}{Z}$

This will get you

$\textcolor{b l u e}{Z} = 86 - 84 = 2$

The unknown particle has a mass number equal to $4$ and an atomic number equal to $2$, which means that you're dealing with an alpha particle. In essence, an alpha particle is simply the nucleus of a helium-4 atom, i.e. it contains $2$ protons and $2$ neutrons.

You can now complete the nuclear equation that describes the alpha decay of radon-222 to polonium-218

${\text{_ (color(white)(1)86)^222"Rn" -> ""_ (color(white)(1)84)^218"Po" + }}_{2}^{4} \alpha$