# Question 5dc62

Mar 27, 2017

Bismuth-214.

#### Explanation:

All you really need to use in order to answer this question is a Periodic Table and the fact that when a radioactive nuclide undergoes alpha decay

• its atomic number decreases by two, $\textcolor{g r e e n}{Z - 2}$
• its mass number decreases by four, $\textcolor{b l u e}{A - 4}$

This is the case because when a radioactive nuclide undergoes alpha decay, its emits an alpha particle, which is essentially the nucleus of a helium-4 atom.

So, grab a Periodic Table and look for astatine, $\text{At}$. You'll find it located in period 6, group 17.

Astatine has an atomic number equal to $\textcolor{g r e e n}{85}$, which means that you can write the starting nuclide as

$\text{_(color(white)(1)color(green)(85))^color(blue)(218)"At}$

Since an alpha particle contains $2$ protons and $2$ neutrons, you can say that

${\text{_ (color(white)(1)color(green)(85))^color(blue)(218)"At" -> ""_ color(green)(Z)^color(blue)(A)"?" + }}_{\textcolor{g r e e n}{2}}^{\textcolor{b l u e}{4}} \alpha$

Your goal here is to find the identity of the unknown nuclide, which we labeled as "?#".

As you know, mass and charge must be conserved in a nuclear reaction. This implies that the total mass number and the total atomic number must be equal on both sides of the equation.

You can thus say that

• conservation of mass $\to \text{ } \textcolor{b l u e}{218 = A + 4}$

• conservation of charge $\to \text{ } \textcolor{g r e e n}{85 = Z + 2}$

This will get you

$\textcolor{b l u e}{A = 218 - 4 = 214} \text{ }$ and $\text{ } \textcolor{g r e e n}{Z = 85 - 2 = 83}$

Check the Periodic Table for the element that has an atomic number equal to $83$. You'll find that bismuth is the match, so

$\text{_ color(green)(Z)^color(blue)(A)"?" = ""_ (color(white)(1)color(green)(83))^color(blue)(214)"Bi}$

Therefore, you can say that the alpha decay of astatine-218 will produce bismuth-214, or $\text{^214"Bi}$.

The complete nuclear equation will be

${\text{_ (color(white)(1)color(green)(85))^color(blue)(218)"At" -> ""_ (color(white)(1)color(green)(83))^color(blue)(214)"Bi" + }}_{\textcolor{g r e e n}{2}}^{\textcolor{b l u e}{4}} \alpha$