Question

Question #5dc62

Answer 1

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, `color(green)(Z - 2)`
• its mass number decreases by four, `color(blue)(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, `"At"`. You'll find it located in period 6, group 17.

Astatine has an atomic number equal to `color(green)(85)`, which means that you can write the starting nuclide as

`""_(color(white)(1)color(green)(85))^color(blue)(218)"At"`

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

`""_ (color(white)(1)color(green)(85))^color(blue)(218)"At" -> ""_ color(green)(Z)^color(blue)(A)"?" + ""_ color(green)(2)^color(blue)(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 `->" "color(blue)(218 = A + 4)`

• conservation of charge `-> " "color(green)(85 = Z + 2)`

This will get you

`color(blue)(A = 218 - 4 = 214)" "` and `" "color(green)(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

`""_ 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 `""^214"Bi"`.

The complete nuclear equation will be

`""_ (color(white)(1)color(green)(85))^color(blue)(218)"At" -> ""_ (color(white)(1)color(green)(83))^color(blue)(214)"Bi" + ""_ color(green)(2)^color(blue)(4)alpha`