# Radiation emitted during radon's decay process was discovered to be useful in cancer therapy. What is the nuclide X formed in the reaction ""_86^222Rn -> ""_Z^A X + _2^4He, in which radon-222 undergoes alpha decay?

Apr 17, 2016

$\text{_84^218"Po}$

#### Explanation:

Radon-222 is said to undergo radioactive decay via alpha decay, which takes place when the nucleus of a radioactive nuclide emits an alpha particle, ""_2^4alpha.

Now, an alpha particle is simply the nucleus of a helium-4 atom, which is why you'll often see the notation $\text{_2^4"He}$ used to denote an alpha particle.

The nucleus of a helium-4 atom contains

• two protons
• two neutrons

Notice that after the alpha particle is emitted from the nucleus, the mass number of the nuclide, which tells you how many protons and neutrons it contains in its nucleus, decreases by $4$.

This happens because the nucleus of the radioactive nuclide loses two protons and two neutrons.

Likewise, the atomic number of the nuclide, which tells you how many protons it contains in its nucleus, decreases by $2$.

This means that the daughter nuclide will have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{A = 222 - 4 = 218} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$ and $\text{ } \textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{Z = 86 - 2 = 84} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$A$ - mass number
$Z$ - atomic number

Grab a periodic table and look for the element that has an atomic number equal to $84$. You'll find located in period 6, group 18, two spots before radon.

This element is called polonium, $\text{Po}$. The alpha decay of radon-222 will thus produce polonium-218, an isotope of polonium that has a mass number equal to $218$.

The nuclear equation looks like this

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{_ (color(white)(a)86)^222"Rn" -> ""_ (color(white)(a)84)^218"Po" + ""_2^4"He}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$